From d3d016e2cb01a3b34ea0a00242a13854f05411b5 Mon Sep 17 00:00:00 2001 From: Claude Date: Wed, 22 Jul 2026 09:09:44 +0000 Subject: [PATCH 1/3] Seed content: Quant exponents & roots lesson + 7 questions MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Add a new Arithmetic / exponents-roots bundle to the content bank — a previously uncovered but high-frequency GMAT Focus Quant topic: - 1 lesson (power rules, negative/zero/fractional exponents, radical simplification, factoring common powers). - 7 problem-solving questions with an easy->hard difficulty progression, each with a worked explanation and hints, and distractors keyed to specific common errors (multiplying vs. adding exponents, merging unlike radicals, treating a sum of powers as a new base). Choice-level math is wrapped in \( \) so the study site renders it via KaTeX. Rebuilt docs/data/content.json (39 questions, 9 lessons). Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_01PjmqN8j1zyySgtsg6qTbXz --- .../quant/quant-arithmetic-exponents-roots.md | 84 +++++++ .../quant-arithmetic-exponents-roots-0017.md | 42 ++++ .../quant-arithmetic-exponents-roots-0018.md | 45 ++++ .../quant-arithmetic-exponents-roots-0019.md | 47 ++++ .../quant-arithmetic-exponents-roots-0020.md | 44 ++++ .../quant-arithmetic-exponents-roots-0021.md | 47 ++++ .../quant-arithmetic-exponents-roots-0022.md | 51 ++++ .../quant-arithmetic-exponents-roots-0023.md | 52 ++++ docs/data/content.json | 234 +++++++++++++++++- 9 files changed, 642 insertions(+), 4 deletions(-) create mode 100644 content/lessons/quant/quant-arithmetic-exponents-roots.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0017.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0018.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0019.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0020.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0021.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0022.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0023.md diff --git a/content/lessons/quant/quant-arithmetic-exponents-roots.md b/content/lessons/quant/quant-arithmetic-exponents-roots.md new file mode 100644 index 0000000..d8640e2 --- /dev/null +++ b/content/lessons/quant/quant-arithmetic-exponents-roots.md @@ -0,0 +1,84 @@ +--- +id: quant-arithmetic-exponents-roots +section: quant +topic: arithmetic +subtopic: exponents-roots +title: "Exponents & Roots" +tags: [exponents, roots, radicals, powers] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Most GMAT Focus exponent questions reward you for **rewriting everything in the same base** and +then just tracking the exponents. Almost nothing here needs a calculator — the whole game is a +handful of rules applied cleanly and, above all, **not confusing multiplication of bases with +addition of exponents**. + +## Core concepts + +**The power rules.** For any nonzero base \(a\): + +\[a^m \cdot a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{mn}\] + +Read them carefully: you **add** exponents when you *multiply* like bases, and you **multiply** +exponents when you *raise a power to a power*. Mixing these two up is the single most common +exponent error on the test. + +**Zero and negative exponents.** + +\[a^0 = 1 \ (a \neq 0), \qquad a^{-n} = \frac{1}{a^n}\] + +A negative exponent means "reciprocal," not "negative number": \(2^{-3} = \tfrac{1}{8}\), not \(-8\). + +**Fractional exponents are roots.** + +\[a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^{m}}\] + +So \(27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9\). Take the root first when it keeps the numbers +small. + +**Roots simplify by pulling out perfect squares.** \(\sqrt{ab} = \sqrt{a}\,\sqrt{b}\), so + +\[\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}\] + +You can only **add or subtract** radicals when the part under the root matches: +\(5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\). You may **never** merge \(\sqrt{a} + \sqrt{b}\) into +\(\sqrt{a+b}\). + +**Factoring out a common power** cracks the hardest questions. Terms like \(2^n + 2^n\) are not a +new base — they collapse: + +\[2^n + 2^n = 2 \cdot 2^n = 2^{n+1}, \qquad 3^{k+1} - 3^{k} = 3^{k}(3 - 1) = 2\cdot 3^{k}\] + +## Worked examples + +**Same-base equation.** Solve \(2^{x+1} = 8^{x-1}\). Rewrite \(8 = 2^3\), so the right side is +\(2^{3(x-1)} = 2^{3x-3}\). Same base ⟹ equal exponents: \(x + 1 = 3x - 3\), giving \(2x = 4\) and +\(x = 2\). + +**Collapse a sum of equal powers.** If \(2^n + 2^n + 2^n + 2^n = 2^{12}\), the left side is +\(4 \cdot 2^n = 2^2 \cdot 2^n = 2^{n+2}\). So \(n + 2 = 12\) and \(n = 10\). + +**Ratio of powers.** With \(x = 3^{20}\) and \(y = 3^{18}\), \(\dfrac{x}{y} = 3^{20-18} = 3^2 = 9\). + +## Common traps + +- **Multiplying exponents when you should add them** (and vice versa). \(2^5 \cdot 2^3 = 2^8\), not + \(2^{15}\). +- **Reading a negative exponent as a negative number.** \(3^{-2} = \tfrac{1}{9}\), and \(3^0 = 1\) + (not \(0\)). +- **Merging unlike radicals.** \(\sqrt{50} + \sqrt{18} \neq \sqrt{68}\); simplify each first, then add. +- **Treating a sum of powers as a new base.** \(2^n + 2^n\) is \(2^{n+1}\), not \(4^n\) or \(2^{2n}\). + +## Key takeaways + +- Get everything into a **common base**, then just manipulate exponents. +- Multiply like bases → **add** exponents; raise a power to a power → **multiply** exponents. +- \(a^0 = 1\), \(a^{-n} = 1/a^n\), and \(a^{m/n} = \sqrt[n]{a^m}\). +- Simplify radicals by pulling out perfect squares; only combine radicals with the same radicand. +- Sums of equal powers **factor**: \(k\) copies of \(a^n\) is \(k \cdot a^n\). diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0017.md b/content/questions/quant/quant-arithmetic-exponents-roots-0017.md new file mode 100644 index 0000000..c7db1d0 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0017.md @@ -0,0 +1,42 @@ +--- +id: quant-arithmetic-exponents-roots-0017 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, power-rules] +choices: + A: "4" + B: "8" + C: "16" + D: "64" + E: "512" +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(\dfrac{2^{5} \cdot 2^{3}}{2^{6}}\)? + +## Explanation + +Keep everything in base 2 and just track exponents. When you multiply like bases you **add** +exponents; when you divide, you **subtract**: + +\[\frac{2^{5} \cdot 2^{3}}{2^{6}} = 2^{\,5 + 3 - 6} = 2^{2} = 4\] + +**The trap:** answer **E (512)** comes from *multiplying* the exponents in the numerator +(\(5 \times 3 = 15\), then \(15 - 6 = 9\), giving \(2^{9} = 512\)). You add exponents when the bases +are multiplied, not multiply them. Answer **D (64)** is \(2^{6}\), what you get if you drop one of +the numerator terms. + +## Hints + +- Rewrite the whole expression as a single power of 2. +- Multiply like bases → add exponents; divide → subtract exponents. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0018.md b/content/questions/quant/quant-arithmetic-exponents-roots-0018.md new file mode 100644 index 0000000..3c7b04e --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0018.md @@ -0,0 +1,45 @@ +--- +id: quant-arithmetic-exponents-roots-0018 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, negative-exponent, zero-exponent] +choices: + A: "0" + B: "1/9" + C: "10/9" + D: "2" + E: "9" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(3^{-2} + 3^{0}\)? + +## Explanation + +Two rules do all the work here. A **negative exponent** means take the reciprocal, and **anything +nonzero raised to the 0 power is 1**: + +\[3^{-2} = \frac{1}{3^{2}} = \frac{1}{9}, \qquad 3^{0} = 1\] + +So the sum is + +\[\frac{1}{9} + 1 = \frac{1}{9} + \frac{9}{9} = \frac{10}{9}\] + +**The traps:** answer **B (1/9)** drops the \(3^{0} = 1\) term (a studier who thinks \(3^0 = 0\)). +Answer **D (2)** comes from mistakenly treating \(3^{-2}\) as \(1\) as well. A negative exponent is +never a negative number, so **A (0)** and **E (9)** are out. + +## Hints + +- A negative exponent means "reciprocal," not "negative": \(3^{-2} = \tfrac{1}{3^2}\). +- Remember that \(3^{0} = 1\), not \(0\). diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0019.md b/content/questions/quant/quant-arithmetic-exponents-roots-0019.md new file mode 100644 index 0000000..1440e55 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0019.md @@ -0,0 +1,47 @@ +--- +id: quant-arithmetic-exponents-roots-0019 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [exponents, common-base, equations] +choices: + A: "1" + B: "2" + C: "3" + D: "4" + E: "5" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(2^{\,x+1} = 8^{\,x-1}\), what is the value of \(x\)? + +## Explanation + +The two sides have different bases, so first rewrite them with the **same base**. Since +\(8 = 2^{3}\): + +\[8^{\,x-1} = \left(2^{3}\right)^{x-1} = 2^{\,3(x-1)} = 2^{\,3x-3}\] + +Now both sides are powers of 2, so the exponents must be equal: + +\[x + 1 = 3x - 3 \;\Rightarrow\; 4 = 2x \;\Rightarrow\; x = 2\] + +Check: \(2^{2+1} = 2^{3} = 8\) and \(8^{2-1} = 8^{1} = 8\). ✓ + +**The trap:** answer **A (1)** comes from forgetting to distribute the 3, i.e. writing the exponent +as \(3x - 1\) instead of \(3(x-1) = 3x - 3\). Distributing carefully is the whole question. + +## Hints + +- Rewrite \(8\) as a power of \(2\) so both sides share a base. +- When two equal powers have the same base, set the exponents equal — and distribute the exponent + fully. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0020.md b/content/questions/quant/quant-arithmetic-exponents-roots-0020.md new file mode 100644 index 0000000..cfc14aa --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0020.md @@ -0,0 +1,44 @@ +--- +id: quant-arithmetic-exponents-roots-0020 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [exponents, fractional-exponents, roots] +choices: + A: "6" + B: "9" + C: "18" + D: "54" + E: "729" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(27^{2/3}\)? + +## Explanation + +A fractional exponent \(m/n\) means "take the \(n\)th root, then raise to the \(m\)th power." The +denominator is the root; the numerator is the power. Take the root first to keep the numbers small: + +\[27^{2/3} = \left(27^{1/3}\right)^{2} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9\] + +**The traps:** + +- **E (729)** ignores the cube-root denominator and just squares: \(27^{2} = 729\). +- **A (6)** multiplies the root by the numerator instead of raising to it: \(3 \times 2 = 6\). +- **C (18)** and **D (54)** treat the exponent as ordinary multiplication (\(27 \times \tfrac{2}{3} = 18\), + or \(27 \times 2 = 54\)) rather than as a root-and-power. + +## Hints + +- The denominator of the exponent is a root; the numerator is a power. +- \(27^{2/3} = (\sqrt[3]{27})^{2}\) — take the cube root first. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0021.md b/content/questions/quant/quant-arithmetic-exponents-roots-0021.md new file mode 100644 index 0000000..5228f38 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0021.md @@ -0,0 +1,47 @@ +--- +id: quant-arithmetic-exponents-roots-0021 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [roots, radicals, simplifying] +choices: + A: "\(2\sqrt{17}\)" + B: "\(15\sqrt{2}\)" + C: "\(8\sqrt{2}\)" + D: "\(4\sqrt{2}\)" + E: "\(34\sqrt{2}\)" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is \(\sqrt{50} + \sqrt{18}\) in simplest form? + +## Explanation + +You can only add radicals once the parts **under** the root match, so simplify each one by pulling +out perfect squares: + +\[\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \qquad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\] + +Now both terms are multiples of \(\sqrt{2}\), so add the coefficients: + +\[5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\] + +**The traps:** + +- **A (\(2\sqrt{17}\))** comes from illegally merging the roots: \(\sqrt{50} + \sqrt{18} \neq \sqrt{68}\). + Square roots do not add that way. +- **B (\(15\sqrt{2}\))** multiplies the coefficients (\(5 \times 3\)) instead of adding them. + +## Hints + +- Simplify each radical first by factoring out the largest perfect square. +- You can add radicals only when the number under the root is the same — then add the coefficients. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0022.md b/content/questions/quant/quant-arithmetic-exponents-roots-0022.md new file mode 100644 index 0000000..5657e2b --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0022.md @@ -0,0 +1,51 @@ +--- +id: quant-arithmetic-exponents-roots-0022 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: hard +tags: [exponents, factoring-powers, common-base] +choices: + A: "3" + B: "6" + C: "10" + D: "12" + E: "48" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(2^{n} + 2^{n} + 2^{n} + 2^{n} = 2^{12}\), what is the value of \(n\)? + +## Explanation + +The left side is **four copies of the same term**, so factor — do not treat it as a new base. +Adding four equal terms is multiplying by 4: + +\[2^{n} + 2^{n} + 2^{n} + 2^{n} = 4 \cdot 2^{n}\] + +Now write the coefficient as a power of 2 so both sides share a base. Since \(4 = 2^{2}\): + +\[4 \cdot 2^{n} = 2^{2} \cdot 2^{n} = 2^{\,n+2}\] + +Setting \(2^{\,n+2} = 2^{12}\) gives \(n + 2 = 12\), so \(n = 10\). + +**The traps:** + +- **A (3)** treats the sum as a *product* of the four terms, \((2^{n})^{4} = 2^{4n} = 2^{12}\), + giving \(4n = 12\). Adding equal powers is not the same as multiplying them. +- **B (6)** correctly writes \(4 \cdot 2^{n}\) but then *multiplies* the exponents (\(2 \cdot n = 12\)) + instead of adding: \(2^{2} \cdot 2^{n} = 2^{n+2}\), not \(2^{2n}\). +- **D (12)** simply ignores the coefficient of 4. + +## Hints + +- Four copies of \(2^{n}\) is \(4 \cdot 2^{n}\), not \((2^{n})^{4}\). +- Rewrite \(4\) as \(2^{2}\), then combine with \(2^{n}\) by **adding** exponents. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0023.md b/content/questions/quant/quant-arithmetic-exponents-roots-0023.md new file mode 100644 index 0000000..9754c91 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0023.md @@ -0,0 +1,52 @@ +--- +id: quant-arithmetic-exponents-roots-0023 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: hard +tags: [exponents, common-base, factoring-powers] +choices: + A: "1/9" + B: "2" + C: "8" + D: "9" + E: "\(3^{38}\)" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(x = 3^{20}\) and \(y = 3^{18}\), what is the value of \(\dfrac{x - y}{y}\)? + +## Explanation + +Do not try to evaluate \(3^{20}\) or \(3^{18}\) — they are enormous. Split the fraction so the huge +powers turn into a ratio you can simplify: + +\[\frac{x - y}{y} = \frac{x}{y} - \frac{y}{y} = \frac{3^{20}}{3^{18}} - 1\] + +Dividing like bases means **subtracting** exponents: + +\[\frac{3^{20}}{3^{18}} = 3^{\,20 - 18} = 3^{2} = 9\] + +So the value is \(9 - 1 = 8\). + +**The traps:** + +- **D (9)** is \(\dfrac{x}{y} = 3^{2}\) — correct so far, but it forgets to subtract the \(\dfrac{y}{y} = 1\) + term. +- **B (2)** stops at the exponent difference \(20 - 18 = 2\), as if the whole expression equaled + \(3^{20-18}\) without the "\(-1\)". +- **E (\(3^{38}\))** comes from *multiplying* the powers (adding exponents \(20 + 18\)) instead of + dividing. + +## Hints + +- Break \(\dfrac{x - y}{y}\) into \(\dfrac{x}{y} - 1\) before touching the exponents. +- Dividing like bases subtracts exponents: \(\dfrac{3^{20}}{3^{18}} = 3^{2}\). diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..617aa4e 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,8 +7,8 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 16, - "lessons": 5 + "questions": 23, + "lessons": 6 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 32, - "lessons": 8 + "questions": 39, + "lessons": 9 }, "questions": [ { @@ -301,6 +301,215 @@ "Or use the coefficient shortcut: sum of roots \\(= -b\\), product \\(= c\\)." ] }, + { + "id": "quant-arithmetic-exponents-roots-0017", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "exponents", + "power-rules" + ], + "choices": { + "A": "4", + "B": "8", + "C": "16", + "D": "64", + "E": "512" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(\\dfrac{2^{5} \\cdot 2^{3}}{2^{6}}\\)?", + "explanation": "Keep everything in base 2 and just track exponents. When you multiply like bases you **add**\nexponents; when you divide, you **subtract**:\n\n\\[\\frac{2^{5} \\cdot 2^{3}}{2^{6}} = 2^{\\,5 + 3 - 6} = 2^{2} = 4\\]\n\n**The trap:** answer **E (512)** comes from *multiplying* the exponents in the numerator\n(\\(5 \\times 3 = 15\\), then \\(15 - 6 = 9\\), giving \\(2^{9} = 512\\)). You add exponents when the bases\nare multiplied, not multiply them. Answer **D (64)** is \\(2^{6}\\), what you get if you drop one of\nthe numerator terms.", + "hints": [ + "Rewrite the whole expression as a single power of 2.", + "Multiply like bases → add exponents; divide → subtract exponents." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0018", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "exponents", + "negative-exponent", + "zero-exponent" + ], + "choices": { + "A": "0", + "B": "1/9", + "C": "10/9", + "D": "2", + "E": "9" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(3^{-2} + 3^{0}\\)?", + "explanation": "Two rules do all the work here. A **negative exponent** means take the reciprocal, and **anything\nnonzero raised to the 0 power is 1**:\n\n\\[3^{-2} = \\frac{1}{3^{2}} = \\frac{1}{9}, \\qquad 3^{0} = 1\\]\n\nSo the sum is\n\n\\[\\frac{1}{9} + 1 = \\frac{1}{9} + \\frac{9}{9} = \\frac{10}{9}\\]\n\n**The traps:** answer **B (1/9)** drops the \\(3^{0} = 1\\) term (a studier who thinks \\(3^0 = 0\\)).\nAnswer **D (2)** comes from mistakenly treating \\(3^{-2}\\) as \\(1\\) as well. A negative exponent is\nnever a negative number, so **A (0)** and **E (9)** are out.", + "hints": [ + "A negative exponent means \"reciprocal,\" not \"negative\": \\(3^{-2} = \\tfrac{1}{3^2}\\).", + "Remember that \\(3^{0} = 1\\), not \\(0\\)." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0019", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "exponents", + "common-base", + "equations" + ], + "choices": { + "A": "1", + "B": "2", + "C": "3", + "D": "4", + "E": "5" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(2^{\\,x+1} = 8^{\\,x-1}\\), what is the value of \\(x\\)?", + "explanation": "The two sides have different bases, so first rewrite them with the **same base**. Since\n\\(8 = 2^{3}\\):\n\n\\[8^{\\,x-1} = \\left(2^{3}\\right)^{x-1} = 2^{\\,3(x-1)} = 2^{\\,3x-3}\\]\n\nNow both sides are powers of 2, so the exponents must be equal:\n\n\\[x + 1 = 3x - 3 \\;\\Rightarrow\\; 4 = 2x \\;\\Rightarrow\\; x = 2\\]\n\nCheck: \\(2^{2+1} = 2^{3} = 8\\) and \\(8^{2-1} = 8^{1} = 8\\). ✓\n\n**The trap:** answer **A (1)** comes from forgetting to distribute the 3, i.e. writing the exponent\nas \\(3x - 1\\) instead of \\(3(x-1) = 3x - 3\\). Distributing carefully is the whole question.", + "hints": [ + "Rewrite \\(8\\) as a power of \\(2\\) so both sides share a base.", + "When two equal powers have the same base, set the exponents equal — and distribute the exponent" + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0020", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "exponents", + "fractional-exponents", + "roots" + ], + "choices": { + "A": "6", + "B": "9", + "C": "18", + "D": "54", + "E": "729" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(27^{2/3}\\)?", + "explanation": "A fractional exponent \\(m/n\\) means \"take the \\(n\\)th root, then raise to the \\(m\\)th power.\" The\ndenominator is the root; the numerator is the power. Take the root first to keep the numbers small:\n\n\\[27^{2/3} = \\left(27^{1/3}\\right)^{2} = \\left(\\sqrt[3]{27}\\right)^{2} = 3^{2} = 9\\]\n\n**The traps:**\n\n- **E (729)** ignores the cube-root denominator and just squares: \\(27^{2} = 729\\).\n- **A (6)** multiplies the root by the numerator instead of raising to it: \\(3 \\times 2 = 6\\).\n- **C (18)** and **D (54)** treat the exponent as ordinary multiplication (\\(27 \\times \\tfrac{2}{3} = 18\\),\n or \\(27 \\times 2 = 54\\)) rather than as a root-and-power.", + "hints": [ + "The denominator of the exponent is a root; the numerator is a power.", + "\\(27^{2/3} = (\\sqrt[3]{27})^{2}\\) — take the cube root first." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0021", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "roots", + "radicals", + "simplifying" + ], + "choices": { + "A": "\\(2\\sqrt{17}\\)", + "B": "\\(15\\sqrt{2}\\)", + "C": "\\(8\\sqrt{2}\\)", + "D": "\\(4\\sqrt{2}\\)", + "E": "\\(34\\sqrt{2}\\)" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is \\(\\sqrt{50} + \\sqrt{18}\\) in simplest form?", + "explanation": "You can only add radicals once the parts **under** the root match, so simplify each one by pulling\nout perfect squares:\n\n\\[\\sqrt{50} = \\sqrt{25 \\cdot 2} = 5\\sqrt{2}, \\qquad \\sqrt{18} = \\sqrt{9 \\cdot 2} = 3\\sqrt{2}\\]\n\nNow both terms are multiples of \\(\\sqrt{2}\\), so add the coefficients:\n\n\\[5\\sqrt{2} + 3\\sqrt{2} = 8\\sqrt{2}\\]\n\n**The traps:**\n\n- **A (\\(2\\sqrt{17}\\))** comes from illegally merging the roots: \\(\\sqrt{50} + \\sqrt{18} \\neq \\sqrt{68}\\).\n Square roots do not add that way.\n- **B (\\(15\\sqrt{2}\\))** multiplies the coefficients (\\(5 \\times 3\\)) instead of adding them.", + "hints": [ + "Simplify each radical first by factoring out the largest perfect square.", + "You can add radicals only when the number under the root is the same — then add the coefficients." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0022", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "exponents", + "factoring-powers", + "common-base" + ], + "choices": { + "A": "3", + "B": "6", + "C": "10", + "D": "12", + "E": "48" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(2^{n} + 2^{n} + 2^{n} + 2^{n} = 2^{12}\\), what is the value of \\(n\\)?", + "explanation": "The left side is **four copies of the same term**, so factor — do not treat it as a new base.\nAdding four equal terms is multiplying by 4:\n\n\\[2^{n} + 2^{n} + 2^{n} + 2^{n} = 4 \\cdot 2^{n}\\]\n\nNow write the coefficient as a power of 2 so both sides share a base. Since \\(4 = 2^{2}\\):\n\n\\[4 \\cdot 2^{n} = 2^{2} \\cdot 2^{n} = 2^{\\,n+2}\\]\n\nSetting \\(2^{\\,n+2} = 2^{12}\\) gives \\(n + 2 = 12\\), so \\(n = 10\\).\n\n**The traps:**\n\n- **A (3)** treats the sum as a *product* of the four terms, \\((2^{n})^{4} = 2^{4n} = 2^{12}\\),\n giving \\(4n = 12\\). Adding equal powers is not the same as multiplying them.\n- **B (6)** correctly writes \\(4 \\cdot 2^{n}\\) but then *multiplies* the exponents (\\(2 \\cdot n = 12\\))\n instead of adding: \\(2^{2} \\cdot 2^{n} = 2^{n+2}\\), not \\(2^{2n}\\).\n- **D (12)** simply ignores the coefficient of 4.", + "hints": [ + "Four copies of \\(2^{n}\\) is \\(4 \\cdot 2^{n}\\), not \\((2^{n})^{4}\\).", + "Rewrite \\(4\\) as \\(2^{2}\\), then combine with \\(2^{n}\\) by **adding** exponents." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0023", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "exponents", + "common-base", + "factoring-powers" + ], + "choices": { + "A": "1/9", + "B": "2", + "C": "8", + "D": "9", + "E": "\\(3^{38}\\)" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(x = 3^{20}\\) and \\(y = 3^{18}\\), what is the value of \\(\\dfrac{x - y}{y}\\)?", + "explanation": "Do not try to evaluate \\(3^{20}\\) or \\(3^{18}\\) — they are enormous. Split the fraction so the huge\npowers turn into a ratio you can simplify:\n\n\\[\\frac{x - y}{y} = \\frac{x}{y} - \\frac{y}{y} = \\frac{3^{20}}{3^{18}} - 1\\]\n\nDividing like bases means **subtracting** exponents:\n\n\\[\\frac{3^{20}}{3^{18}} = 3^{\\,20 - 18} = 3^{2} = 9\\]\n\nSo the value is \\(9 - 1 = 8\\).\n\n**The traps:**\n\n- **D (9)** is \\(\\dfrac{x}{y} = 3^{2}\\) — correct so far, but it forgets to subtract the \\(\\dfrac{y}{y} = 1\\)\n term.\n- **B (2)** stops at the exponent difference \\(20 - 18 = 2\\), as if the whole expression equaled\n \\(3^{20-18}\\) without the \"\\(-1\\)\".\n- **E (\\(3^{38}\\))** comes from *multiplying* the powers (adding exponents \\(20 + 18\\)) instead of\n dividing.", + "hints": [ + "Break \\(\\dfrac{x - y}{y}\\) into \\(\\dfrac{x}{y} - 1\\) before touching the exponents.", + "Dividing like bases subtracts exponents: \\(\\dfrac{3^{20}}{3^{18}} = 3^{2}\\)." + ] + }, { "id": "quant-arithmetic-fractions-decimals-0015", "section": "quant", @@ -968,6 +1177,23 @@ "status": "in-review", "body": "## Overview\n\nA quadratic equation has the form \\(ax^2 + bx + c = 0\\). On GMAT Focus Quant, most quadratics are\nbuilt to **factor cleanly**, so factoring — not the quadratic formula — is usually the fast route.\n\n## Core concepts\n\n**Factoring \\(x^2 + bx + c\\).** Find two numbers that **multiply to \\(c\\)** and **add to \\(b\\)**.\nThen \\(x^2 + bx + c = (x + p)(x + q)\\), and the solutions make each factor zero.\n\nExample: \\(x^2 - 5x + 6\\). Two numbers multiplying to \\(+6\\) and adding to \\(-5\\) are \\(-2\\) and\n\\(-3\\), so \\((x-2)(x-3) = 0\\) and \\(x = 2\\) or \\(x = 3\\).\n\n**Sum and product shortcut.** For \\(x^2 + bx + c = 0\\):\n\n\\[\\text{sum of roots} = -b, \\qquad \\text{product of roots} = c\\]\n\nThis lets you answer \"sum/product of solutions\" questions without fully solving.\n\n**Special products** worth recognizing instantly:\n\n\\[a^2 - b^2 = (a-b)(a+b), \\quad (a+b)^2 = a^2 + 2ab + b^2, \\quad (a-b)^2 = a^2 - 2ab + b^2\\]\n\n**The quadratic formula** (fallback when factoring is ugly):\n\n\\[x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}\\]\n\n## Worked examples\n\n**Factor and solve.** \\(x^2 + 2x - 15 = 0 \\Rightarrow (x+5)(x-3) = 0 \\Rightarrow x = -5 \\text{ or } 3\\).\n\n**Difference of squares.** \\(x^2 - 49 = 0 \\Rightarrow (x-7)(x+7) = 0 \\Rightarrow x = \\pm 7\\).\n\n## Common traps\n\n- **Confusing sum and product of roots.** Sum is \\(-b\\), product is \\(c\\). Read which the question wants.\n- **Dropping a solution.** A quadratic usually has *two* roots — check whether the question wants both, their sum, or only positive values.\n- **Sign errors in factoring.** Verify by expanding your factors back out.\n\n## Key takeaways\n\n- Factor into \\((x + p)(x + q)\\) where \\(pq = c\\) and \\(p + q = b\\); each factor set to zero gives a root.\n- Sum of roots \\(= -b\\), product \\(= c\\) — a fast shortcut.\n- Recognize the difference of squares and perfect-square forms on sight." }, + { + "id": "quant-arithmetic-exponents-roots", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "title": "Exponents & Roots", + "tags": [ + "exponents", + "roots", + "radicals", + "powers" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nMost GMAT Focus exponent questions reward you for **rewriting everything in the same base** and\nthen just tracking the exponents. Almost nothing here needs a calculator — the whole game is a\nhandful of rules applied cleanly and, above all, **not confusing multiplication of bases with\naddition of exponents**.\n\n## Core concepts\n\n**The power rules.** For any nonzero base \\(a\\):\n\n\\[a^m \\cdot a^n = a^{m+n}, \\qquad \\frac{a^m}{a^n} = a^{m-n}, \\qquad (a^m)^n = a^{mn}\\]\n\nRead them carefully: you **add** exponents when you *multiply* like bases, and you **multiply**\nexponents when you *raise a power to a power*. Mixing these two up is the single most common\nexponent error on the test.\n\n**Zero and negative exponents.**\n\n\\[a^0 = 1 \\ (a \\neq 0), \\qquad a^{-n} = \\frac{1}{a^n}\\]\n\nA negative exponent means \"reciprocal,\" not \"negative number\": \\(2^{-3} = \\tfrac{1}{8}\\), not \\(-8\\).\n\n**Fractional exponents are roots.**\n\n\\[a^{1/n} = \\sqrt[n]{a}, \\qquad a^{m/n} = \\left(\\sqrt[n]{a}\\right)^{m} = \\sqrt[n]{a^{m}}\\]\n\nSo \\(27^{2/3} = \\left(\\sqrt[3]{27}\\right)^2 = 3^2 = 9\\). Take the root first when it keeps the numbers\nsmall.\n\n**Roots simplify by pulling out perfect squares.** \\(\\sqrt{ab} = \\sqrt{a}\\,\\sqrt{b}\\), so\n\n\\[\\sqrt{50} = \\sqrt{25 \\cdot 2} = 5\\sqrt{2}\\]\n\nYou can only **add or subtract** radicals when the part under the root matches:\n\\(5\\sqrt{2} + 3\\sqrt{2} = 8\\sqrt{2}\\). You may **never** merge \\(\\sqrt{a} + \\sqrt{b}\\) into\n\\(\\sqrt{a+b}\\).\n\n**Factoring out a common power** cracks the hardest questions. Terms like \\(2^n + 2^n\\) are not a\nnew base — they collapse:\n\n\\[2^n + 2^n = 2 \\cdot 2^n = 2^{n+1}, \\qquad 3^{k+1} - 3^{k} = 3^{k}(3 - 1) = 2\\cdot 3^{k}\\]\n\n## Worked examples\n\n**Same-base equation.** Solve \\(2^{x+1} = 8^{x-1}\\). Rewrite \\(8 = 2^3\\), so the right side is\n\\(2^{3(x-1)} = 2^{3x-3}\\). Same base ⟹ equal exponents: \\(x + 1 = 3x - 3\\), giving \\(2x = 4\\) and\n\\(x = 2\\).\n\n**Collapse a sum of equal powers.** If \\(2^n + 2^n + 2^n + 2^n = 2^{12}\\), the left side is\n\\(4 \\cdot 2^n = 2^2 \\cdot 2^n = 2^{n+2}\\). So \\(n + 2 = 12\\) and \\(n = 10\\).\n\n**Ratio of powers.** With \\(x = 3^{20}\\) and \\(y = 3^{18}\\), \\(\\dfrac{x}{y} = 3^{20-18} = 3^2 = 9\\).\n\n## Common traps\n\n- **Multiplying exponents when you should add them** (and vice versa). \\(2^5 \\cdot 2^3 = 2^8\\), not\n \\(2^{15}\\).\n- **Reading a negative exponent as a negative number.** \\(3^{-2} = \\tfrac{1}{9}\\), and \\(3^0 = 1\\)\n (not \\(0\\)).\n- **Merging unlike radicals.** \\(\\sqrt{50} + \\sqrt{18} \\neq \\sqrt{68}\\); simplify each first, then add.\n- **Treating a sum of powers as a new base.** \\(2^n + 2^n\\) is \\(2^{n+1}\\), not \\(4^n\\) or \\(2^{2n}\\).\n\n## Key takeaways\n\n- Get everything into a **common base**, then just manipulate exponents.\n- Multiply like bases → **add** exponents; raise a power to a power → **multiply** exponents.\n- \\(a^0 = 1\\), \\(a^{-n} = 1/a^n\\), and \\(a^{m/n} = \\sqrt[n]{a^m}\\).\n- Simplify radicals by pulling out perfect squares; only combine radicals with the same radicand.\n- Sums of equal powers **factor**: \\(k\\) copies of \\(a^n\\) is \\(k \\cdot a^n\\)." + }, { "id": "quant-arithmetic-percents-ratios", "section": "quant", From 6a49c64ced99299265bf9039ec68022f3bc68536 Mon Sep 17 00:00:00 2001 From: Claude Date: Fri, 24 Jul 2026 09:10:22 +0000 Subject: [PATCH 2/3] Seed content: Quant number-properties remainders lesson + 7 questions Adds an under-covered subtopic (quant/number-properties/remainders): - 1 lesson covering the division identity, combining remainders across sums/products, power cyclicity, and simultaneous-condition problems. - 7 original questions (0024-0030) with an easy->hard progression, each with a worked explanation, two hints, and distractors keyed to specific common errors. - Rebuilt docs/data/content.json (46 questions, 10 lessons; zero warnings). All questions status: in-review; author: openmat; original; CC-BY-SA-4.0. Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_013yxWoWPkgojs7Afgr6Pw6L --- .../quant-number-properties-remainders.md | 90 +++++++ ...quant-number-properties-remainders-0024.md | 46 ++++ ...quant-number-properties-remainders-0025.md | 47 ++++ ...quant-number-properties-remainders-0026.md | 47 ++++ ...quant-number-properties-remainders-0027.md | 47 ++++ ...quant-number-properties-remainders-0028.md | 48 ++++ ...quant-number-properties-remainders-0029.md | 47 ++++ ...quant-number-properties-remainders-0030.md | 49 ++++ docs/data/content.json | 232 +++++++++++++++++- 9 files changed, 649 insertions(+), 4 deletions(-) create mode 100644 content/lessons/quant/quant-number-properties-remainders.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0024.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0025.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0026.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0027.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0028.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0029.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0030.md diff --git a/content/lessons/quant/quant-number-properties-remainders.md b/content/lessons/quant/quant-number-properties-remainders.md new file mode 100644 index 0000000..e54efc3 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-remainders.md @@ -0,0 +1,90 @@ +--- +id: quant-number-properties-remainders +section: quant +topic: number-properties +subtopic: remainders +title: "Remainders" +tags: [remainders, divisibility, modular-arithmetic, number-properties] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Remainder questions look like they need long division, but almost none of them do. The whole +topic rests on one identity and a few rules for how remainders behave when you **add**, +**multiply**, and take **powers**. Master those and you can answer most remainder questions in +under a minute — often just by picking the smallest number that fits and testing it. + +## Core concepts + +**The division identity.** When a positive integer \(n\) is divided by a positive divisor \(d\), +there is a unique quotient \(q\) and remainder \(r\) with + +\[n = dq + r, \qquad 0 \le r < d.\] + +The remainder is always **less than the divisor** (a remainder of 7 is impossible when dividing +by 5). Reading a word problem, "quotient 7, remainder 4, divisor 6" becomes +\(n = 6\cdot 7 + 4 = 46\) — nothing more. + +**Remainders add and multiply, then reduce.** If you only care about the remainder mod \(d\), you +may replace any number by its remainder at every step, as long as you **reduce again at the end**: + +\[\text{rem}(a+b) = \text{rem}\big(\text{rem}(a) + \text{rem}(b)\big), \qquad +\text{rem}(ab) = \text{rem}\big(\text{rem}(a)\cdot \text{rem}(b)\big).\] + +So if \(a\) leaves remainder 5 and \(b\) leaves remainder 8 when divided by 12, then \(ab\) leaves +the remainder of \(5\times 8 = 40\), which is \(40 - 36 = 4\). The single most common mistake is +stopping at 40 (or at \(5+8=13\)) and forgetting the final reduction. + +**Powers cycle.** Remainders of \(a^1, a^2, a^3, \dots\) repeat in a short cycle. Dividing by 10 +is the same as asking for the **units digit**, and units digits cycle with period at most 4: + +\[7^1 \to 7,\quad 7^2 \to 9,\quad 7^3 \to 3,\quad 7^4 \to 1,\quad 7^5 \to 7,\ \dots\] + +To find \(7^{20} \bmod 10\), note the cycle length is 4 and \(20\) is a multiple of 4, so \(7^{20}\) +lands on the **last** entry of the cycle: remainder **1**. Match the exponent to its position with +"exponent mod cycle-length," and remember that a remainder of 0 means the *end* of the cycle, not +the start. + +**Pick the smallest number that fits.** "\(n\) leaves remainder 2 when divided by 5" just means +\(n\) is one of \(2, 7, 12, 17, \dots\). For two conditions at once, list each and find the first +overlap; the pattern then repeats every \(\text{lcm}\) of the divisors. + +## Worked examples + +**Translate and compute.** A positive integer divided by 6 gives quotient 7 and remainder 4. +The integer is \(6\cdot 7 + 4 = 46\). (Swapping quotient and remainder — \(6\cdot 4 + 7 = 31\) — is +the classic trap.) + +**Combine two conditions.** Find the smallest integer above 10 that leaves remainder 2 mod 5 and +remainder 1 mod 3. Numbers \(\equiv 2 \pmod 5\): \(2, 7, 12, 17, 22, \dots\); of these, the ones +\(\equiv 1 \pmod 3\) are \(7, 22, 37, \dots\) (they repeat every \(\text{lcm}(5,3)=15\)). The first +one above 10 is **22**. + +**Reason with squares.** If \(n^2\) leaves remainder 4 when divided by 7, what can \(n\) leave? +Test each possible remainder \(r = 0,1,\dots,6\): \(r^2 \bmod 7\) gives \(0,1,4,2,2,4,1\). The value +4 appears at \(r = 2\) and \(r = 5\), so \(n\) leaves remainder **2 or 5** — *not* 4. The remainder of +\(n\) is generally **not** the same as the remainder of \(n^2\). + +## Common traps + +- **Forgetting to reduce at the end.** \(5\times 8 = 40\) is not the answer when dividing by 12; the + remainder is \(40 - 36 = 4\). +- **Adding remainders when the numbers are multiplied** (or vice versa). Combine remainders the + same way you combine the numbers, then reduce. +- **Swapping quotient and remainder** in \(n = dq + r\). +- **A remainder as large as the divisor.** Dividing by \(d\), the remainder must be \(0\) to \(d-1\). +- **Assuming \(n\) and \(n^2\) (or \(n\) and \(2n\)) share a remainder.** They usually don't — test it. + +## Key takeaways + +- Everything flows from \(n = dq + r\) with \(0 \le r < d\). +- To combine numbers, combine their remainders **and reduce again**: works for sums and products. +- Powers of a number cycle; use "exponent mod cycle-length," and treat a result of 0 as the end of + the cycle. +- For "which could be" and two-condition problems, list the smallest values that fit and look for + the overlap; it repeats every lcm of the divisors. diff --git a/content/questions/quant/quant-number-properties-remainders-0024.md b/content/questions/quant/quant-number-properties-remainders-0024.md new file mode 100644 index 0000000..fc015b2 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0024.md @@ -0,0 +1,46 @@ +--- +id: quant-number-properties-remainders-0024 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: easy +tags: [remainders, sums] +choices: + A: "2" + B: "3" + C: "5" + D: "7" + E: "12" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +When the positive integer \(n\) is divided by 9, the remainder is 5. What is the remainder when +\(n + 7\) is divided by 9? + +## Explanation + +You don't need to know \(n\) itself. Adding 7 to \(n\) adds 7 to the remainder, and then you reduce +mod 9: + +\[\text{remainder of } (n+7) = \text{remainder of } (5 + 7) = \text{remainder of } 12 = 12 - 9 = 3.\] + +You can confirm with the smallest such \(n\): \(n = 5\) gives \(n + 7 = 12\), and \(12 \div 9\) leaves +remainder **3**. + +**Traps:** answer **E (12)** is \(5 + 7\) with the final reduction forgotten. Answer **D (7)** is +just the number being added, as if the original remainder didn't matter. Answer **C (5)** assumes +adding 7 leaves the remainder unchanged. Answer **A (2)** comes from *subtracting* the remainders +(\(7 - 5\)) instead of adding. + +## Hints + +- You never need the actual value of \(n\) — work with its remainder. +- Add the remainders, then reduce mod 9 one more time. diff --git a/content/questions/quant/quant-number-properties-remainders-0025.md b/content/questions/quant/quant-number-properties-remainders-0025.md new file mode 100644 index 0000000..575851f --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0025.md @@ -0,0 +1,47 @@ +--- +id: quant-number-properties-remainders-0025 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: easy +tags: [remainders, division-identity] +choices: + A: "31" + B: "42" + C: "46" + D: "52" + E: "66" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +When a positive integer is divided by 6, the quotient is 7 and the remainder is 4. What is the +integer? + +## Explanation + +Every division fits the identity + +\[n = (\text{divisor})(\text{quotient}) + (\text{remainder}).\] + +Here the divisor is 6, the quotient is 7, and the remainder is 4, so + +\[n = 6 \times 7 + 4 = 42 + 4 = 46.\] + +Check: \(46 \div 6 = 7\) with \(46 - 42 = 4\) left over. ✓ + +**Traps:** answer **A (31)** swaps the quotient and remainder (\(6 \times 4 + 7\)). Answer +**B (42)** is \(6 \times 7\) with the remainder dropped. Answer **D (52)** multiplies by one too many +(\(6 \times 8 + 4\)). Answer **E (66)** treats the problem as \(6 \times (7 + 4)\). + +## Hints + +- Use \(n = \text{divisor} \times \text{quotient} + \text{remainder}\). +- Keep the quotient and the remainder in their correct roles — one gets multiplied, one gets added. diff --git a/content/questions/quant/quant-number-properties-remainders-0026.md b/content/questions/quant/quant-number-properties-remainders-0026.md new file mode 100644 index 0000000..526a202 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0026.md @@ -0,0 +1,47 @@ +--- +id: quant-number-properties-remainders-0026 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: medium +tags: [remainders, products] +choices: + A: "1" + B: "2" + C: "4" + D: "8" + E: "11" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +When the positive integers \(a\) and \(b\) are each divided by 12, the remainders are 5 and 8 +respectively. What is the remainder when the product \(ab\) is divided by 12? + +## Explanation + +To get the remainder of a product, multiply the remainders and then reduce mod 12: + +\[\text{remainder of } ab = \text{remainder of } (5 \times 8) = \text{remainder of } 40.\] + +Since \(40 = 12 \times 3 + 4\), the remainder is **4**. + +Confirm with the smallest values: \(a = 5\), \(b = 8\), so \(ab = 40\), and \(40 \div 12\) leaves +remainder **4**. ✓ + +**Traps:** answer **A (1)** *adds* the remainders (\(5 + 8 = 13\), then \(13 - 12 = 1\)) instead of +multiplying them. Answer **D (8)** and answer **E (11)** come from botching the final reduction of +40 (subtracting 32 or 29 rather than a multiple of 12). The key discipline: combine the remainders +the same way the numbers are combined — here, multiply — then reduce one last time. + +## Hints + +- The remainder of a product is the remainder of (remainder \(\times\) remainder). +- \(5 \times 8 = 40\) is not the final answer — reduce 40 modulo 12. diff --git a/content/questions/quant/quant-number-properties-remainders-0027.md b/content/questions/quant/quant-number-properties-remainders-0027.md new file mode 100644 index 0000000..f792de4 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0027.md @@ -0,0 +1,47 @@ +--- +id: quant-number-properties-remainders-0027 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: medium +tags: [remainders, powers, cyclicity, units-digit] +choices: + A: "1" + B: "3" + C: "7" + D: "9" + E: "0" +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the remainder when \(7^{20}\) is divided by 10? + +## Explanation + +Dividing by 10 asks for the **units digit**. The units digit of powers of 7 repeats in a cycle of +length 4: + +\[7^1 \to 7, \quad 7^2 \to 9, \quad 7^3 \to 3, \quad 7^4 \to 1, \quad 7^5 \to 7, \ \dots\] + +To find where \(7^{20}\) lands, divide the exponent by the cycle length: \(20 = 4 \times 5\), so 20 is +a multiple of 4 and \(7^{20}\) sits at the **end** of a cycle — the same spot as \(7^4\), whose units +digit is **1**. The remainder when \(7^{20}\) is divided by 10 is **1**. + +**Traps:** answer **C (7)** treats "\(20 \div 4\) has remainder 0" as if it pointed to the *first* +entry (7) — but a remainder of 0 means the *last* entry of the cycle, not the first. Answers +**D (9)** and **B (3)** are the units digits of \(7^2\) and \(7^3\), chosen by miscounting the +position. Answer **E (0)** wrongly assumes any large power is divisible by 10. + +## Hints + +- Remainder mod 10 is just the units digit. +- Units digits of \(7^n\) cycle 7, 9, 3, 1; use "exponent mod 4," and treat a result of 0 as the end + of the cycle. diff --git a/content/questions/quant/quant-number-properties-remainders-0028.md b/content/questions/quant/quant-number-properties-remainders-0028.md new file mode 100644 index 0000000..be78272 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0028.md @@ -0,0 +1,48 @@ +--- +id: quant-number-properties-remainders-0028 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: medium +tags: [remainders, simultaneous-conditions] +choices: + A: "12" + B: "16" + C: "17" + D: "22" + E: "31" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +The positive integer \(n\) leaves a remainder of 2 when divided by 5 and a remainder of 1 when +divided by 3. What is the least possible value of \(n\) that is greater than 10? + +## Explanation + +List the numbers that satisfy each condition and find the overlap. + +- Remainder 2 when divided by 5: \(2, 7, 12, 17, 22, 27, \dots\) +- Remainder 1 when divided by 3: \(1, 4, 7, 10, 13, 16, 19, 22, 25, \dots\) + +The values in **both** lists are \(7, 22, 37, \dots\) — they repeat every \(\text{lcm}(5, 3) = 15\). +The smallest one greater than 10 is **22**. + +Check: \(22 = 5 \times 4 + 2\) (remainder 2) and \(22 = 3 \times 7 + 1\) (remainder 1). ✓ + +**Traps:** answer **A (12)** satisfies only the mod-5 condition (\(12 \div 3\) leaves remainder 0). +Answer **B (16)** satisfies only the mod-3 condition (\(16 \div 5\) leaves remainder 1). Answer +**C (17)** leaves remainder 2 mod 5 but remainder 2 mod 3, so it fails the second condition. Answer +**E (31)** leaves remainder 1 mod 5, satisfying neither cleanly. Only 22 meets both. + +## Hints + +- Write out the numbers that fit each condition separately, then look for the first common value. +- Above the first overlap, solutions repeat every lcm of the two divisors. diff --git a/content/questions/quant/quant-number-properties-remainders-0029.md b/content/questions/quant/quant-number-properties-remainders-0029.md new file mode 100644 index 0000000..034c27f --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0029.md @@ -0,0 +1,47 @@ +--- +id: quant-number-properties-remainders-0029 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: hard +tags: [remainders, squares, could-be] +choices: + A: "2" + B: "3" + C: "4" + D: "6" + E: "0" +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +The positive integer \(n\) is such that \(n^2\) leaves a remainder of 4 when divided by 7. Which of +the following could be the remainder when \(n\) is divided by 7? + +## Explanation + +The remainder of \(n^2\) depends only on the remainder of \(n\), so test every possible remainder +\(r = 0, 1, 2, \dots, 6\) and see which give \(r^2 \equiv 4 \pmod 7\): + +\[0^2 \to 0,\quad 1^2 \to 1,\quad 2^2 \to 4,\quad 3^2 \to 2,\quad 4^2 \to 2,\quad 5^2 \to 4,\quad 6^2 \to 1.\] + +(For example, \(5^2 = 25 = 7 \times 3 + 4\).) The value 4 appears when \(r = 2\) or \(r = 5\). Of the +answer choices, only **2** is possible. + +Check: if \(n = 2\), then \(n^2 = 4\), and \(4 \div 7\) leaves remainder 4. ✓ + +**Traps:** answer **C (4)** assumes \(n\) leaves the *same* remainder as \(n^2\) — but a number and +its square rarely share a remainder (here \(4^2 = 16\) leaves remainder 2, not 4). Answers **B (3)**, +**D (6)**, and **E (0)** all fail: their squares leave remainders 2, 1, and 0 respectively. + +## Hints + +- The remainder of \(n^2\) depends only on the remainder of \(n\), so just test \(r = 0\) through 6. +- Don't assume \(n\) and \(n^2\) leave the same remainder — square each candidate and check. diff --git a/content/questions/quant/quant-number-properties-remainders-0030.md b/content/questions/quant/quant-number-properties-remainders-0030.md new file mode 100644 index 0000000..c002c22 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0030.md @@ -0,0 +1,49 @@ +--- +id: quant-number-properties-remainders-0030 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: hard +tags: [remainders, simultaneous-conditions, lcm] +choices: + A: "1" + B: "5" + C: "7" + D: "9" + E: "11" +answer: E +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +When the positive integer \(n\) is divided by 4, the remainder is 3, and when \(n\) is divided by 6, +the remainder is 5. What is the remainder when \(n\) is divided by 12? + +## Explanation + +Find a number that satisfies both conditions, then reduce mod 12. List each: + +- Remainder 3 when divided by 4: \(3, 7, 11, 15, 19, 23, \dots\) +- Remainder 5 when divided by 6: \(5, 11, 17, 23, 29, \dots\) + +The first common value is **11**. Both divisors go into \(\text{lcm}(4, 6) = 12\), so the solutions +repeat every 12: \(11, 23, 35, \dots\), all of which leave remainder **11** when divided by 12. + +Check: \(11 = 4 \times 2 + 3\) (remainder 3) and \(11 = 6 \times 1 + 5\) (remainder 5). ✓ + +**Traps:** answers **B (5)** and **C (7)** grab one of the given remainders (or their sum minus one) +without combining the conditions. Answer **D (9)** and answer **A (1)** come from guessing a value +that fits only one of the two divisions. A shortcut worth noticing: each condition says \(n\) is one +short of a multiple (\(n + 1\) is divisible by 4 **and** by 6, hence by 12), so \(n\) is one less than +a multiple of 12 — remainder **11**. + +## Hints + +- Find the smallest \(n\) satisfying both conditions, then take its remainder mod 12. +- Notice that \(n + 1\) is divisible by both 4 and 6 — what does that say about \(n + 1\) and 12? diff --git a/docs/data/content.json b/docs/data/content.json index 617aa4e..e3e130b 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,8 +7,8 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 23, - "lessons": 6 + "questions": 30, + "lessons": 7 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 39, - "lessons": 9 + "questions": 46, + "lessons": 10 }, "questions": [ { @@ -627,6 +627,213 @@ "If \\(N = p^a q^b\\), the divisor count is \\((a+1)(b+1)\\). Remember the \"+1\" accounts for using the prime zero times." ] }, + { + "id": "quant-number-properties-remainders-0024", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "remainders", + "sums" + ], + "choices": { + "A": "2", + "B": "3", + "C": "5", + "D": "7", + "E": "12" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "When the positive integer \\(n\\) is divided by 9, the remainder is 5. What is the remainder when\n\\(n + 7\\) is divided by 9?", + "explanation": "You don't need to know \\(n\\) itself. Adding 7 to \\(n\\) adds 7 to the remainder, and then you reduce\nmod 9:\n\n\\[\\text{remainder of } (n+7) = \\text{remainder of } (5 + 7) = \\text{remainder of } 12 = 12 - 9 = 3.\\]\n\nYou can confirm with the smallest such \\(n\\): \\(n = 5\\) gives \\(n + 7 = 12\\), and \\(12 \\div 9\\) leaves\nremainder **3**.\n\n**Traps:** answer **E (12)** is \\(5 + 7\\) with the final reduction forgotten. Answer **D (7)** is\njust the number being added, as if the original remainder didn't matter. Answer **C (5)** assumes\nadding 7 leaves the remainder unchanged. Answer **A (2)** comes from *subtracting* the remainders\n(\\(7 - 5\\)) instead of adding.", + "hints": [ + "You never need the actual value of \\(n\\) — work with its remainder.", + "Add the remainders, then reduce mod 9 one more time." + ] + }, + { + "id": "quant-number-properties-remainders-0025", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "remainders", + "division-identity" + ], + "choices": { + "A": "31", + "B": "42", + "C": "46", + "D": "52", + "E": "66" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "When a positive integer is divided by 6, the quotient is 7 and the remainder is 4. What is the\ninteger?", + "explanation": "Every division fits the identity\n\n\\[n = (\\text{divisor})(\\text{quotient}) + (\\text{remainder}).\\]\n\nHere the divisor is 6, the quotient is 7, and the remainder is 4, so\n\n\\[n = 6 \\times 7 + 4 = 42 + 4 = 46.\\]\n\nCheck: \\(46 \\div 6 = 7\\) with \\(46 - 42 = 4\\) left over. ✓\n\n**Traps:** answer **A (31)** swaps the quotient and remainder (\\(6 \\times 4 + 7\\)). Answer\n**B (42)** is \\(6 \\times 7\\) with the remainder dropped. Answer **D (52)** multiplies by one too many\n(\\(6 \\times 8 + 4\\)). Answer **E (66)** treats the problem as \\(6 \\times (7 + 4)\\).", + "hints": [ + "Use \\(n = \\text{divisor} \\times \\text{quotient} + \\text{remainder}\\).", + "Keep the quotient and the remainder in their correct roles — one gets multiplied, one gets added." + ] + }, + { + "id": "quant-number-properties-remainders-0026", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "remainders", + "products" + ], + "choices": { + "A": "1", + "B": "2", + "C": "4", + "D": "8", + "E": "11" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "When the positive integers \\(a\\) and \\(b\\) are each divided by 12, the remainders are 5 and 8\nrespectively. What is the remainder when the product \\(ab\\) is divided by 12?", + "explanation": "To get the remainder of a product, multiply the remainders and then reduce mod 12:\n\n\\[\\text{remainder of } ab = \\text{remainder of } (5 \\times 8) = \\text{remainder of } 40.\\]\n\nSince \\(40 = 12 \\times 3 + 4\\), the remainder is **4**.\n\nConfirm with the smallest values: \\(a = 5\\), \\(b = 8\\), so \\(ab = 40\\), and \\(40 \\div 12\\) leaves\nremainder **4**. ✓\n\n**Traps:** answer **A (1)** *adds* the remainders (\\(5 + 8 = 13\\), then \\(13 - 12 = 1\\)) instead of\nmultiplying them. Answer **D (8)** and answer **E (11)** come from botching the final reduction of\n40 (subtracting 32 or 29 rather than a multiple of 12). The key discipline: combine the remainders\nthe same way the numbers are combined — here, multiply — then reduce one last time.", + "hints": [ + "The remainder of a product is the remainder of (remainder \\(\\times\\) remainder).", + "\\(5 \\times 8 = 40\\) is not the final answer — reduce 40 modulo 12." + ] + }, + { + "id": "quant-number-properties-remainders-0027", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "remainders", + "powers", + "cyclicity", + "units-digit" + ], + "choices": { + "A": "1", + "B": "3", + "C": "7", + "D": "9", + "E": "0" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the remainder when \\(7^{20}\\) is divided by 10?", + "explanation": "Dividing by 10 asks for the **units digit**. The units digit of powers of 7 repeats in a cycle of\nlength 4:\n\n\\[7^1 \\to 7, \\quad 7^2 \\to 9, \\quad 7^3 \\to 3, \\quad 7^4 \\to 1, \\quad 7^5 \\to 7, \\ \\dots\\]\n\nTo find where \\(7^{20}\\) lands, divide the exponent by the cycle length: \\(20 = 4 \\times 5\\), so 20 is\na multiple of 4 and \\(7^{20}\\) sits at the **end** of a cycle — the same spot as \\(7^4\\), whose units\ndigit is **1**. The remainder when \\(7^{20}\\) is divided by 10 is **1**.\n\n**Traps:** answer **C (7)** treats \"\\(20 \\div 4\\) has remainder 0\" as if it pointed to the *first*\nentry (7) — but a remainder of 0 means the *last* entry of the cycle, not the first. Answers\n**D (9)** and **B (3)** are the units digits of \\(7^2\\) and \\(7^3\\), chosen by miscounting the\nposition. Answer **E (0)** wrongly assumes any large power is divisible by 10.", + "hints": [ + "Remainder mod 10 is just the units digit.", + "Units digits of \\(7^n\\) cycle 7, 9, 3, 1; use \"exponent mod 4,\" and treat a result of 0 as the end" + ] + }, + { + "id": "quant-number-properties-remainders-0028", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "remainders", + "simultaneous-conditions" + ], + "choices": { + "A": "12", + "B": "16", + "C": "17", + "D": "22", + "E": "31" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "The positive integer \\(n\\) leaves a remainder of 2 when divided by 5 and a remainder of 1 when\ndivided by 3. What is the least possible value of \\(n\\) that is greater than 10?", + "explanation": "List the numbers that satisfy each condition and find the overlap.\n\n- Remainder 2 when divided by 5: \\(2, 7, 12, 17, 22, 27, \\dots\\)\n- Remainder 1 when divided by 3: \\(1, 4, 7, 10, 13, 16, 19, 22, 25, \\dots\\)\n\nThe values in **both** lists are \\(7, 22, 37, \\dots\\) — they repeat every \\(\\text{lcm}(5, 3) = 15\\).\nThe smallest one greater than 10 is **22**.\n\nCheck: \\(22 = 5 \\times 4 + 2\\) (remainder 2) and \\(22 = 3 \\times 7 + 1\\) (remainder 1). ✓\n\n**Traps:** answer **A (12)** satisfies only the mod-5 condition (\\(12 \\div 3\\) leaves remainder 0).\nAnswer **B (16)** satisfies only the mod-3 condition (\\(16 \\div 5\\) leaves remainder 1). Answer\n**C (17)** leaves remainder 2 mod 5 but remainder 2 mod 3, so it fails the second condition. Answer\n**E (31)** leaves remainder 1 mod 5, satisfying neither cleanly. Only 22 meets both.", + "hints": [ + "Write out the numbers that fit each condition separately, then look for the first common value.", + "Above the first overlap, solutions repeat every lcm of the two divisors." + ] + }, + { + "id": "quant-number-properties-remainders-0029", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "remainders", + "squares", + "could-be" + ], + "choices": { + "A": "2", + "B": "3", + "C": "4", + "D": "6", + "E": "0" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "The positive integer \\(n\\) is such that \\(n^2\\) leaves a remainder of 4 when divided by 7. Which of\nthe following could be the remainder when \\(n\\) is divided by 7?", + "explanation": "The remainder of \\(n^2\\) depends only on the remainder of \\(n\\), so test every possible remainder\n\\(r = 0, 1, 2, \\dots, 6\\) and see which give \\(r^2 \\equiv 4 \\pmod 7\\):\n\n\\[0^2 \\to 0,\\quad 1^2 \\to 1,\\quad 2^2 \\to 4,\\quad 3^2 \\to 2,\\quad 4^2 \\to 2,\\quad 5^2 \\to 4,\\quad 6^2 \\to 1.\\]\n\n(For example, \\(5^2 = 25 = 7 \\times 3 + 4\\).) The value 4 appears when \\(r = 2\\) or \\(r = 5\\). Of the\nanswer choices, only **2** is possible.\n\nCheck: if \\(n = 2\\), then \\(n^2 = 4\\), and \\(4 \\div 7\\) leaves remainder 4. ✓\n\n**Traps:** answer **C (4)** assumes \\(n\\) leaves the *same* remainder as \\(n^2\\) — but a number and\nits square rarely share a remainder (here \\(4^2 = 16\\) leaves remainder 2, not 4). Answers **B (3)**,\n**D (6)**, and **E (0)** all fail: their squares leave remainders 2, 1, and 0 respectively.", + "hints": [ + "The remainder of \\(n^2\\) depends only on the remainder of \\(n\\), so just test \\(r = 0\\) through 6.", + "Don't assume \\(n\\) and \\(n^2\\) leave the same remainder — square each candidate and check." + ] + }, + { + "id": "quant-number-properties-remainders-0030", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "remainders", + "simultaneous-conditions", + "lcm" + ], + "choices": { + "A": "1", + "B": "5", + "C": "7", + "D": "9", + "E": "11" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "When the positive integer \\(n\\) is divided by 4, the remainder is 3, and when \\(n\\) is divided by 6,\nthe remainder is 5. What is the remainder when \\(n\\) is divided by 12?", + "explanation": "Find a number that satisfies both conditions, then reduce mod 12. List each:\n\n- Remainder 3 when divided by 4: \\(3, 7, 11, 15, 19, 23, \\dots\\)\n- Remainder 5 when divided by 6: \\(5, 11, 17, 23, 29, \\dots\\)\n\nThe first common value is **11**. Both divisors go into \\(\\text{lcm}(4, 6) = 12\\), so the solutions\nrepeat every 12: \\(11, 23, 35, \\dots\\), all of which leave remainder **11** when divided by 12.\n\nCheck: \\(11 = 4 \\times 2 + 3\\) (remainder 3) and \\(11 = 6 \\times 1 + 5\\) (remainder 5). ✓\n\n**Traps:** answers **B (5)** and **C (7)** grab one of the given remainders (or their sum minus one)\nwithout combining the conditions. Answer **D (9)** and answer **A (1)** come from guessing a value\nthat fits only one of the two divisions. A shortcut worth noticing: each condition says \\(n\\) is one\nshort of a multiple (\\(n + 1\\) is divisible by 4 **and** by 6, hence by 12), so \\(n\\) is one less than\na multiple of 12 — remainder **11**.", + "hints": [ + "Find the smallest \\(n\\) satisfying both conditions, then take its remainder mod 12.", + "Notice that \\(n + 1\\) is divisible by both 4 and 6 — what does that say about \\(n + 1\\) and 12?" + ] + }, { "id": "quant-statistics-descriptive-0009", "section": "quant", @@ -1210,6 +1417,23 @@ "status": "in-review", "body": "## Overview\n\nPercents and ratios are the most frequently tested arithmetic ideas on GMAT Focus Quant, and\nthey hide inside word problems everywhere. Master three moves: converting between forms, handling\npercent *change*, and scaling ratios.\n\n## Core concepts\n\n**Percent as a factor.** A percent is just a number over 100. The fastest way to apply a percent\nchange is to turn it into a **multiplier**:\n\n- Increase by \\(r\\%\\): multiply by \\(1 + \\tfrac{r}{100}\\). (+25% → \\(\\times 1.25\\))\n- Decrease by \\(r\\%\\): multiply by \\(1 - \\tfrac{r}{100}\\). (−20% → \\(\\times 0.80\\))\n\n**Successive changes multiply.** Two changes in a row are multiplied, never added:\n\n\\[(+25\\%)\\text{ then }(-20\\%): \\quad 1.25 \\times 0.80 = 1.00 \\quad (\\text{no net change})\\]\n\n**Percent change formula.**\n\n\\[\\text{percent change} = \\frac{\\text{new} - \\text{old}}{\\text{old}} \\times 100\\%\\]\n\nAlways divide by the **original** value, not the new one.\n\n**Ratios scale together.** A ratio \\(a : b\\) means the quantities are \\(ak\\) and \\(bk\\) for some\nmultiplier \\(k\\). If boys : girls \\(= 3 : 5\\) and there are 24 boys, then \\(k = 8\\), so there are\n\\(5 \\times 8 = 40\\) girls.\n\n## Worked examples\n\n**Percent change.** A stock rises from $80 to $100. Percent increase \\(= \\tfrac{100 - 80}{80} = \\tfrac{20}{80} = 25\\%\\).\nNote it later falls from $100 back to $80: that's \\(\\tfrac{-20}{100} = -20\\%\\) — a *smaller* percent, because the base is now larger.\n\n**Ratio scaling.** A recipe uses flour : sugar \\(= 7 : 2\\). To use 21 cups of flour, \\(k = 3\\), so\nyou need \\(2 \\times 3 = 6\\) cups of sugar.\n\n## Common traps\n\n- **Adding successive percents.** +25% then −20% is *not* +5%; it's \\(1.25 \\times 0.80 = 1.00\\).\n- **Wrong base.** Percent change always divides by the original amount. A rise then an equal-percent fall does not return to the start.\n- **Ratio ≠ actual count.** \\(3 : 5\\) does not mean 3 and 5 — it means \\(3k\\) and \\(5k\\). Find \\(k\\) first.\n\n## Key takeaways\n\n- Convert percent changes to multipliers and multiply them for successive changes.\n- Percent change = (new − old) / old.\n- A ratio \\(a : b\\) represents \\(ak\\) and \\(bk\\); solve for the multiplier \\(k\\), then scale." }, + { + "id": "quant-number-properties-remainders", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "title": "Remainders", + "tags": [ + "remainders", + "divisibility", + "modular-arithmetic", + "number-properties" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nRemainder questions look like they need long division, but almost none of them do. The whole\ntopic rests on one identity and a few rules for how remainders behave when you **add**,\n**multiply**, and take **powers**. Master those and you can answer most remainder questions in\nunder a minute — often just by picking the smallest number that fits and testing it.\n\n## Core concepts\n\n**The division identity.** When a positive integer \\(n\\) is divided by a positive divisor \\(d\\),\nthere is a unique quotient \\(q\\) and remainder \\(r\\) with\n\n\\[n = dq + r, \\qquad 0 \\le r < d.\\]\n\nThe remainder is always **less than the divisor** (a remainder of 7 is impossible when dividing\nby 5). Reading a word problem, \"quotient 7, remainder 4, divisor 6\" becomes\n\\(n = 6\\cdot 7 + 4 = 46\\) — nothing more.\n\n**Remainders add and multiply, then reduce.** If you only care about the remainder mod \\(d\\), you\nmay replace any number by its remainder at every step, as long as you **reduce again at the end**:\n\n\\[\\text{rem}(a+b) = \\text{rem}\\big(\\text{rem}(a) + \\text{rem}(b)\\big), \\qquad\n\\text{rem}(ab) = \\text{rem}\\big(\\text{rem}(a)\\cdot \\text{rem}(b)\\big).\\]\n\nSo if \\(a\\) leaves remainder 5 and \\(b\\) leaves remainder 8 when divided by 12, then \\(ab\\) leaves\nthe remainder of \\(5\\times 8 = 40\\), which is \\(40 - 36 = 4\\). The single most common mistake is\nstopping at 40 (or at \\(5+8=13\\)) and forgetting the final reduction.\n\n**Powers cycle.** Remainders of \\(a^1, a^2, a^3, \\dots\\) repeat in a short cycle. Dividing by 10\nis the same as asking for the **units digit**, and units digits cycle with period at most 4:\n\n\\[7^1 \\to 7,\\quad 7^2 \\to 9,\\quad 7^3 \\to 3,\\quad 7^4 \\to 1,\\quad 7^5 \\to 7,\\ \\dots\\]\n\nTo find \\(7^{20} \\bmod 10\\), note the cycle length is 4 and \\(20\\) is a multiple of 4, so \\(7^{20}\\)\nlands on the **last** entry of the cycle: remainder **1**. Match the exponent to its position with\n\"exponent mod cycle-length,\" and remember that a remainder of 0 means the *end* of the cycle, not\nthe start.\n\n**Pick the smallest number that fits.** \"\\(n\\) leaves remainder 2 when divided by 5\" just means\n\\(n\\) is one of \\(2, 7, 12, 17, \\dots\\). For two conditions at once, list each and find the first\noverlap; the pattern then repeats every \\(\\text{lcm}\\) of the divisors.\n\n## Worked examples\n\n**Translate and compute.** A positive integer divided by 6 gives quotient 7 and remainder 4.\nThe integer is \\(6\\cdot 7 + 4 = 46\\). (Swapping quotient and remainder — \\(6\\cdot 4 + 7 = 31\\) — is\nthe classic trap.)\n\n**Combine two conditions.** Find the smallest integer above 10 that leaves remainder 2 mod 5 and\nremainder 1 mod 3. Numbers \\(\\equiv 2 \\pmod 5\\): \\(2, 7, 12, 17, 22, \\dots\\); of these, the ones\n\\(\\equiv 1 \\pmod 3\\) are \\(7, 22, 37, \\dots\\) (they repeat every \\(\\text{lcm}(5,3)=15\\)). The first\none above 10 is **22**.\n\n**Reason with squares.** If \\(n^2\\) leaves remainder 4 when divided by 7, what can \\(n\\) leave?\nTest each possible remainder \\(r = 0,1,\\dots,6\\): \\(r^2 \\bmod 7\\) gives \\(0,1,4,2,2,4,1\\). The value\n4 appears at \\(r = 2\\) and \\(r = 5\\), so \\(n\\) leaves remainder **2 or 5** — *not* 4. The remainder of\n\\(n\\) is generally **not** the same as the remainder of \\(n^2\\).\n\n## Common traps\n\n- **Forgetting to reduce at the end.** \\(5\\times 8 = 40\\) is not the answer when dividing by 12; the\n remainder is \\(40 - 36 = 4\\).\n- **Adding remainders when the numbers are multiplied** (or vice versa). Combine remainders the\n same way you combine the numbers, then reduce.\n- **Swapping quotient and remainder** in \\(n = dq + r\\).\n- **A remainder as large as the divisor.** Dividing by \\(d\\), the remainder must be \\(0\\) to \\(d-1\\).\n- **Assuming \\(n\\) and \\(n^2\\) (or \\(n\\) and \\(2n\\)) share a remainder.** They usually don't — test it.\n\n## Key takeaways\n\n- Everything flows from \\(n = dq + r\\) with \\(0 \\le r < d\\).\n- To combine numbers, combine their remainders **and reduce again**: works for sums and products.\n- Powers of a number cycle; use \"exponent mod cycle-length,\" and treat a result of 0 as the end of\n the cycle.\n- For \"which could be\" and two-condition problems, list the smallest values that fit and look for\n the overlap; it repeats every lcm of the divisors." + }, { "id": "quant-statistics-descriptive", "section": "quant", From ee080e4aace29188292a01e720917b3156172ceb Mon Sep 17 00:00:00 2001 From: Claude Date: Sun, 26 Jul 2026 09:07:15 +0000 Subject: [PATCH 3/3] Seed content: Quant counting-probability combinatorics lesson + 7 questions Add coverage for the previously-empty counting-probability/combinatorics subtopic: one concept lesson plus 7 original problem-solving questions (0031-0037) with an easy->hard progression. Covers combinations vs permutations, the multiplication principle, identical-item arrangements, circular seating, the at-least-one complement, and the block method for adjacency restrictions. Each question has a worked explanation, hints, and distractors keyed to specific common errors. Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_01C26sxojAipA1Nh7q9EGGty --- ...uant-counting-probability-combinatorics.md | 94 +++++++ ...counting-probability-combinatorics-0031.md | 44 ++++ ...counting-probability-combinatorics-0032.md | 42 +++ ...counting-probability-combinatorics-0033.md | 47 ++++ ...counting-probability-combinatorics-0034.md | 43 ++++ ...counting-probability-combinatorics-0035.md | 44 ++++ ...counting-probability-combinatorics-0036.md | 48 ++++ ...counting-probability-combinatorics-0037.md | 48 ++++ docs/data/content.json | 242 +++++++++++++++++- 9 files changed, 648 insertions(+), 4 deletions(-) create mode 100644 content/lessons/quant/quant-counting-probability-combinatorics.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0031.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0032.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0033.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0034.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0035.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0036.md create mode 100644 content/questions/quant/quant-counting-probability-combinatorics-0037.md diff --git a/content/lessons/quant/quant-counting-probability-combinatorics.md b/content/lessons/quant/quant-counting-probability-combinatorics.md new file mode 100644 index 0000000..db313f0 --- /dev/null +++ b/content/lessons/quant/quant-counting-probability-combinatorics.md @@ -0,0 +1,94 @@ +--- +id: quant-counting-probability-combinatorics +section: quant +topic: counting-probability +subtopic: combinatorics +title: "Combinatorics: Counting Arrangements and Selections" +tags: [combinatorics, permutations, combinations, factorial, counting] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Combinatorics is the art of counting outcomes **without listing them all**. On GMAT Focus +Quant the whole game is deciding a single question: **does order matter?** Get that right and +pick the matching tool, and most counting problems collapse into one line of arithmetic. + +## Core concepts + +**The factorial.** The number of ways to arrange \(n\) distinct objects in a row is +\(n!\) (read "\(n\) factorial"): + +\[n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\] + +So \(5! = 5\cdot4\cdot3\cdot2\cdot1 = 120\). By convention \(0! = 1\). + +**Permutations — order matters.** The number of ways to arrange \(k\) objects chosen from +\(n\) distinct objects, where the order of the chosen objects matters, is + +\[P(n,k) = \frac{n!}{(n-k)!} = n(n-1)\cdots(n-k+1)\] + +Use permutations for rankings, seatings, passwords, or codes — anything where "A then B" +differs from "B then A." + +**Combinations — order does NOT matter.** The number of ways to *choose* \(k\) objects from +\(n\), ignoring order, is + +\[C(n,k) = \binom{n}{k} = \frac{n!}{k!\,(n-k)!}\] + +Use combinations for committees, teams, handshakes, or any unordered selection. Notice +\(C(n,k) = \dfrac{P(n,k)}{k!}\): a combination is a permutation with the \(k!\) reorderings of +the same group divided out. + +**Two governing principles.** + +- **Multiplication (AND).** If a task is a sequence of independent stages, multiply the counts. + Choosing 3 seniors *and* 2 juniors: multiply \(C(5,3)\times C(4,2)\). +- **Addition (OR).** If outcomes fall into mutually exclusive cases, add the counts. + +**Identical items.** When some objects repeat, divide by the factorial of each repeat count. +Distinct arrangements of a word with letter counts \(n_1, n_2, \dots\) totalling \(n\) letters: + +\[\frac{n!}{n_1!\,n_2!\cdots}\] + +**Circular arrangements.** Seating \(n\) people around a round table gives \((n-1)!\) +arrangements, because rotations of the same circle are not distinct — fix one person, then +arrange the rest. + +## Worked examples + +**Combination.** A book club picks 3 of its 8 members to lead a discussion. Order does not +matter, so \(C(8,3) = \dfrac{8\cdot7\cdot6}{3!} = \dfrac{336}{6} = 56\) ways. + +**Permutation.** Those same 8 members award a gold, silver, and bronze badge — now order +matters, so \(P(8,3) = 8\cdot7\cdot6 = 336\) ways. Same people, \(3! = 6\) times as many +outcomes because the badges are distinguishable. + +**Complement ("at least").** From 4 men and 3 women, choose a 3-person team with at least one +woman. Count the total and subtract the unwanted case: \(C(7,3) - C(4,3) = 35 - 4 = 31\). + +## Common traps + +- **Order confusion.** The single biggest error. A committee is unordered (combination); a + lineup or ranking is ordered (permutation). Using \(P\) where \(C\) belongs overcounts by + exactly \(k!\). +- **Adding instead of multiplying.** "3 seniors AND 2 juniors" multiplies: \(C(5,3)\times C(4,2)\), + not \(C(5,3)+C(4,2)\). +- **Forgetting repeated items.** Arranging the letters of a word with repeats requires dividing + \(n!\) by the factorial of *each* repeated letter's count — miss one and you overcount. +- **"At least one" head-on.** Counting cases directly often double-counts. Use the complement: + total minus "none." +- **Blocks that must stay together.** Glue the block into one unit, arrange the units, then + multiply by the internal arrangements of the block. + +## Key takeaways + +- Ask first: **does order matter?** Yes → permutation \(P(n,k)\); no → combination \(C(n,k)\). +- \(C(n,k) = \dfrac{P(n,k)}{k!}\) — dividing out the reorderings is the whole difference. +- Independent stages **multiply**; mutually exclusive cases **add**. +- Repeated identical items → divide by each repeat's factorial; round tables → \((n-1)!\). +- For "at least one," count the complement and subtract. diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0031.md b/content/questions/quant/quant-counting-probability-combinatorics-0031.md new file mode 100644 index 0000000..e503783 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0031.md @@ -0,0 +1,44 @@ +--- +id: quant-counting-probability-combinatorics-0031 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: easy +tags: [combinatorics, combinations, committee] +choices: + A: "21" + B: "24" + C: "56" + D: "336" + E: "512" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A book club will choose a 3-person panel from its 8 members. In how many different ways can the +panel be chosen? + +## Explanation + +A panel is an **unordered** selection — being picked "first" versus "third" makes no difference, +so this is a combination: + +\[C(8,3) = \frac{8 \cdot 7 \cdot 6}{3!} = \frac{336}{6} = 56\] + +**The trap:** answer **D (336)** is \(P(8,3) = 8\cdot7\cdot6\), the *ordered* count. That treats +"Ann, then Bob, then Cara" as different from "Cara, then Bob, then Ann," but a panel is the same +group either way — so you must divide by \(3! = 6\). Answer **B (24)** comes from \(8 \times 3\) +and **A (21)** from \(8 + 7 + 6\); neither reflects how counting works. Answer **E (512)** is +\(8^3\), which would allow the same person to be picked three times. + +## Hints + +- Does the order in which the three people are chosen change who is on the panel? +- If order does not matter, use \(C(n,k) = \dfrac{n!}{k!\,(n-k)!}\). diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0032.md b/content/questions/quant/quant-counting-probability-combinatorics-0032.md new file mode 100644 index 0000000..aabadf9 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0032.md @@ -0,0 +1,42 @@ +--- +id: quant-counting-probability-combinatorics-0032 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: easy +tags: [combinatorics, permutations, factorial, arrangement] +choices: + A: "20" + B: "25" + C: "60" + D: "120" + E: "3125" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +In how many different orders can 5 distinct books be arranged on a shelf? + +## Explanation + +Every book is distinct and every position is distinct, so this is a straight arrangement of all +5 objects: + +\[5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\] + +**The trap:** answer **E (3125)** is \(5^5\), which would count arrangements as if each of the 5 +slots could independently hold any of the 5 books — but once a book is placed it cannot be reused, +so the choices shrink \(5 \to 4 \to 3 \to \dots\). Answer **A (20)** is \(5 \times 4\) (arranging +only 2 of the books), and **C (60)** is \(\tfrac{5!}{2}\). Answer **B (25)** is \(5^2\). + +## Hints + +- Placing the first book leaves 4 choices for the next slot, then 3, and so on. +- Multiply the shrinking number of choices: that product is \(5!\). diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0033.md b/content/questions/quant/quant-counting-probability-combinatorics-0033.md new file mode 100644 index 0000000..3fa4182 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0033.md @@ -0,0 +1,47 @@ +--- +id: quant-counting-probability-combinatorics-0033 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: medium +tags: [combinatorics, combinations, multiplication-principle] +choices: + A: "16" + B: "60" + C: "40" + D: "126" + E: "3600" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A committee of 5 is to be formed from a group of 5 seniors and 4 juniors. If the committee must +contain exactly 3 seniors and exactly 2 juniors, how many different committees are possible? + +## Explanation + +Choosing the seniors and choosing the juniors are two independent stages, and within each stage +order does not matter — so use combinations and **multiply** (the "AND" principle): + +\[C(5,3) \times C(4,2) = 10 \times 6 = 60\] + +Here \(C(5,3) = \dfrac{5\cdot4\cdot3}{3!} = 10\) and \(C(4,2) = \dfrac{4\cdot3}{2!} = 6\). + +**The trap:** answer **A (16)** *adds* the two counts (\(10 + 6\)) instead of multiplying — but you +need a group of seniors **and** a group of juniors together, so the counts multiply. Answer +**C (40)** swaps the roles, computing \(C(5,2)\times C(4,3) = 10 \times 4\) (2 seniors, 3 juniors). +Answer **D (126)** is \(C(9,5)\), which ignores the required senior/junior split. Answer +**E (3600)** treats each selection as ordered (\(P(5,3)\times P(4,2)\)). + +## Hints + +- Count the ways to pick the seniors and the ways to pick the juniors separately. +- "3 seniors AND 2 juniors" — do the two counts add or multiply? +- Within each group, does the order you pick people in matter? diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0034.md b/content/questions/quant/quant-counting-probability-combinatorics-0034.md new file mode 100644 index 0000000..943c315 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0034.md @@ -0,0 +1,43 @@ +--- +id: quant-counting-probability-combinatorics-0034 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: medium +tags: [combinatorics, arrangements, identical-items, permutations] +choices: + A: "120" + B: "60" + C: "30" + D: "20" + E: "10" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +How many distinct arrangements can be made using all the letters of the word **LEVEL**? + +## Explanation + +LEVEL has 5 letters, but they are not all distinct: **L** appears twice, **E** appears twice, and +**V** appears once. When items repeat, divide \(n!\) by the factorial of each repeat count so that +swapping identical letters is not counted as a new arrangement: + +\[\frac{5!}{2!\,\cdot\,2!\,\cdot\,1!} = \frac{120}{2 \cdot 2} = \frac{120}{4} = 30\] + +**The trap:** answer **A (120)** is \(5!\), which counts the two L's (and the two E's) as though +they were distinguishable — but "L\(_1\)EVEL\(_2\)" looks identical to "L\(_2\)EVEL\(_1\)," so those +duplicates must be divided out. Answer **B (60)** is \(\tfrac{5!}{2!}\), correcting for only *one* +of the two repeated letters. Answers **D** and **E** under-count further. + +## Hints + +- Count how many times each letter appears: which letters repeat? +- Divide \(5!\) by the factorial of *each* repeated letter's count. diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0035.md b/content/questions/quant/quant-counting-probability-combinatorics-0035.md new file mode 100644 index 0000000..4ef8447 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0035.md @@ -0,0 +1,44 @@ +--- +id: quant-counting-probability-combinatorics-0035 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: medium +tags: [combinatorics, circular-arrangement, permutations] +choices: + A: "24" + B: "60" + C: "360" + D: "720" + E: "120" +answer: E +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +In how many distinct ways can 6 people be seated around a circular table, if two seatings are +considered the same when one can be obtained from the other by rotating the whole table? + +## Explanation + +Around a circle there is no fixed "first" seat: rotating everyone one seat to the left produces the +same arrangement of neighbors. Fix one person's seat as a reference point and arrange the remaining +\(6 - 1 = 5\) people in the other seats: + +\[(6-1)! = 5! = 120\] + +**The trap:** answer **D (720)** is \(6!\), the count for a *straight row* of 6 seats. In a row all +6 positions are distinct, but a circle's rotations collapse every arrangement into a group of 6 +equivalent rotations, so you divide \(6!\) by 6 — equivalently, use \((n-1)!\). Answer **C (360)** +is \(\tfrac{6!}{2}\), an incorrect partial correction. + +## Hints + +- Around a round table, is there a fixed "first" chair, or do rotations look identical? +- Pin one person down as a reference, then arrange the rest: that gives \((n-1)!\). diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0036.md b/content/questions/quant/quant-counting-probability-combinatorics-0036.md new file mode 100644 index 0000000..7ab8c5e --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0036.md @@ -0,0 +1,48 @@ +--- +id: quant-counting-probability-combinatorics-0036 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: hard +tags: [combinatorics, combinations, complement, at-least] +choices: + A: "31" + B: "4" + C: "18" + D: "35" + E: "45" +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A 3-person team is to be selected from a pool of 4 men and 3 women. How many different teams +include **at least one woman**? + +## Explanation + +"At least one" is fastest to count with the **complement**: take every possible team and subtract +the teams that violate the condition (the all-men teams). + +\[\underbrace{C(7,3)}_{\text{all teams}} - \underbrace{C(4,3)}_{\text{all men}} = 35 - 4 = 31\] + +**The trap:** answer **E (45)** comes from "pick 1 woman, then any 2 of the remaining 6 people": +\(C(3,1)\times C(6,2) = 3 \times 15 = 45\). That double-counts — a team with two women gets counted +once for each woman chosen as "the guaranteed one." Answer **C (18)** counts only *exactly one* +woman, \(C(3,1)\times C(4,2) = 3 \times 6 = 18\), missing the two-woman and three-woman teams. +Answer **D (35)** forgets to remove the all-men teams, and **B (4)** is just the all-men count you +were supposed to subtract. + +You can confirm 31 the direct way: exactly 1 woman \(= 18\), exactly 2 \(= C(3,2)C(4,1) = 12\), +exactly 3 \(= C(3,3)C(4,0) = 1\); \(18 + 12 + 1 = 31\). + +## Hints + +- Counting "at least one" directly is error-prone — try counting the opposite. +- Total teams minus all-men teams leaves exactly the teams with at least one woman. diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0037.md b/content/questions/quant/quant-counting-probability-combinatorics-0037.md new file mode 100644 index 0000000..07d6a24 --- /dev/null +++ b/content/questions/quant/quant-counting-probability-combinatorics-0037.md @@ -0,0 +1,48 @@ +--- +id: quant-counting-probability-combinatorics-0037 +section: quant +topic: counting-probability +subtopic: combinatorics +type: problem-solving +difficulty: hard +tags: [combinatorics, permutations, arrangements, restrictions, block-method] +choices: + A: "24" + B: "48" + C: "72" + D: "120" + E: "240" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +Five students are to be arranged in a row for a photograph. Two of them, Priya and Quinn, insist on +standing next to each other. In how many different arrangements can the five students stand? + +## Explanation + +Use the **block method** for "must be together." Glue Priya and Quinn into a single unit. That +leaves 4 items to arrange in the row — the block plus the other 3 students: + +\[4! = 24 \text{ ways to order the units.}\] + +Within the block, Priya and Quinn can stand in 2 orders (Priya–Quinn or Quinn–Priya), so multiply: + +\[4! \times 2! = 24 \times 2 = 48\] + +**The trap:** answer **C (72)** is the count for the *opposite* condition — arrangements where the +two are **not** adjacent: \(5! - 48 = 120 - 48 = 72\). If you solve for the wrong condition (or +subtract when you should not), you land on 72. Answer **A (24)** forgets the internal \(2!\) for the +block's two orders, and **D (120)** is \(5!\), ignoring the restriction entirely. Answer +**E (240)** is \(5! \times 2\), which double-applies the arrangement. + +## Hints + +- Treat the two who must be adjacent as one combined block, then count the arrangements of the blocks. +- Don't forget the two people inside the block can switch places. diff --git a/docs/data/content.json b/docs/data/content.json index e3e130b..ca4c368 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,8 +7,8 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 30, - "lessons": 7 + "questions": 37, + "lessons": 8 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 46, - "lessons": 10 + "questions": 53, + "lessons": 11 }, "questions": [ { @@ -568,6 +568,222 @@ "Multiply the factors together and compare to 1." ] }, + { + "id": "quant-counting-probability-combinatorics-0031", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "combinatorics", + "combinations", + "committee" + ], + "choices": { + "A": "21", + "B": "24", + "C": "56", + "D": "336", + "E": "512" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A book club will choose a 3-person panel from its 8 members. In how many different ways can the\npanel be chosen?", + "explanation": "A panel is an **unordered** selection — being picked \"first\" versus \"third\" makes no difference,\nso this is a combination:\n\n\\[C(8,3) = \\frac{8 \\cdot 7 \\cdot 6}{3!} = \\frac{336}{6} = 56\\]\n\n**The trap:** answer **D (336)** is \\(P(8,3) = 8\\cdot7\\cdot6\\), the *ordered* count. That treats\n\"Ann, then Bob, then Cara\" as different from \"Cara, then Bob, then Ann,\" but a panel is the same\ngroup either way — so you must divide by \\(3! = 6\\). Answer **B (24)** comes from \\(8 \\times 3\\)\nand **A (21)** from \\(8 + 7 + 6\\); neither reflects how counting works. Answer **E (512)** is\n\\(8^3\\), which would allow the same person to be picked three times.", + "hints": [ + "Does the order in which the three people are chosen change who is on the panel?", + "If order does not matter, use \\(C(n,k) = \\dfrac{n!}{k!\\,(n-k)!}\\)." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0032", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "combinatorics", + "permutations", + "factorial", + "arrangement" + ], + "choices": { + "A": "20", + "B": "25", + "C": "60", + "D": "120", + "E": "3125" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "In how many different orders can 5 distinct books be arranged on a shelf?", + "explanation": "Every book is distinct and every position is distinct, so this is a straight arrangement of all\n5 objects:\n\n\\[5! = 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120\\]\n\n**The trap:** answer **E (3125)** is \\(5^5\\), which would count arrangements as if each of the 5\nslots could independently hold any of the 5 books — but once a book is placed it cannot be reused,\nso the choices shrink \\(5 \\to 4 \\to 3 \\to \\dots\\). Answer **A (20)** is \\(5 \\times 4\\) (arranging\nonly 2 of the books), and **C (60)** is \\(\\tfrac{5!}{2}\\). Answer **B (25)** is \\(5^2\\).", + "hints": [ + "Placing the first book leaves 4 choices for the next slot, then 3, and so on.", + "Multiply the shrinking number of choices: that product is \\(5!\\)." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0033", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "combinatorics", + "combinations", + "multiplication-principle" + ], + "choices": { + "A": "16", + "B": "60", + "C": "40", + "D": "126", + "E": "3600" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A committee of 5 is to be formed from a group of 5 seniors and 4 juniors. If the committee must\ncontain exactly 3 seniors and exactly 2 juniors, how many different committees are possible?", + "explanation": "Choosing the seniors and choosing the juniors are two independent stages, and within each stage\norder does not matter — so use combinations and **multiply** (the \"AND\" principle):\n\n\\[C(5,3) \\times C(4,2) = 10 \\times 6 = 60\\]\n\nHere \\(C(5,3) = \\dfrac{5\\cdot4\\cdot3}{3!} = 10\\) and \\(C(4,2) = \\dfrac{4\\cdot3}{2!} = 6\\).\n\n**The trap:** answer **A (16)** *adds* the two counts (\\(10 + 6\\)) instead of multiplying — but you\nneed a group of seniors **and** a group of juniors together, so the counts multiply. Answer\n**C (40)** swaps the roles, computing \\(C(5,2)\\times C(4,3) = 10 \\times 4\\) (2 seniors, 3 juniors).\nAnswer **D (126)** is \\(C(9,5)\\), which ignores the required senior/junior split. Answer\n**E (3600)** treats each selection as ordered (\\(P(5,3)\\times P(4,2)\\)).", + "hints": [ + "Count the ways to pick the seniors and the ways to pick the juniors separately.", + "\"3 seniors AND 2 juniors\" — do the two counts add or multiply?", + "Within each group, does the order you pick people in matter?" + ] + }, + { + "id": "quant-counting-probability-combinatorics-0034", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "combinatorics", + "arrangements", + "identical-items", + "permutations" + ], + "choices": { + "A": "120", + "B": "60", + "C": "30", + "D": "20", + "E": "10" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "How many distinct arrangements can be made using all the letters of the word **LEVEL**?", + "explanation": "LEVEL has 5 letters, but they are not all distinct: **L** appears twice, **E** appears twice, and\n**V** appears once. When items repeat, divide \\(n!\\) by the factorial of each repeat count so that\nswapping identical letters is not counted as a new arrangement:\n\n\\[\\frac{5!}{2!\\,\\cdot\\,2!\\,\\cdot\\,1!} = \\frac{120}{2 \\cdot 2} = \\frac{120}{4} = 30\\]\n\n**The trap:** answer **A (120)** is \\(5!\\), which counts the two L's (and the two E's) as though\nthey were distinguishable — but \"L\\(_1\\)EVEL\\(_2\\)\" looks identical to \"L\\(_2\\)EVEL\\(_1\\),\" so those\nduplicates must be divided out. Answer **B (60)** is \\(\\tfrac{5!}{2!}\\), correcting for only *one*\nof the two repeated letters. Answers **D** and **E** under-count further.", + "hints": [ + "Count how many times each letter appears: which letters repeat?", + "Divide \\(5!\\) by the factorial of *each* repeated letter's count." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0035", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "combinatorics", + "circular-arrangement", + "permutations" + ], + "choices": { + "A": "24", + "B": "60", + "C": "360", + "D": "720", + "E": "120" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "In how many distinct ways can 6 people be seated around a circular table, if two seatings are\nconsidered the same when one can be obtained from the other by rotating the whole table?", + "explanation": "Around a circle there is no fixed \"first\" seat: rotating everyone one seat to the left produces the\nsame arrangement of neighbors. Fix one person's seat as a reference point and arrange the remaining\n\\(6 - 1 = 5\\) people in the other seats:\n\n\\[(6-1)! = 5! = 120\\]\n\n**The trap:** answer **D (720)** is \\(6!\\), the count for a *straight row* of 6 seats. In a row all\n6 positions are distinct, but a circle's rotations collapse every arrangement into a group of 6\nequivalent rotations, so you divide \\(6!\\) by 6 — equivalently, use \\((n-1)!\\). Answer **C (360)**\nis \\(\\tfrac{6!}{2}\\), an incorrect partial correction.", + "hints": [ + "Around a round table, is there a fixed \"first\" chair, or do rotations look identical?", + "Pin one person down as a reference, then arrange the rest: that gives \\((n-1)!\\)." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0036", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "combinatorics", + "combinations", + "complement", + "at-least" + ], + "choices": { + "A": "31", + "B": "4", + "C": "18", + "D": "35", + "E": "45" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A 3-person team is to be selected from a pool of 4 men and 3 women. How many different teams\ninclude **at least one woman**?", + "explanation": "\"At least one\" is fastest to count with the **complement**: take every possible team and subtract\nthe teams that violate the condition (the all-men teams).\n\n\\[\\underbrace{C(7,3)}_{\\text{all teams}} - \\underbrace{C(4,3)}_{\\text{all men}} = 35 - 4 = 31\\]\n\n**The trap:** answer **E (45)** comes from \"pick 1 woman, then any 2 of the remaining 6 people\":\n\\(C(3,1)\\times C(6,2) = 3 \\times 15 = 45\\). That double-counts — a team with two women gets counted\nonce for each woman chosen as \"the guaranteed one.\" Answer **C (18)** counts only *exactly one*\nwoman, \\(C(3,1)\\times C(4,2) = 3 \\times 6 = 18\\), missing the two-woman and three-woman teams.\nAnswer **D (35)** forgets to remove the all-men teams, and **B (4)** is just the all-men count you\nwere supposed to subtract.\n\nYou can confirm 31 the direct way: exactly 1 woman \\(= 18\\), exactly 2 \\(= C(3,2)C(4,1) = 12\\),\nexactly 3 \\(= C(3,3)C(4,0) = 1\\); \\(18 + 12 + 1 = 31\\).", + "hints": [ + "Counting \"at least one\" directly is error-prone — try counting the opposite.", + "Total teams minus all-men teams leaves exactly the teams with at least one woman." + ] + }, + { + "id": "quant-counting-probability-combinatorics-0037", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "combinatorics", + "permutations", + "arrangements", + "restrictions", + "block-method" + ], + "choices": { + "A": "24", + "B": "48", + "C": "72", + "D": "120", + "E": "240" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "Five students are to be arranged in a row for a photograph. Two of them, Priya and Quinn, insist on\nstanding next to each other. In how many different arrangements can the five students stand?", + "explanation": "Use the **block method** for \"must be together.\" Glue Priya and Quinn into a single unit. That\nleaves 4 items to arrange in the row — the block plus the other 3 students:\n\n\\[4! = 24 \\text{ ways to order the units.}\\]\n\nWithin the block, Priya and Quinn can stand in 2 orders (Priya–Quinn or Quinn–Priya), so multiply:\n\n\\[4! \\times 2! = 24 \\times 2 = 48\\]\n\n**The trap:** answer **C (72)** is the count for the *opposite* condition — arrangements where the\ntwo are **not** adjacent: \\(5! - 48 = 120 - 48 = 72\\). If you solve for the wrong condition (or\nsubtract when you should not), you land on 72. Answer **A (24)** forgets the internal \\(2!\\) for the\nblock's two orders, and **D (120)** is \\(5!\\), ignoring the restriction entirely. Answer\n**E (240)** is \\(5! \\times 2\\), which double-applies the arrangement.", + "hints": [ + "Treat the two who must be adjacent as one combined block, then count the arrangements of the blocks.", + "Don't forget the two people inside the block can switch places." + ] + }, { "id": "quant-counting-probability-probability-0007", "section": "quant", @@ -1417,6 +1633,24 @@ "status": "in-review", "body": "## Overview\n\nPercents and ratios are the most frequently tested arithmetic ideas on GMAT Focus Quant, and\nthey hide inside word problems everywhere. Master three moves: converting between forms, handling\npercent *change*, and scaling ratios.\n\n## Core concepts\n\n**Percent as a factor.** A percent is just a number over 100. The fastest way to apply a percent\nchange is to turn it into a **multiplier**:\n\n- Increase by \\(r\\%\\): multiply by \\(1 + \\tfrac{r}{100}\\). (+25% → \\(\\times 1.25\\))\n- Decrease by \\(r\\%\\): multiply by \\(1 - \\tfrac{r}{100}\\). (−20% → \\(\\times 0.80\\))\n\n**Successive changes multiply.** Two changes in a row are multiplied, never added:\n\n\\[(+25\\%)\\text{ then }(-20\\%): \\quad 1.25 \\times 0.80 = 1.00 \\quad (\\text{no net change})\\]\n\n**Percent change formula.**\n\n\\[\\text{percent change} = \\frac{\\text{new} - \\text{old}}{\\text{old}} \\times 100\\%\\]\n\nAlways divide by the **original** value, not the new one.\n\n**Ratios scale together.** A ratio \\(a : b\\) means the quantities are \\(ak\\) and \\(bk\\) for some\nmultiplier \\(k\\). If boys : girls \\(= 3 : 5\\) and there are 24 boys, then \\(k = 8\\), so there are\n\\(5 \\times 8 = 40\\) girls.\n\n## Worked examples\n\n**Percent change.** A stock rises from $80 to $100. Percent increase \\(= \\tfrac{100 - 80}{80} = \\tfrac{20}{80} = 25\\%\\).\nNote it later falls from $100 back to $80: that's \\(\\tfrac{-20}{100} = -20\\%\\) — a *smaller* percent, because the base is now larger.\n\n**Ratio scaling.** A recipe uses flour : sugar \\(= 7 : 2\\). To use 21 cups of flour, \\(k = 3\\), so\nyou need \\(2 \\times 3 = 6\\) cups of sugar.\n\n## Common traps\n\n- **Adding successive percents.** +25% then −20% is *not* +5%; it's \\(1.25 \\times 0.80 = 1.00\\).\n- **Wrong base.** Percent change always divides by the original amount. A rise then an equal-percent fall does not return to the start.\n- **Ratio ≠ actual count.** \\(3 : 5\\) does not mean 3 and 5 — it means \\(3k\\) and \\(5k\\). Find \\(k\\) first.\n\n## Key takeaways\n\n- Convert percent changes to multipliers and multiply them for successive changes.\n- Percent change = (new − old) / old.\n- A ratio \\(a : b\\) represents \\(ak\\) and \\(bk\\); solve for the multiplier \\(k\\), then scale." }, + { + "id": "quant-counting-probability-combinatorics", + "section": "quant", + "topic": "counting-probability", + "subtopic": "combinatorics", + "title": "Combinatorics: Counting Arrangements and Selections", + "tags": [ + "combinatorics", + "permutations", + "combinations", + "factorial", + "counting" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nCombinatorics is the art of counting outcomes **without listing them all**. On GMAT Focus\nQuant the whole game is deciding a single question: **does order matter?** Get that right and\npick the matching tool, and most counting problems collapse into one line of arithmetic.\n\n## Core concepts\n\n**The factorial.** The number of ways to arrange \\(n\\) distinct objects in a row is\n\\(n!\\) (read \"\\(n\\) factorial\"):\n\n\\[n! = n \\times (n-1) \\times (n-2) \\times \\cdots \\times 2 \\times 1\\]\n\nSo \\(5! = 5\\cdot4\\cdot3\\cdot2\\cdot1 = 120\\). By convention \\(0! = 1\\).\n\n**Permutations — order matters.** The number of ways to arrange \\(k\\) objects chosen from\n\\(n\\) distinct objects, where the order of the chosen objects matters, is\n\n\\[P(n,k) = \\frac{n!}{(n-k)!} = n(n-1)\\cdots(n-k+1)\\]\n\nUse permutations for rankings, seatings, passwords, or codes — anything where \"A then B\"\ndiffers from \"B then A.\"\n\n**Combinations — order does NOT matter.** The number of ways to *choose* \\(k\\) objects from\n\\(n\\), ignoring order, is\n\n\\[C(n,k) = \\binom{n}{k} = \\frac{n!}{k!\\,(n-k)!}\\]\n\nUse combinations for committees, teams, handshakes, or any unordered selection. Notice\n\\(C(n,k) = \\dfrac{P(n,k)}{k!}\\): a combination is a permutation with the \\(k!\\) reorderings of\nthe same group divided out.\n\n**Two governing principles.**\n\n- **Multiplication (AND).** If a task is a sequence of independent stages, multiply the counts.\n Choosing 3 seniors *and* 2 juniors: multiply \\(C(5,3)\\times C(4,2)\\).\n- **Addition (OR).** If outcomes fall into mutually exclusive cases, add the counts.\n\n**Identical items.** When some objects repeat, divide by the factorial of each repeat count.\nDistinct arrangements of a word with letter counts \\(n_1, n_2, \\dots\\) totalling \\(n\\) letters:\n\n\\[\\frac{n!}{n_1!\\,n_2!\\cdots}\\]\n\n**Circular arrangements.** Seating \\(n\\) people around a round table gives \\((n-1)!\\)\narrangements, because rotations of the same circle are not distinct — fix one person, then\narrange the rest.\n\n## Worked examples\n\n**Combination.** A book club picks 3 of its 8 members to lead a discussion. Order does not\nmatter, so \\(C(8,3) = \\dfrac{8\\cdot7\\cdot6}{3!} = \\dfrac{336}{6} = 56\\) ways.\n\n**Permutation.** Those same 8 members award a gold, silver, and bronze badge — now order\nmatters, so \\(P(8,3) = 8\\cdot7\\cdot6 = 336\\) ways. Same people, \\(3! = 6\\) times as many\noutcomes because the badges are distinguishable.\n\n**Complement (\"at least\").** From 4 men and 3 women, choose a 3-person team with at least one\nwoman. Count the total and subtract the unwanted case: \\(C(7,3) - C(4,3) = 35 - 4 = 31\\).\n\n## Common traps\n\n- **Order confusion.** The single biggest error. A committee is unordered (combination); a\n lineup or ranking is ordered (permutation). Using \\(P\\) where \\(C\\) belongs overcounts by\n exactly \\(k!\\).\n- **Adding instead of multiplying.** \"3 seniors AND 2 juniors\" multiplies: \\(C(5,3)\\times C(4,2)\\),\n not \\(C(5,3)+C(4,2)\\).\n- **Forgetting repeated items.** Arranging the letters of a word with repeats requires dividing\n \\(n!\\) by the factorial of *each* repeated letter's count — miss one and you overcount.\n- **\"At least one\" head-on.** Counting cases directly often double-counts. Use the complement:\n total minus \"none.\"\n- **Blocks that must stay together.** Glue the block into one unit, arrange the units, then\n multiply by the internal arrangements of the block.\n\n## Key takeaways\n\n- Ask first: **does order matter?** Yes → permutation \\(P(n,k)\\); no → combination \\(C(n,k)\\).\n- \\(C(n,k) = \\dfrac{P(n,k)}{k!}\\) — dividing out the reorderings is the whole difference.\n- Independent stages **multiply**; mutually exclusive cases **add**.\n- Repeated identical items → divide by each repeat's factorial; round tables → \\((n-1)!\\).\n- For \"at least one,\" count the complement and subtract." + }, { "id": "quant-number-properties-remainders", "section": "quant",