From c66d12a378d92600cb336a6994d6fb3503bf5691 Mon Sep 17 00:00:00 2001 From: Claude Date: Fri, 17 Jul 2026 09:09:28 +0000 Subject: [PATCH] Seed content: Number Properties & Exponents (+6 questions, +2 lessons) Add a coherent Quant batch on number properties and exponents, filling several previously uncovered subtopics, plus a Verbal CR paradox question: Lessons - quant-arithmetic-exponents-roots (Exponents and Roots) - quant-number-properties-remainders (Remainders) Questions - exponents-roots: substitution (easy) and factoring equal powers (medium) - number-properties: remainders, odds/evens parity, factors via inclusion-exclusion - verbal critical-reasoning: paradox (resolve/explain) Rebuilds docs/data/content.json (38 questions, 10 lessons). Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_01UKcsGFztmGMUBUYfPSzfhW --- .../quant/quant-arithmetic-exponents-roots.md | 74 ++++++ .../quant-number-properties-remainders.md | 77 ++++++ .../quant-arithmetic-exponents-roots-0017.md | 45 ++++ .../quant-arithmetic-exponents-roots-0018.md | 50 ++++ ...umber-properties-factors-multiples-0021.md | 52 ++++ ...number-properties-odds-evens-signs-0020.md | 49 ++++ ...quant-number-properties-remainders-0019.md | 54 +++++ .../verbal-critical-reasoning-paradox-0009.md | 60 +++++ docs/data/content.json | 224 +++++++++++++++++- 9 files changed, 680 insertions(+), 5 deletions(-) create mode 100644 content/lessons/quant/quant-arithmetic-exponents-roots.md create mode 100644 content/lessons/quant/quant-number-properties-remainders.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-number-properties-factors-multiples-0021.md create mode 100644 content/questions/quant/quant-number-properties-odds-evens-signs-0020.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0019.md create mode 100644 content/questions/verbal/verbal-critical-reasoning-paradox-0009.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..d86add7 --- /dev/null +++ b/content/lessons/quant/quant-arithmetic-exponents-roots.md @@ -0,0 +1,74 @@ +--- +id: quant-arithmetic-exponents-roots +section: quant +topic: arithmetic +subtopic: exponents-roots +title: "Exponents and Roots" +tags: [exponents, roots, powers, rules-of-exponents] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Exponent questions on the GMAT rarely test heavy computation. They test whether you can +**rewrite** expressions into a common base and apply a small set of rules cleanly. Get the base +consistent, and most problems collapse in one or two steps. + +## Core concepts + +For any nonzero base \(a\) and integers \(m, n\): + +- **Product rule** — same base, add exponents: \(a^m \cdot a^n = a^{m+n}\). +- **Quotient rule** — same base, subtract exponents: \(\dfrac{a^m}{a^n} = a^{m-n}\). +- **Power of a power** — multiply exponents: \((a^m)^n = a^{mn}\). +- **Zero exponent** — \(a^0 = 1\) for any \(a \neq 0\). +- **Negative exponent** — a reciprocal: \(a^{-n} = \dfrac{1}{a^n}\). +- **Fractional exponent** — a root: \(a^{1/n} = \sqrt[n]{a}\), and \(a^{m/n} = \sqrt[n]{a^m}\). + +The single most useful move is **matching bases**. Since \(4 = 2^2\), \(8 = 2^3\), \(9 = 3^2\), +\(27 = 3^3\), you can turn a messy fraction of different bases into one base and just subtract +exponents. + +Two rules that do **not** exist (the classic traps): + +- \(a^m + a^n \neq a^{m+n}\). Addition of powers is **not** a rule — you must factor instead. +- \((a + b)^2 \neq a^2 + b^2\). Expand it: \((a+b)^2 = a^2 + 2ab + b^2\). + +## Worked examples + +**Example 1 — match the base.** Simplify \(\dfrac{4^{10}}{2^{15}}\). + +Rewrite \(4^{10} = (2^2)^{10} = 2^{20}\). Then \(\dfrac{2^{20}}{2^{15}} = 2^{20-15} = 2^5 = 32\). + +**Example 2 — factor a sum of equal powers.** If \(3^n + 3^n + 3^n = 3^7\), find \(n\). + +You cannot add the exponents. Instead, three copies of \(3^n\) is \(3 \cdot 3^n = 3^{1+n}\). +So \(3^{n+1} = 3^7\), which gives \(n + 1 = 7\), so \(n = 6\). + +**Example 3 — negative and fractional exponents.** Evaluate \(27^{-2/3}\). + +Work outward: \(27^{1/3} = 3\), then \(3^2 = 9\), and the negative sign takes the reciprocal: +\(27^{-2/3} = \dfrac{1}{9}\). + +## Common traps + +- **Adding powers as if it were a rule.** When you see \(2^x + 2^x\), factor: it equals + \(2 \cdot 2^x = 2^{x+1}\), not \(2^{2x}\) or \(4^x\)... well, \(2^{x+1}\) *is* correct; the + point is to factor rather than invent an addition rule. +- **Forgetting to convert to a common base** before applying the quotient rule. \(\tfrac{4^{10}}{2^{15}}\) + is not \(2^{10-15}\) — the numerator's base must become 2 first. +- **Mishandling the negative sign vs. the exponent.** \(-2^4 = -16\) (the exponent binds before + the sign), but \((-2)^4 = 16\). +- **Answering the wrong quantity.** If they ask for \(2^{x+3}\) and you find \(x\), remember to + finish the substitution. + +## Key takeaways + +- Convert everything to a **common base**, then add/subtract/multiply exponents by the rules. +- **Sums of powers are factored, not combined** — there is no addition rule for exponents. +- A fractional exponent is a root; a negative exponent is a reciprocal. Apply them one at a time. +- Memorize small powers of 2 and 3 (\(2^1\ldots2^{10}\), \(3^1\ldots3^4\)) to move quickly. 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..88393a5 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-remainders.md @@ -0,0 +1,77 @@ +--- +id: quant-number-properties-remainders +section: quant +topic: number-properties +subtopic: remainders +title: "Remainders" +tags: [remainders, divisibility, modular-arithmetic, cyclicity] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +A remainder question is really a question about **structure**: what an integer looks like when +you write it in terms of the divisor. Once you write \(n = dq + r\), most remainder problems +become simple substitution or a short search over cases. + +## 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 at least 0 and strictly less than the divisor. "Divided by 5" means the +remainder is one of \(0, 1, 2, 3, 4\). + +**Remainders travel through addition and multiplication.** If you only care about the remainder +of a sum or product, you can replace each number by its own remainder first: + +- remainder of \((a + b)\) = remainder of (remainder of \(a\) + remainder of \(b\)); +- remainder of \((a \cdot b)\) = remainder of (remainder of \(a\) \(\times\) remainder of \(b\)). + +Then reduce again if the result is still \(\ge d\). This lets you handle huge numbers without +ever computing them. + +**Units digits cycle.** The units digit of a power repeats on a short cycle, which is a +remainder-mod-10 fact. For example, powers of 3 cycle \(3, 9, 7, 1, 3, 9, 7, 1, \ldots\) with +period 4. + +## Worked examples + +**Example 1 — substitute the form.** When \(n\) is divided by 5 the remainder is 3. What is the +remainder when \(3n\) is divided by 5? + +Write \(n = 5k + 3\). Then \(3n = 15k + 9\). Since \(15k\) is a multiple of 5, only the 9 +matters, and \(9 = 5 + 4\), so the remainder is **4**. (Shortcut with the multiplication rule: +remainder of \(3 \times 3 = 9\), reduce mod 5 to get 4.) + +**Example 2 — combine two conditions.** A number leaves remainder 2 when divided by 3 and +remainder 2 when divided by 4. What is the smallest such positive integer greater than 2? + +Numbers that are 2 more than a multiple of both 3 and 4 are 2 more than a multiple of 12: +\(12m + 2\). The smallest above 2 is \(14\). + +**Example 3 — units digit via cyclicity.** What is the units digit of \(3^{23}\)? + +The cycle of the units digit of powers of 3 has length 4: \(3, 9, 7, 1\). Divide the exponent by +4: \(23 = 4 \cdot 5 + 3\), so we are at position 3 in the cycle, which is **7**. + +## Common traps + +- **Remainder of at least the divisor.** A remainder when dividing by 5 can never be 5 or more. + If your arithmetic gives 9, reduce it (\(9 \bmod 5 = 4\)). +- **Assuming the remainder is unchanged when you scale.** Multiplying \(n\) by 3 does *not* keep + the remainder at 3 — you must recompute (Example 1). +- **Zero is a valid remainder.** "Divisible by \(d\)" is exactly "remainder 0." +- **Off-by-one on cyclicity.** A remainder of 0 in the exponent lands on the *last* term of the + cycle, not the first. + +## Key takeaways + +- Write \(n = dq + r\) with \(0 \le r < d\); substitute and simplify. +- Remainders pass cleanly through addition and multiplication — reduce early, reduce often. +- For units digits of powers, find the cycle length and take the exponent mod that length. +- Always sanity-check that your final remainder is less than the divisor. 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..ad67ee3 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0017.md @@ -0,0 +1,45 @@ +--- +id: quant-arithmetic-exponents-roots-0017 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, powers-of-two, substitution] +choices: + A: "35" + B: "64" + C: "128" + D: "256" + E: "512" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(2^x = 32\), what is the value of \(2^{x+3}\)? + +## Explanation + +You don't even need to solve for \(x\). Use the product rule to split the exponent: +\[2^{x+3} = 2^x \cdot 2^3.\] + +You're told \(2^x = 32\), and \(2^3 = 8\), so +\[2^{x+3} = 32 \cdot 8 = 256.\] + +(If you prefer to find \(x\): \(2^x = 32 = 2^5\), so \(x = 5\), and \(2^{5+3} = 2^8 = 256\).) +**Correct: (D).** + +**The trap:** choice **(A) 35** comes from adding \(32 + 3\) — treating the exponent's "+3" as +if it were added to the value. Exponents add when powers of the same base are *multiplied*, so +the "+3" multiplies by \(2^3 = 8\). + +## Hints + +- \(2^{x+3}\) can be rewritten as \(2^x \cdot 2^3\) — and you already know \(2^x\). +- "+3" in the exponent is not "+3" in the value; it means one more factor of \(2^3\). 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..0d47a44 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0018.md @@ -0,0 +1,50 @@ +--- +id: quant-arithmetic-exponents-roots-0018 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [exponents, factoring, same-base] +choices: + A: "5" + B: "6" + C: "7" + D: "8" + E: "21" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(3^n + 3^n + 3^n = 3^7\), what is the value of \(n\)? + +## Explanation + +There is no rule that lets you add exponents when powers are *added*, so don't touch the 7 yet. +Instead, notice the left side is three identical copies of \(3^n\): +\[3^n + 3^n + 3^n = 3 \cdot 3^n.\] + +Now use the product rule, since \(3 = 3^1\): +\[3 \cdot 3^n = 3^{1} \cdot 3^{n} = 3^{\,n+1}.\] + +So the equation becomes \(3^{\,n+1} = 3^7\). Same base means equal exponents: +\[n + 1 = 7 \implies n = 6.\] +**Correct: (B).** + +Why the others fail: + +- **(C) 7** treats \(3^n + 3^n + 3^n\) as if it were just \(3^n\), ignoring the coefficient of 3. +- **(E) 21** comes from multiplying the "3 copies" into the exponent (\(3 \times 7\)) — combining + the count of terms with the exponent, which no rule permits. +- **(A) 5** and **(D) 8** are off-by-one slips on \(n + 1 = 7\). + +## Hints + +- You can't add the exponents. Factor the left side: how many copies of \(3^n\) are there? +- Three copies of \(3^n\) is \(3 \cdot 3^n\), and \(3 = 3^1\). diff --git a/content/questions/quant/quant-number-properties-factors-multiples-0021.md b/content/questions/quant/quant-number-properties-factors-multiples-0021.md new file mode 100644 index 0000000..fb3ee20 --- /dev/null +++ b/content/questions/quant/quant-number-properties-factors-multiples-0021.md @@ -0,0 +1,52 @@ +--- +id: quant-number-properties-factors-multiples-0021 +section: quant +topic: number-properties +subtopic: factors-multiples +type: problem-solving +difficulty: hard +tags: [factors-multiples, counting, inclusion-exclusion] +choices: + A: "30" + B: "33" + C: "34" + D: "50" + E: "66" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +How many positive integers less than 100 are divisible by neither 2 nor 3? + +## Explanation + +The integers in play are \(1, 2, \ldots, 99\) — that's 99 numbers. It's easier to **count what to +throw away** (multiples of 2 or 3) and subtract, using inclusion–exclusion. + +- Divisible by 2: \(\left\lfloor \tfrac{99}{2} \right\rfloor = 49\). +- Divisible by 3: \(\left\lfloor \tfrac{99}{3} \right\rfloor = 33\). +- Divisible by both 2 and 3 (i.e. by 6): \(\left\lfloor \tfrac{99}{6} \right\rfloor = 16\). + +By inclusion–exclusion, the count divisible by 2 **or** 3 is +\[49 + 33 - 16 = 66.\] + +So the count divisible by **neither** is +\[99 - 66 = 33.\] +**Correct: (B).** + +**Traps:** + +- **(E) 66** is the number divisible by 2 or 3 — the complement of what's asked. +- **Forgetting to subtract the overlap (16)** double-counts multiples of 6 and gives \(99 - 82 = 17\), + or leads to other wrong totals; inclusion–exclusion prevents this. + +## Hints + +- Count the "bad" numbers (multiples of 2 or 3) and subtract from 99. +- Don't double-count: multiples of 6 are in both the "÷2" and "÷3" groups, so subtract them once. diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0020.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0020.md new file mode 100644 index 0000000..62adb76 --- /dev/null +++ b/content/questions/quant/quant-number-properties-odds-evens-signs-0020.md @@ -0,0 +1,49 @@ +--- +id: quant-number-properties-odds-evens-signs-0020 +section: quant +topic: number-properties +subtopic: odds-evens-signs +type: problem-solving +difficulty: medium +tags: [odd-even, parity, consecutive-integers] +choices: + A: "n²" + B: "n + 2" + C: "3n" + D: "n² + n" + E: "2n + 1" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(n\) is an odd integer, which of the following must be even? + +## Explanation + +Track the parity of each expression when \(n\) is odd. + +- **(A)** \(n^2\): odd \(\times\) odd \(=\) odd. Not even. +- **(B)** \(n + 2\): odd \(+\) even \(=\) odd. Not even. +- **(C)** \(3n\): odd \(\times\) odd \(=\) odd. Not even. +- **(D)** \(n^2 + n = n(n + 1)\): this is a product of **two consecutive integers**, \(n\) and + \(n + 1\). Of any two consecutive integers, exactly one is even, so their product is always + even. **Even — correct.** +- **(E)** \(2n + 1\): even \(+\) odd \(=\) odd. Not even. + +So only (D) is guaranteed even. **Correct: (D).** + +Note the reasoning for (D) doesn't even require \(n\) to be odd — \(n^2 + n = n(n+1)\) is even for +*every* integer \(n\), because consecutive integers always include one even number. Recognizing +\(n^2 + n\) as \(n(n+1)\) is the key move. + +## Hints + +- Test each choice with a small odd number like \(n = 3\), then confirm it always holds. +- Factor (D): \(n^2 + n = n(n+1)\). What's special about two consecutive integers? + diff --git a/content/questions/quant/quant-number-properties-remainders-0019.md b/content/questions/quant/quant-number-properties-remainders-0019.md new file mode 100644 index 0000000..7048811 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0019.md @@ -0,0 +1,54 @@ +--- +id: quant-number-properties-remainders-0019 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: medium +tags: [remainders, divisibility, substitution] +choices: + A: "1" + B: "2" + C: "3" + D: "4" + E: "9" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +When the positive integer \(n\) is divided by 5, the remainder is 3. What is the remainder when +\(3n\) is divided by 5? + +## Explanation + +Write \(n\) in the form that captures "remainder 3 when divided by 5": +\[n = 5k + 3\] +for some non-negative integer \(k\). Then +\[3n = 3(5k + 3) = 15k + 9.\] + +The term \(15k\) is a multiple of 5, so it contributes remainder 0. That leaves the 9: +\[9 = 5 + 4,\] +so the remainder when \(3n\) is divided by 5 is **4**. **Correct: (D).** + +**Shortcut:** remainders pass through multiplication. The remainder of \(3 \times 3 = 9\), +reduced mod 5, is 4. + +**Traps:** + +- **(E) 9** forgets the final step — a remainder when dividing by 5 must be less than 5, so 9 + must be reduced. +- **(C) 3** assumes multiplying \(n\) by 3 leaves the remainder unchanged. It does not. + +You can confirm with a concrete value: \(n = 8\) leaves remainder 3 over 5, and +\(3n = 24 = 5 \cdot 4 + 4\) — remainder 4. + +## Hints + +- Represent \(n\) as \(5k + 3\) and multiply through by 3. +- Any final remainder over 5 has to be one of \(0,1,2,3,4\) — reduce if you land on 9. diff --git a/content/questions/verbal/verbal-critical-reasoning-paradox-0009.md b/content/questions/verbal/verbal-critical-reasoning-paradox-0009.md new file mode 100644 index 0000000..83294d7 --- /dev/null +++ b/content/questions/verbal/verbal-critical-reasoning-paradox-0009.md @@ -0,0 +1,60 @@ +--- +id: verbal-critical-reasoning-paradox-0009 +section: verbal +topic: critical-reasoning +subtopic: paradox +type: critical-reasoning +difficulty: medium +tags: [paradox, resolve-explain, critical-reasoning] +choices: + A: "The chain's downtown locations have higher rent than its suburban locations." + B: "Most customers who buy the loyalty subscription would have made the same purchases without it." + C: "Customers with the loyalty subscription visit the stores far more often and buy additional, full-price items while there." + D: "The loyalty subscription was introduced at the same time as a new advertising campaign." + E: "Competing coffee chains have not introduced similar loyalty subscriptions." +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A coffee chain sells a $20 annual loyalty subscription that entitles the holder to a free drink +every day. A typical drink costs $4, so a subscriber who redeems the offer even once a week +receives far more in free drinks than the subscription costs. Yet since introducing the +subscription, the chain's overall profit has **increased**. + +Which of the following, if true, best helps to explain the increase in profit? + +## Explanation + +This is a **paradox (resolve/explain)** question: two facts seem to clash — the chain gives away +drinks worth far more than the $20 fee, yet profit went *up*. The correct answer is the one that +lets **both** facts be true at once by supplying a missing piece of the picture. + +**(C)** does exactly that. If subscribers now visit far more often and buy **additional +full-price items** on those visits, the extra revenue from those purchases can more than offset +the cost of the free drinks. The give-away drives traffic that generates its own profit — both +facts hold. **Correct.** + +Why the others fail: + +- **(A)** Rent differences between locations don't connect the subscription to higher profit; it's + off-topic. +- **(B)** This makes the paradox *worse* — if subscribers would have bought the same things + anyway, the free drinks are pure lost margin, deepening the puzzle rather than resolving it. +- **(D)** A simultaneous ad campaign is a tempting alternate cause, but it doesn't explain how the + chain profits *despite* giving away drinks worth more than the fee; it just raises a new + question about attribution. +- **(E)** What competitors do or don't do says nothing about why *this* chain's profit rose given + the give-away. + +## Hints + +- A paradox answer must let **both** surprising facts be true — don't pick something that only + restates or worsens the conflict. +- The puzzle is "gives away more than $20 in drinks, yet earns more." What else might a frequent + visitor do while picking up a free drink? diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..95e2a4e 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,12 +7,12 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 16, - "lessons": 5 + "questions": 21, + "lessons": 7 }, "verbal": { "label": "Verbal", - "questions": 8, + "questions": 9, "lessons": 2 }, "data-insights": { @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 32, - "lessons": 8 + "questions": 38, + "lessons": 10 }, "questions": [ { @@ -301,6 +301,66 @@ "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", + "powers-of-two", + "substitution" + ], + "choices": { + "A": "35", + "B": "64", + "C": "128", + "D": "256", + "E": "512" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(2^x = 32\\), what is the value of \\(2^{x+3}\\)?", + "explanation": "You don't even need to solve for \\(x\\). Use the product rule to split the exponent:\n\\[2^{x+3} = 2^x \\cdot 2^3.\\]\n\nYou're told \\(2^x = 32\\), and \\(2^3 = 8\\), so\n\\[2^{x+3} = 32 \\cdot 8 = 256.\\]\n\n(If you prefer to find \\(x\\): \\(2^x = 32 = 2^5\\), so \\(x = 5\\), and \\(2^{5+3} = 2^8 = 256\\).)\n**Correct: (D).**\n\n**The trap:** choice **(A) 35** comes from adding \\(32 + 3\\) — treating the exponent's \"+3\" as\nif it were added to the value. Exponents add when powers of the same base are *multiplied*, so\nthe \"+3\" multiplies by \\(2^3 = 8\\).", + "hints": [ + "\\(2^{x+3}\\) can be rewritten as \\(2^x \\cdot 2^3\\) — and you already know \\(2^x\\).", + "\"+3\" in the exponent is not \"+3\" in the value; it means one more factor of \\(2^3\\)." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0018", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "exponents", + "factoring", + "same-base" + ], + "choices": { + "A": "5", + "B": "6", + "C": "7", + "D": "8", + "E": "21" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(3^n + 3^n + 3^n = 3^7\\), what is the value of \\(n\\)?", + "explanation": "There is no rule that lets you add exponents when powers are *added*, so don't touch the 7 yet.\nInstead, notice the left side is three identical copies of \\(3^n\\):\n\\[3^n + 3^n + 3^n = 3 \\cdot 3^n.\\]\n\nNow use the product rule, since \\(3 = 3^1\\):\n\\[3 \\cdot 3^n = 3^{1} \\cdot 3^{n} = 3^{\\,n+1}.\\]\n\nSo the equation becomes \\(3^{\\,n+1} = 3^7\\). Same base means equal exponents:\n\\[n + 1 = 7 \\implies n = 6.\\]\n**Correct: (B).**\n\nWhy the others fail:\n\n- **(C) 7** treats \\(3^n + 3^n + 3^n\\) as if it were just \\(3^n\\), ignoring the coefficient of 3.\n- **(E) 21** comes from multiplying the \"3 copies\" into the exponent (\\(3 \\times 7\\)) — combining\n the count of terms with the exponent, which no rule permits.\n- **(A) 5** and **(D) 8** are off-by-one slips on \\(n + 1 = 7\\).", + "hints": [ + "You can't add the exponents. Factor the left side: how many copies of \\(3^n\\) are there?", + "Three copies of \\(3^n\\) is \\(3 \\cdot 3^n\\), and \\(3 = 3^1\\)." + ] + }, { "id": "quant-arithmetic-fractions-decimals-0015", "section": "quant", @@ -418,6 +478,96 @@ "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-factors-multiples-0021", + "section": "quant", + "topic": "number-properties", + "subtopic": "factors-multiples", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "factors-multiples", + "counting", + "inclusion-exclusion" + ], + "choices": { + "A": "30", + "B": "33", + "C": "34", + "D": "50", + "E": "66" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "How many positive integers less than 100 are divisible by neither 2 nor 3?", + "explanation": "The integers in play are \\(1, 2, \\ldots, 99\\) — that's 99 numbers. It's easier to **count what to\nthrow away** (multiples of 2 or 3) and subtract, using inclusion–exclusion.\n\n- Divisible by 2: \\(\\left\\lfloor \\tfrac{99}{2} \\right\\rfloor = 49\\).\n- Divisible by 3: \\(\\left\\lfloor \\tfrac{99}{3} \\right\\rfloor = 33\\).\n- Divisible by both 2 and 3 (i.e. by 6): \\(\\left\\lfloor \\tfrac{99}{6} \\right\\rfloor = 16\\).\n\nBy inclusion–exclusion, the count divisible by 2 **or** 3 is\n\\[49 + 33 - 16 = 66.\\]\n\nSo the count divisible by **neither** is\n\\[99 - 66 = 33.\\]\n**Correct: (B).**\n\n**Traps:**\n\n- **(E) 66** is the number divisible by 2 or 3 — the complement of what's asked.\n- **Forgetting to subtract the overlap (16)** double-counts multiples of 6 and gives \\(99 - 82 = 17\\),\n or leads to other wrong totals; inclusion–exclusion prevents this.", + "hints": [ + "Count the \"bad\" numbers (multiples of 2 or 3) and subtract from 99.", + "Don't double-count: multiples of 6 are in both the \"÷2\" and \"÷3\" groups, so subtract them once." + ] + }, + { + "id": "quant-number-properties-odds-evens-signs-0020", + "section": "quant", + "topic": "number-properties", + "subtopic": "odds-evens-signs", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "odd-even", + "parity", + "consecutive-integers" + ], + "choices": { + "A": "n²", + "B": "n + 2", + "C": "3n", + "D": "n² + n", + "E": "2n + 1" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(n\\) is an odd integer, which of the following must be even?", + "explanation": "Track the parity of each expression when \\(n\\) is odd.\n\n- **(A)** \\(n^2\\): odd \\(\\times\\) odd \\(=\\) odd. Not even.\n- **(B)** \\(n + 2\\): odd \\(+\\) even \\(=\\) odd. Not even.\n- **(C)** \\(3n\\): odd \\(\\times\\) odd \\(=\\) odd. Not even.\n- **(D)** \\(n^2 + n = n(n + 1)\\): this is a product of **two consecutive integers**, \\(n\\) and\n \\(n + 1\\). Of any two consecutive integers, exactly one is even, so their product is always\n even. **Even — correct.**\n- **(E)** \\(2n + 1\\): even \\(+\\) odd \\(=\\) odd. Not even.\n\nSo only (D) is guaranteed even. **Correct: (D).**\n\nNote the reasoning for (D) doesn't even require \\(n\\) to be odd — \\(n^2 + n = n(n+1)\\) is even for\n*every* integer \\(n\\), because consecutive integers always include one even number. Recognizing\n\\(n^2 + n\\) as \\(n(n+1)\\) is the key move.", + "hints": [ + "Test each choice with a small odd number like \\(n = 3\\), then confirm it always holds.", + "Factor (D): \\(n^2 + n = n(n+1)\\). What's special about two consecutive integers?" + ] + }, + { + "id": "quant-number-properties-remainders-0019", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "remainders", + "divisibility", + "substitution" + ], + "choices": { + "A": "1", + "B": "2", + "C": "3", + "D": "4", + "E": "9" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "When the positive integer \\(n\\) is divided by 5, the remainder is 3. What is the remainder when\n\\(3n\\) is divided by 5?", + "explanation": "Write \\(n\\) in the form that captures \"remainder 3 when divided by 5\":\n\\[n = 5k + 3\\]\nfor some non-negative integer \\(k\\). Then\n\\[3n = 3(5k + 3) = 15k + 9.\\]\n\nThe term \\(15k\\) is a multiple of 5, so it contributes remainder 0. That leaves the 9:\n\\[9 = 5 + 4,\\]\nso the remainder when \\(3n\\) is divided by 5 is **4**. **Correct: (D).**\n\n**Shortcut:** remainders pass through multiplication. The remainder of \\(3 \\times 3 = 9\\),\nreduced mod 5, is 4.\n\n**Traps:**\n\n- **(E) 9** forgets the final step — a remainder when dividing by 5 must be less than 5, so 9\n must be reduced.\n- **(C) 3** assumes multiplying \\(n\\) by 3 leaves the remainder unchanged. It does not.\n\nYou can confirm with a concrete value: \\(n = 8\\) leaves remainder 3 over 5, and\n\\(3n = 24 = 5 \\cdot 4 + 4\\) — remainder 4.", + "hints": [ + "Represent \\(n\\) as \\(5k + 3\\) and multiply through by 3.", + "Any final remainder over 5 has to be one of \\(0,1,2,3,4\\) — reduce if you land on 9." + ] + }, { "id": "quant-statistics-descriptive-0009", "section": "quant", @@ -800,6 +950,36 @@ "A valid inference *must* follow; eliminate any choice that merely *could* be true or that reverses a premise." ] }, + { + "id": "verbal-critical-reasoning-paradox-0009", + "section": "verbal", + "topic": "critical-reasoning", + "subtopic": "paradox", + "type": "critical-reasoning", + "difficulty": "medium", + "tags": [ + "paradox", + "resolve-explain", + "critical-reasoning" + ], + "choices": { + "A": "The chain's downtown locations have higher rent than its suburban locations.", + "B": "Most customers who buy the loyalty subscription would have made the same purchases without it.", + "C": "Customers with the loyalty subscription visit the stores far more often and buy additional, full-price items while there.", + "D": "The loyalty subscription was introduced at the same time as a new advertising campaign.", + "E": "Competing coffee chains have not introduced similar loyalty subscriptions." + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A coffee chain sells a $20 annual loyalty subscription that entitles the holder to a free drink\nevery day. A typical drink costs $4, so a subscriber who redeems the offer even once a week\nreceives far more in free drinks than the subscription costs. Yet since introducing the\nsubscription, the chain's overall profit has **increased**.\n\nWhich of the following, if true, best helps to explain the increase in profit?", + "explanation": "This is a **paradox (resolve/explain)** question: two facts seem to clash — the chain gives away\ndrinks worth far more than the $20 fee, yet profit went *up*. The correct answer is the one that\nlets **both** facts be true at once by supplying a missing piece of the picture.\n\n**(C)** does exactly that. If subscribers now visit far more often and buy **additional\nfull-price items** on those visits, the extra revenue from those purchases can more than offset\nthe cost of the free drinks. The give-away drives traffic that generates its own profit — both\nfacts hold. **Correct.**\n\nWhy the others fail:\n\n- **(A)** Rent differences between locations don't connect the subscription to higher profit; it's\n off-topic.\n- **(B)** This makes the paradox *worse* — if subscribers would have bought the same things\n anyway, the free drinks are pure lost margin, deepening the puzzle rather than resolving it.\n- **(D)** A simultaneous ad campaign is a tempting alternate cause, but it doesn't explain how the\n chain profits *despite* giving away drinks worth more than the fee; it just raises a new\n question about attribution.\n- **(E)** What competitors do or don't do says nothing about why *this* chain's profit rose given\n the give-away.", + "hints": [ + "A paradox answer must let **both** surprising facts be true — don't pick something that only", + "The puzzle is \"gives away more than $20 in drinks, yet earns more.\" What else might a frequent" + ] + }, { "id": "verbal-critical-reasoning-strengthen-weaken-0002", "section": "verbal", @@ -968,6 +1148,23 @@ "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 and Roots", + "tags": [ + "exponents", + "roots", + "powers", + "rules-of-exponents" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nExponent questions on the GMAT rarely test heavy computation. They test whether you can\n**rewrite** expressions into a common base and apply a small set of rules cleanly. Get the base\nconsistent, and most problems collapse in one or two steps.\n\n## Core concepts\n\nFor any nonzero base \\(a\\) and integers \\(m, n\\):\n\n- **Product rule** — same base, add exponents: \\(a^m \\cdot a^n = a^{m+n}\\).\n- **Quotient rule** — same base, subtract exponents: \\(\\dfrac{a^m}{a^n} = a^{m-n}\\).\n- **Power of a power** — multiply exponents: \\((a^m)^n = a^{mn}\\).\n- **Zero exponent** — \\(a^0 = 1\\) for any \\(a \\neq 0\\).\n- **Negative exponent** — a reciprocal: \\(a^{-n} = \\dfrac{1}{a^n}\\).\n- **Fractional exponent** — a root: \\(a^{1/n} = \\sqrt[n]{a}\\), and \\(a^{m/n} = \\sqrt[n]{a^m}\\).\n\nThe single most useful move is **matching bases**. Since \\(4 = 2^2\\), \\(8 = 2^3\\), \\(9 = 3^2\\),\n\\(27 = 3^3\\), you can turn a messy fraction of different bases into one base and just subtract\nexponents.\n\nTwo rules that do **not** exist (the classic traps):\n\n- \\(a^m + a^n \\neq a^{m+n}\\). Addition of powers is **not** a rule — you must factor instead.\n- \\((a + b)^2 \\neq a^2 + b^2\\). Expand it: \\((a+b)^2 = a^2 + 2ab + b^2\\).\n\n## Worked examples\n\n**Example 1 — match the base.** Simplify \\(\\dfrac{4^{10}}{2^{15}}\\).\n\nRewrite \\(4^{10} = (2^2)^{10} = 2^{20}\\). Then \\(\\dfrac{2^{20}}{2^{15}} = 2^{20-15} = 2^5 = 32\\).\n\n**Example 2 — factor a sum of equal powers.** If \\(3^n + 3^n + 3^n = 3^7\\), find \\(n\\).\n\nYou cannot add the exponents. Instead, three copies of \\(3^n\\) is \\(3 \\cdot 3^n = 3^{1+n}\\).\nSo \\(3^{n+1} = 3^7\\), which gives \\(n + 1 = 7\\), so \\(n = 6\\).\n\n**Example 3 — negative and fractional exponents.** Evaluate \\(27^{-2/3}\\).\n\nWork outward: \\(27^{1/3} = 3\\), then \\(3^2 = 9\\), and the negative sign takes the reciprocal:\n\\(27^{-2/3} = \\dfrac{1}{9}\\).\n\n## Common traps\n\n- **Adding powers as if it were a rule.** When you see \\(2^x + 2^x\\), factor: it equals\n \\(2 \\cdot 2^x = 2^{x+1}\\), not \\(2^{2x}\\) or \\(4^x\\)... well, \\(2^{x+1}\\) *is* correct; the\n point is to factor rather than invent an addition rule.\n- **Forgetting to convert to a common base** before applying the quotient rule. \\(\\tfrac{4^{10}}{2^{15}}\\)\n is not \\(2^{10-15}\\) — the numerator's base must become 2 first.\n- **Mishandling the negative sign vs. the exponent.** \\(-2^4 = -16\\) (the exponent binds before\n the sign), but \\((-2)^4 = 16\\).\n- **Answering the wrong quantity.** If they ask for \\(2^{x+3}\\) and you find \\(x\\), remember to\n finish the substitution.\n\n## Key takeaways\n\n- Convert everything to a **common base**, then add/subtract/multiply exponents by the rules.\n- **Sums of powers are factored, not combined** — there is no addition rule for exponents.\n- A fractional exponent is a root; a negative exponent is a reciprocal. Apply them one at a time.\n- Memorize small powers of 2 and 3 (\\(2^1\\ldots2^{10}\\), \\(3^1\\ldots3^4\\)) to move quickly." + }, { "id": "quant-arithmetic-percents-ratios", "section": "quant", @@ -984,6 +1181,23 @@ "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", + "cyclicity" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nA remainder question is really a question about **structure**: what an integer looks like when\nyou write it in terms of the divisor. Once you write \\(n = dq + r\\), most remainder problems\nbecome simple substitution or a short search over cases.\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 = dq + r, \\qquad 0 \\le r < d.\\]\nThe remainder is always at least 0 and strictly less than the divisor. \"Divided by 5\" means the\nremainder is one of \\(0, 1, 2, 3, 4\\).\n\n**Remainders travel through addition and multiplication.** If you only care about the remainder\nof a sum or product, you can replace each number by its own remainder first:\n\n- remainder of \\((a + b)\\) = remainder of (remainder of \\(a\\) + remainder of \\(b\\));\n- remainder of \\((a \\cdot b)\\) = remainder of (remainder of \\(a\\) \\(\\times\\) remainder of \\(b\\)).\n\nThen reduce again if the result is still \\(\\ge d\\). This lets you handle huge numbers without\never computing them.\n\n**Units digits cycle.** The units digit of a power repeats on a short cycle, which is a\nremainder-mod-10 fact. For example, powers of 3 cycle \\(3, 9, 7, 1, 3, 9, 7, 1, \\ldots\\) with\nperiod 4.\n\n## Worked examples\n\n**Example 1 — substitute the form.** When \\(n\\) is divided by 5 the remainder is 3. What is the\nremainder when \\(3n\\) is divided by 5?\n\nWrite \\(n = 5k + 3\\). Then \\(3n = 15k + 9\\). Since \\(15k\\) is a multiple of 5, only the 9\nmatters, and \\(9 = 5 + 4\\), so the remainder is **4**. (Shortcut with the multiplication rule:\nremainder of \\(3 \\times 3 = 9\\), reduce mod 5 to get 4.)\n\n**Example 2 — combine two conditions.** A number leaves remainder 2 when divided by 3 and\nremainder 2 when divided by 4. What is the smallest such positive integer greater than 2?\n\nNumbers that are 2 more than a multiple of both 3 and 4 are 2 more than a multiple of 12:\n\\(12m + 2\\). The smallest above 2 is \\(14\\).\n\n**Example 3 — units digit via cyclicity.** What is the units digit of \\(3^{23}\\)?\n\nThe cycle of the units digit of powers of 3 has length 4: \\(3, 9, 7, 1\\). Divide the exponent by\n4: \\(23 = 4 \\cdot 5 + 3\\), so we are at position 3 in the cycle, which is **7**.\n\n## Common traps\n\n- **Remainder of at least the divisor.** A remainder when dividing by 5 can never be 5 or more.\n If your arithmetic gives 9, reduce it (\\(9 \\bmod 5 = 4\\)).\n- **Assuming the remainder is unchanged when you scale.** Multiplying \\(n\\) by 3 does *not* keep\n the remainder at 3 — you must recompute (Example 1).\n- **Zero is a valid remainder.** \"Divisible by \\(d\\)\" is exactly \"remainder 0.\"\n- **Off-by-one on cyclicity.** A remainder of 0 in the exponent lands on the *last* term of the\n cycle, not the first.\n\n## Key takeaways\n\n- Write \\(n = dq + r\\) with \\(0 \\le r < d\\); substitute and simplify.\n- Remainders pass cleanly through addition and multiplication — reduce early, reduce often.\n- For units digits of powers, find the cycle length and take the exponent mod that length.\n- Always sanity-check that your final remainder is less than the divisor." + }, { "id": "quant-statistics-descriptive", "section": "quant",