From d8f5e6bf9b535f0cd1876dfb49eef97e88cdd569 Mon Sep 17 00:00:00 2001 From: Claude Date: Thu, 9 Jul 2026 09:08:20 +0000 Subject: [PATCH] Day 3: Seed Number Properties topic (+3 lessons, +6 questions) Adds full coverage of the Quant number-properties topic, the last Foundations topic with no lessons: - Lessons: factors/multiples/primes & divisibility, odds/evens & sign rules, and remainders & modular thinking. - Questions (0017-0022): two each on factors-multiples, odds-evens-signs, and remainders, spanning medium and hard difficulty. Rebuilt docs/data/content.json (38 questions, 11 lessons). Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_019vb2R14JLMiDc788ZZfc3W --- ...ant-number-properties-factors-multiples.md | 75 ++++++ ...uant-number-properties-odds-evens-signs.md | 62 +++++ .../quant-number-properties-remainders.md | 64 +++++ ...umber-properties-factors-multiples-0017.md | 50 ++++ ...umber-properties-factors-multiples-0018.md | 47 ++++ ...number-properties-odds-evens-signs-0019.md | 44 ++++ ...number-properties-odds-evens-signs-0020.md | 46 ++++ ...quant-number-properties-remainders-0021.md | 46 ++++ ...quant-number-properties-remainders-0022.md | 47 ++++ docs/data/content.json | 239 +++++++++++++++++- 10 files changed, 716 insertions(+), 4 deletions(-) create mode 100644 content/lessons/quant/quant-number-properties-factors-multiples.md create mode 100644 content/lessons/quant/quant-number-properties-odds-evens-signs.md create mode 100644 content/lessons/quant/quant-number-properties-remainders.md create mode 100644 content/questions/quant/quant-number-properties-factors-multiples-0017.md create mode 100644 content/questions/quant/quant-number-properties-factors-multiples-0018.md create mode 100644 content/questions/quant/quant-number-properties-odds-evens-signs-0019.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-0021.md create mode 100644 content/questions/quant/quant-number-properties-remainders-0022.md diff --git a/content/lessons/quant/quant-number-properties-factors-multiples.md b/content/lessons/quant/quant-number-properties-factors-multiples.md new file mode 100644 index 0000000..adbcf68 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-factors-multiples.md @@ -0,0 +1,75 @@ +--- +id: quant-number-properties-factors-multiples +section: quant +topic: number-properties +subtopic: factors-multiples +title: "Factors, Multiples, Primes & Divisibility" +tags: [factors, multiples, primes, gcd, lcm, divisibility] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Almost every Number Properties question rests on one skill: **breaking an integer into its +prime factors**. Once you have the prime factorization, divisor counts, greatest common divisors, +least common multiples, and divisibility all follow mechanically. Get comfortable factoring fast +and this whole topic gets easier. + +## Core concepts + +**Prime factorization.** Every integer greater than 1 is a unique product of primes. Split the +number into any two factors and keep splitting until everything is prime. + +\[360 = 36 \times 10 = (2^2 \cdot 3^2)(2 \cdot 5) = 2^3 \cdot 3^2 \cdot 5\] + +**Counting divisors.** If \(N = p^a \cdot q^b \cdot r^c \dots\), the number of positive divisors is + +\[(a+1)(b+1)(c+1)\dots\] + +The "+1" is there because each divisor can use prime \(p\) to a power from \(0\) up to \(a\). +For \(360 = 2^3 \cdot 3^2 \cdot 5^1\): \((3+1)(2+1)(1+1) = 24\) divisors. + +**GCD (greatest common divisor).** Take each prime the two numbers **share**, using the **lower** +exponent. **LCM (least common multiple).** Take **every** prime that appears in either number, +using the **higher** exponent. + +\[12 = 2^2 \cdot 3, \quad 18 = 2 \cdot 3^2 \;\Rightarrow\; \gcd = 2^1 \cdot 3^1 = 6, \quad \operatorname{lcm} = 2^2 \cdot 3^2 = 36\] + +A useful identity for any two positive integers: \(\gcd(a,b) \times \operatorname{lcm}(a,b) = a \times b\). + +**Perfect squares** have only **even** exponents in their prime factorization (e.g. +\(900 = 2^2 \cdot 3^2 \cdot 5^2\)). To turn a number into a perfect square, supply whatever primes +are needed to make every exponent even. + +**Quick divisibility rules.** Divisible by 3 if the digit sum is divisible by 3; by 9 if the digit +sum is divisible by 9; by 4 if the last two digits form a multiple of 4; by 6 if it passes the 2 +**and** 3 tests. + +## Worked examples + +**LCM in disguise.** Two signals flash every 12 and 18 seconds and just flashed together. When do +they next coincide? That is \(\operatorname{lcm}(12,18) = 36\) seconds — *not* their product. + +**Smallest multiplier for a perfect square.** What is the smallest \(n\) making \(90n\) a perfect +square? \(90 = 2 \cdot 3^2 \cdot 5\). The 2 and the 5 have odd exponents, so \(n = 2 \cdot 5 = 10\) +and \(90 \cdot 10 = 900 = 30^2\). + +## Common traps + +- **Confusing GCD with LCM.** GCD is the *small* shared part (lower exponents); LCM is the *big* + covering number (higher exponents). "How often do things line up again?" is almost always LCM. +- **Multiplying instead of taking the LCM.** \(12 \times 18 = 216\) is a common multiple, but not + the *least* one. +- **Forgetting 1 and the number itself** when counting divisors — the \((a+1)(b+1)\) formula + already includes them. + +## Key takeaways + +- Factor into primes first; divisor count, GCD, and LCM all read straight off the exponents. +- Divisor count \(= (a+1)(b+1)\dots\); GCD uses shared primes at the lower power; LCM uses all + primes at the higher power. +- \(\gcd \times \operatorname{lcm} = a \times b\), and perfect squares have all-even exponents. diff --git a/content/lessons/quant/quant-number-properties-odds-evens-signs.md b/content/lessons/quant/quant-number-properties-odds-evens-signs.md new file mode 100644 index 0000000..ad06668 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-odds-evens-signs.md @@ -0,0 +1,62 @@ +--- +id: quant-number-properties-odds-evens-signs +section: quant +topic: number-properties +subtopic: odds-evens-signs +title: "Odds, Evens & Sign Rules" +tags: [parity, odd, even, signs, positive, negative] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Many "which of the following **must** be true?" questions never require you to find a value — only +to track **parity** (odd/even) and **sign** (positive/negative). Memorize a handful of rules and +these become fast, reliable points. + +## Core concepts + +**Parity of sums.** odd \(+\) odd \(=\) even; even \(+\) even \(=\) even; odd \(+\) even \(=\) odd. +A sum is **odd exactly when an odd number of its terms are odd.** + +**Parity of products.** A product is **odd only if every factor is odd.** If even one factor is +even, the product is even. So \(xy\) odd \(\Rightarrow\) both \(x\) and \(y\) are odd. + +**Consecutive integers.** Among any two consecutive integers, exactly one is even, so their product +is always even. Among any three consecutive integers, the product is divisible by \(3! = 6\). + +**Sign rules for products.** A product is **negative when it has an odd number of negative +factors**, and **positive when it has an even number** (zero counts as even). So \(xyz < 0\) means +one or three of the factors are negative. + +**A "must be true" strategy.** When a question asks what *must* be true, try to build a +counterexample. If you can find even one case where a choice fails, eliminate it. What survives +every test is the answer. + +## Worked examples + +**Product forces parity.** If \(xy\) is odd, both \(x\) and \(y\) are odd, so \(x + y\) is even and +\(x^2 + y^2 = \text{odd} + \text{odd}\) is even. Any choice claiming \(x+y\) is odd is out. + +**Chaining signs.** If \(xyz < 0\) and \(xy > 0\): \(xy > 0\) means \(x\) and \(y\) share a sign, so +their product is positive; for the triple product to be negative, \(z\) must be **negative**. Note +you still can't pin down the signs of \(x\) and \(y\) individually. + +## Common traps + +- **Assuming "integer" means positive.** Integers include 0 and negatives. Test a negative and 0 + against every "must be true" choice. +- **Forgetting 0.** Zero is even, is neither positive nor negative, and makes any product it + touches equal 0. +- **Reading "could be true" as "must be true."** One counterexample kills a *must*-be-true choice, + even if it's true most of the time. + +## Key takeaways + +- Product is odd only if **all** factors are odd; a sum is odd only with an **odd count** of odd terms. +- Product sign flips with each negative factor: odd number of negatives → negative. +- For "must be true," hunt for a counterexample — and always test negatives and 0. 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..6b85f45 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-remainders.md @@ -0,0 +1,64 @@ +--- +id: quant-number-properties-remainders +section: quant +topic: number-properties +subtopic: remainders +title: "Remainders & Modular Thinking" +tags: [remainders, modular, divisibility, cycles] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +A remainder question is really a statement about form. "\(n\) leaves remainder 4 when divided by 7" +just means \(n = 7k + 4\) for some integer \(k\). Write that form down and most remainder problems +turn into ordinary algebra or a short list. + +## Core concepts + +**The division form.** "\(n\) divided by \(d\) leaves remainder \(r\)" means + +\[n = dq + r, \qquad 0 \le r < d\] + +The remainder is always **less than the divisor** — a remainder of 7 (mod 7) is impossible; it is 0. + +**Remainders add and multiply.** To find the remainder of a sum or product, you may reduce each +piece to its remainder first. If \(n\) leaves remainder 4 mod 7, then \(3n\) leaves the remainder of +\(3 \times 4 = 12\), which is \(12 - 7 = 5\). + +**Testing the smallest case.** When only the remainder is given, plug in the smallest value that +fits (here \(n = 4\)) — the remainder of any sum/product won't depend on which valid \(n\) you pick. + +**Two conditions at once.** "\(n\) leaves remainder 3 mod 5 **and** remainder 2 mod 3": list each, + +\[n \equiv 3 \pmod 5: \; 3, 8, 13, 18, 23, \dots \qquad n \equiv 2 \pmod 3: \; 2, 5, 8, 11, 14, 17, 20, 23, \dots\] + +The first common value is 8, and solutions then repeat every \(\operatorname{lcm}(5,3) = 15\): +\(8, 23, 38, \dots\) + +## Worked examples + +**Scaling a remainder.** \(n\) leaves remainder 4 mod 7. Remainder of \(3n\) mod 7? Reduce: +\(3 \cdot 4 = 12 \equiv 5\). So the remainder is **5**. + +**Combining two moduli.** Smallest \(n > 10\) with remainder 3 mod 5 and remainder 2 mod 3? From the +lists above the pattern is \(8, 23, 38, \dots\); the first one over 10 is **23**. + +## Common traps + +- **Remainder \(\ge\) divisor.** The remainder must be strictly smaller than what you divide by. + Always reduce (e.g. \(12 \bmod 7 = 5\)). +- **Forgetting the small cases.** \(n\) could be the remainder itself (e.g. \(n = 4\)) — don't + assume \(n\) is large. +- **Checking only one condition.** With two divisors, verify a candidate against **both** before + choosing it. + +## Key takeaways + +- Rewrite "remainder \(r\) mod \(d\)" as \(n = dq + r\) with \(0 \le r < d\). +- Reduce sums and products to their remainders, then reduce again if the result exceeds the divisor. +- Two simultaneous conditions repeat every LCM of the divisors — list a few terms and find the overlap. diff --git a/content/questions/quant/quant-number-properties-factors-multiples-0017.md b/content/questions/quant/quant-number-properties-factors-multiples-0017.md new file mode 100644 index 0000000..3b5f69b --- /dev/null +++ b/content/questions/quant/quant-number-properties-factors-multiples-0017.md @@ -0,0 +1,50 @@ +--- +id: quant-number-properties-factors-multiples-0017 +section: quant +topic: number-properties +subtopic: factors-multiples +type: problem-solving +difficulty: medium +tags: [lcm, cycles, multiples] +choices: + A: "6" + B: "36" + C: "54" + D: "72" + E: "216" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A red beacon flashes every 12 seconds and a green beacon flashes every 18 seconds. If both beacons +flash at the same instant right now, how many seconds from now will they next flash at the same +instant? + +## Explanation + +"Flash together again" means the first time that is a multiple of **both** 12 and 18 — the **least +common multiple**. + +Factor each interval into primes: + +\[12 = 2^2 \cdot 3, \qquad 18 = 2 \cdot 3^2\] + +The LCM takes each prime at its **highest** power: + +\[\operatorname{lcm}(12, 18) = 2^2 \cdot 3^2 = 36\] + +So the beacons next coincide after **36** seconds. + +Watch the two classic traps: \(6\) is the *greatest common divisor* (lower powers), not the LCM, +and \(12 \times 18 = 216\) is *a* common multiple but not the *least* one. + +## Hints + +- "Line up again" points to a common multiple — and you want the *least* one. +- Factor both numbers into primes; the LCM uses the higher power of each prime, the GCD the lower. diff --git a/content/questions/quant/quant-number-properties-factors-multiples-0018.md b/content/questions/quant/quant-number-properties-factors-multiples-0018.md new file mode 100644 index 0000000..37accb2 --- /dev/null +++ b/content/questions/quant/quant-number-properties-factors-multiples-0018.md @@ -0,0 +1,47 @@ +--- +id: quant-number-properties-factors-multiples-0018 +section: quant +topic: number-properties +subtopic: factors-multiples +type: problem-solving +difficulty: hard +tags: [prime-factorization, perfect-square] +choices: + A: "2" + B: "5" + C: "10" + D: "15" + E: "90" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the smallest positive integer \(n\) such that \(90n\) is the square of an integer? + +## Explanation + +A perfect square has **only even exponents** in its prime factorization. Factor 90: + +\[90 = 2 \cdot 3^2 \cdot 5\] + +The exponent of 3 is already even (2), but the exponents of 2 and 5 are each **odd** (1). To make +every exponent even, \(n\) must supply one more 2 and one more 5: + +\[n = 2 \cdot 5 = 10\] + +Check: \(90 \cdot 10 = 900 = 2^2 \cdot 3^2 \cdot 5^2 = 30^2\). ✓ + +Smaller candidates fail because they fix only one offending prime: \(n = 2\) leaves the 5 odd, and +\(n = 5\) leaves the 2 odd. The answer is **10**. + +## Hints + +- Factor 90 into primes and look at which exponents are odd. +- A perfect square needs every prime exponent to be even — multiply in exactly the primes required + to fix the odd ones. diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0019.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0019.md new file mode 100644 index 0000000..18dee1f --- /dev/null +++ b/content/questions/quant/quant-number-properties-odds-evens-signs-0019.md @@ -0,0 +1,44 @@ +--- +id: quant-number-properties-odds-evens-signs-0019 +section: quant +topic: number-properties +subtopic: odds-evens-signs +type: problem-solving +difficulty: medium +tags: [parity, must-be-true] +choices: + A: "\(x + y\) is odd" + B: "\(x + y\) is even" + C: "\(x\) is even" + D: "\(x - y\) is odd" + E: "\(x^2 + y^2\) is odd" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(x\) and \(y\) are integers and the product \(xy\) is odd, which of the following must be true? + +## Explanation + +A product is odd **only when every factor is odd**. So \(xy\) odd forces **both** \(x\) and \(y\) to +be odd. Now test each choice with "odd, odd": + +- **A.** odd \(+\) odd \(=\) even, so \(x + y\) is *not* odd. Eliminate. +- **B.** odd \(+\) odd \(=\) **even**. This must be true. ✓ +- **C.** \(x\) is odd, not even. Eliminate. +- **D.** odd \(-\) odd \(=\) even, so \(x - y\) is not odd. Eliminate. +- **E.** \(x^2\) and \(y^2\) are both odd, and odd \(+\) odd \(=\) even, so \(x^2 + y^2\) is even, not + odd. Eliminate. + +Only **B** holds. The answer is **\(x + y\) is even**. + +## Hints + +- What does an *odd product* tell you about each of the two factors? +- Once you know the parity of \(x\) and \(y\), just apply odd/even sum rules to each choice. 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..936d85a --- /dev/null +++ b/content/questions/quant/quant-number-properties-odds-evens-signs-0020.md @@ -0,0 +1,46 @@ +--- +id: quant-number-properties-odds-evens-signs-0020 +section: quant +topic: number-properties +subtopic: odds-evens-signs +type: problem-solving +difficulty: medium +tags: [signs, must-be-true] +choices: + A: "\(x < 0\)" + B: "\(y > 0\)" + C: "\(z < 0\)" + D: "\(xz > 0\)" + E: "\(yz > 0\)" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(x\), \(y\), and \(z\) are nonzero numbers with \(xyz < 0\) and \(xy > 0\), which of the +following must be true? + +## Explanation + +Work with the sign of each product. + +Since \(xy > 0\), \(x\) and \(y\) have the **same sign**, so their product is positive. Then + +\[xyz = (xy) \cdot z < 0 \quad\text{with}\quad xy > 0 \;\Rightarrow\; z < 0.\] + +So \(z\) must be negative — choice **C**. + +The other choices aren't forced: \(x\) and \(y\) could **both** be positive or **both** be negative +(either keeps \(xy > 0\)), so A and B fail. And because \(x\)'s sign is unknown while \(z < 0\), the +product \(xz\) could be positive or negative, so D fails; the same reasoning kills E. The answer is +**\(z < 0\)**. + +## Hints + +- Treat \(xy\) as a single quantity — its sign is given directly. +- A product is negative when it has an odd number of negative factors; use that to pin down \(z\). diff --git a/content/questions/quant/quant-number-properties-remainders-0021.md b/content/questions/quant/quant-number-properties-remainders-0021.md new file mode 100644 index 0000000..4f77ce6 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0021.md @@ -0,0 +1,46 @@ +--- +id: quant-number-properties-remainders-0021 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: medium +tags: [remainders, scaling] +choices: + A: "1" + B: "2" + C: "4" + D: "5" + E: "6" +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 7, the remainder is 4. What is the remainder when +\(3n\) is divided by 7? + +## Explanation + +To find the remainder of a product, you can reduce each part to its remainder first. Here \(n\) +leaves remainder 4, so \(3n\) leaves the same remainder as \(3 \times 4 = 12\). + +But a remainder must be **less than the divisor**, and \(12 > 7\), so reduce once more: + +\[12 = 7 \cdot 1 + 5 \quad\Rightarrow\quad 12 \equiv 5 \pmod 7.\] + +So the remainder is **5**. + +Confirm with a concrete value: the smallest such \(n\) is 4, and \(3 \cdot 4 = 12\), which divided by +7 leaves remainder 5. (Any valid \(n\), like 11 or 18, gives the same result.) A common slip is to +answer 4 — the *original* remainder — but multiplying by 3 changes it. + +## Hints + +- You don't need to know \(n\); reduce \(3 \times (\text{remainder})\) modulo 7. +- If your intermediate result is 7 or more, subtract 7 until it drops below 7. diff --git a/content/questions/quant/quant-number-properties-remainders-0022.md b/content/questions/quant/quant-number-properties-remainders-0022.md new file mode 100644 index 0000000..feaa6cf --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0022.md @@ -0,0 +1,47 @@ +--- +id: quant-number-properties-remainders-0022 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: hard +tags: [remainders, simultaneous-conditions, lcm] +choices: + A: "13" + B: "18" + C: "23" + D: "28" + E: "38" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +A positive integer \(n\) leaves a remainder of 3 when divided by 5 and a remainder of 2 when divided +by 3. What is the least value of \(n\) that is greater than 10? + +## Explanation + +List the numbers satisfying each condition and find where they first overlap. + +Remainder 3 mod 5: \(3, 8, 13, 18, 23, 28, \dots\) +Remainder 2 mod 3: \(2, 5, 8, 11, 14, 17, 20, 23, \dots\) + +The values common to both are \(8, 23, 38, \dots\) (they repeat every +\(\operatorname{lcm}(5, 3) = 15\)). The first one **greater than 10** is **23**. + +Check 23: \(23 = 5 \cdot 4 + 3\) (remainder 3 ✓) and \(23 = 3 \cdot 7 + 2\) (remainder 2 ✓). + +The distractors satisfy only one condition: 13 and 18 both leave remainder 3 mod 5 but fail the +mod-3 test (\(13 \equiv 1\), \(18 \equiv 0\)), and 28 leaves remainder 3 mod 5 but \(28 \equiv 1 +\pmod 3\). (38 satisfies both but isn't the *least* one over 10.) The answer is **23**. + +## Hints + +- Write out the first several numbers that fit each condition separately. +- The values satisfying both repeat every LCM of the two divisors — find the first overlap above 10. diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..2c09f8d 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,8 +7,8 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 16, - "lessons": 5 + "questions": 22, + "lessons": 8 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 32, - "lessons": 8 + "questions": 38, + "lessons": 11 }, "questions": [ { @@ -418,6 +418,182 @@ "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-0017", + "section": "quant", + "topic": "number-properties", + "subtopic": "factors-multiples", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "lcm", + "cycles", + "multiples" + ], + "choices": { + "A": "6", + "B": "36", + "C": "54", + "D": "72", + "E": "216" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A red beacon flashes every 12 seconds and a green beacon flashes every 18 seconds. If both beacons\nflash at the same instant right now, how many seconds from now will they next flash at the same\ninstant?", + "explanation": "\"Flash together again\" means the first time that is a multiple of **both** 12 and 18 — the **least\ncommon multiple**.\n\nFactor each interval into primes:\n\n\\[12 = 2^2 \\cdot 3, \\qquad 18 = 2 \\cdot 3^2\\]\n\nThe LCM takes each prime at its **highest** power:\n\n\\[\\operatorname{lcm}(12, 18) = 2^2 \\cdot 3^2 = 36\\]\n\nSo the beacons next coincide after **36** seconds.\n\nWatch the two classic traps: \\(6\\) is the *greatest common divisor* (lower powers), not the LCM,\nand \\(12 \\times 18 = 216\\) is *a* common multiple but not the *least* one.", + "hints": [ + "\"Line up again\" points to a common multiple — and you want the *least* one.", + "Factor both numbers into primes; the LCM uses the higher power of each prime, the GCD the lower." + ] + }, + { + "id": "quant-number-properties-factors-multiples-0018", + "section": "quant", + "topic": "number-properties", + "subtopic": "factors-multiples", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "prime-factorization", + "perfect-square" + ], + "choices": { + "A": "2", + "B": "5", + "C": "10", + "D": "15", + "E": "90" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the smallest positive integer \\(n\\) such that \\(90n\\) is the square of an integer?", + "explanation": "A perfect square has **only even exponents** in its prime factorization. Factor 90:\n\n\\[90 = 2 \\cdot 3^2 \\cdot 5\\]\n\nThe exponent of 3 is already even (2), but the exponents of 2 and 5 are each **odd** (1). To make\nevery exponent even, \\(n\\) must supply one more 2 and one more 5:\n\n\\[n = 2 \\cdot 5 = 10\\]\n\nCheck: \\(90 \\cdot 10 = 900 = 2^2 \\cdot 3^2 \\cdot 5^2 = 30^2\\). ✓\n\nSmaller candidates fail because they fix only one offending prime: \\(n = 2\\) leaves the 5 odd, and\n\\(n = 5\\) leaves the 2 odd. The answer is **10**.", + "hints": [ + "Factor 90 into primes and look at which exponents are odd.", + "A perfect square needs every prime exponent to be even — multiply in exactly the primes required" + ] + }, + { + "id": "quant-number-properties-odds-evens-signs-0019", + "section": "quant", + "topic": "number-properties", + "subtopic": "odds-evens-signs", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "parity", + "must-be-true" + ], + "choices": { + "A": "\\(x + y\\) is odd", + "B": "\\(x + y\\) is even", + "C": "\\(x\\) is even", + "D": "\\(x - y\\) is odd", + "E": "\\(x^2 + y^2\\) is odd" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(x\\) and \\(y\\) are integers and the product \\(xy\\) is odd, which of the following must be true?", + "explanation": "A product is odd **only when every factor is odd**. So \\(xy\\) odd forces **both** \\(x\\) and \\(y\\) to\nbe odd. Now test each choice with \"odd, odd\":\n\n- **A.** odd \\(+\\) odd \\(=\\) even, so \\(x + y\\) is *not* odd. Eliminate.\n- **B.** odd \\(+\\) odd \\(=\\) **even**. This must be true. ✓\n- **C.** \\(x\\) is odd, not even. Eliminate.\n- **D.** odd \\(-\\) odd \\(=\\) even, so \\(x - y\\) is not odd. Eliminate.\n- **E.** \\(x^2\\) and \\(y^2\\) are both odd, and odd \\(+\\) odd \\(=\\) even, so \\(x^2 + y^2\\) is even, not\n odd. Eliminate.\n\nOnly **B** holds. The answer is **\\(x + y\\) is even**.", + "hints": [ + "What does an *odd product* tell you about each of the two factors?", + "Once you know the parity of \\(x\\) and \\(y\\), just apply odd/even sum rules to each choice." + ] + }, + { + "id": "quant-number-properties-odds-evens-signs-0020", + "section": "quant", + "topic": "number-properties", + "subtopic": "odds-evens-signs", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "signs", + "must-be-true" + ], + "choices": { + "A": "\\(x < 0\\)", + "B": "\\(y > 0\\)", + "C": "\\(z < 0\\)", + "D": "\\(xz > 0\\)", + "E": "\\(yz > 0\\)" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(x\\), \\(y\\), and \\(z\\) are nonzero numbers with \\(xyz < 0\\) and \\(xy > 0\\), which of the\nfollowing must be true?", + "explanation": "Work with the sign of each product.\n\nSince \\(xy > 0\\), \\(x\\) and \\(y\\) have the **same sign**, so their product is positive. Then\n\n\\[xyz = (xy) \\cdot z < 0 \\quad\\text{with}\\quad xy > 0 \\;\\Rightarrow\\; z < 0.\\]\n\nSo \\(z\\) must be negative — choice **C**.\n\nThe other choices aren't forced: \\(x\\) and \\(y\\) could **both** be positive or **both** be negative\n(either keeps \\(xy > 0\\)), so A and B fail. And because \\(x\\)'s sign is unknown while \\(z < 0\\), the\nproduct \\(xz\\) could be positive or negative, so D fails; the same reasoning kills E. The answer is\n**\\(z < 0\\)**.", + "hints": [ + "Treat \\(xy\\) as a single quantity — its sign is given directly.", + "A product is negative when it has an odd number of negative factors; use that to pin down \\(z\\)." + ] + }, + { + "id": "quant-number-properties-remainders-0021", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "remainders", + "scaling" + ], + "choices": { + "A": "1", + "B": "2", + "C": "4", + "D": "5", + "E": "6" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "When the positive integer \\(n\\) is divided by 7, the remainder is 4. What is the remainder when\n\\(3n\\) is divided by 7?", + "explanation": "To find the remainder of a product, you can reduce each part to its remainder first. Here \\(n\\)\nleaves remainder 4, so \\(3n\\) leaves the same remainder as \\(3 \\times 4 = 12\\).\n\nBut a remainder must be **less than the divisor**, and \\(12 > 7\\), so reduce once more:\n\n\\[12 = 7 \\cdot 1 + 5 \\quad\\Rightarrow\\quad 12 \\equiv 5 \\pmod 7.\\]\n\nSo the remainder is **5**.\n\nConfirm with a concrete value: the smallest such \\(n\\) is 4, and \\(3 \\cdot 4 = 12\\), which divided by\n7 leaves remainder 5. (Any valid \\(n\\), like 11 or 18, gives the same result.) A common slip is to\nanswer 4 — the *original* remainder — but multiplying by 3 changes it.", + "hints": [ + "You don't need to know \\(n\\); reduce \\(3 \\times (\\text{remainder})\\) modulo 7.", + "If your intermediate result is 7 or more, subtract 7 until it drops below 7." + ] + }, + { + "id": "quant-number-properties-remainders-0022", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "remainders", + "simultaneous-conditions", + "lcm" + ], + "choices": { + "A": "13", + "B": "18", + "C": "23", + "D": "28", + "E": "38" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "A positive integer \\(n\\) leaves a remainder of 3 when divided by 5 and a remainder of 2 when divided\nby 3. What is the least value of \\(n\\) that is greater than 10?", + "explanation": "List the numbers satisfying each condition and find where they first overlap.\n\nRemainder 3 mod 5: \\(3, 8, 13, 18, 23, 28, \\dots\\)\nRemainder 2 mod 3: \\(2, 5, 8, 11, 14, 17, 20, 23, \\dots\\)\n\nThe values common to both are \\(8, 23, 38, \\dots\\) (they repeat every\n\\(\\operatorname{lcm}(5, 3) = 15\\)). The first one **greater than 10** is **23**.\n\nCheck 23: \\(23 = 5 \\cdot 4 + 3\\) (remainder 3 ✓) and \\(23 = 3 \\cdot 7 + 2\\) (remainder 2 ✓).\n\nThe distractors satisfy only one condition: 13 and 18 both leave remainder 3 mod 5 but fail the\nmod-3 test (\\(13 \\equiv 1\\), \\(18 \\equiv 0\\)), and 28 leaves remainder 3 mod 5 but \\(28 \\equiv 1\n\\pmod 3\\). (38 satisfies both but isn't the *least* one over 10.) The answer is **23**.", + "hints": [ + "Write out the first several numbers that fit each condition separately.", + "The values satisfying both repeat every LCM of the two divisors — find the first overlap above 10." + ] + }, { "id": "quant-statistics-descriptive-0009", "section": "quant", @@ -984,6 +1160,61 @@ "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-factors-multiples", + "section": "quant", + "topic": "number-properties", + "subtopic": "factors-multiples", + "title": "Factors, Multiples, Primes & Divisibility", + "tags": [ + "factors", + "multiples", + "primes", + "gcd", + "lcm", + "divisibility" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nAlmost every Number Properties question rests on one skill: **breaking an integer into its\nprime factors**. Once you have the prime factorization, divisor counts, greatest common divisors,\nleast common multiples, and divisibility all follow mechanically. Get comfortable factoring fast\nand this whole topic gets easier.\n\n## Core concepts\n\n**Prime factorization.** Every integer greater than 1 is a unique product of primes. Split the\nnumber into any two factors and keep splitting until everything is prime.\n\n\\[360 = 36 \\times 10 = (2^2 \\cdot 3^2)(2 \\cdot 5) = 2^3 \\cdot 3^2 \\cdot 5\\]\n\n**Counting divisors.** If \\(N = p^a \\cdot q^b \\cdot r^c \\dots\\), the number of positive divisors is\n\n\\[(a+1)(b+1)(c+1)\\dots\\]\n\nThe \"+1\" is there because each divisor can use prime \\(p\\) to a power from \\(0\\) up to \\(a\\).\nFor \\(360 = 2^3 \\cdot 3^2 \\cdot 5^1\\): \\((3+1)(2+1)(1+1) = 24\\) divisors.\n\n**GCD (greatest common divisor).** Take each prime the two numbers **share**, using the **lower**\nexponent. **LCM (least common multiple).** Take **every** prime that appears in either number,\nusing the **higher** exponent.\n\n\\[12 = 2^2 \\cdot 3, \\quad 18 = 2 \\cdot 3^2 \\;\\Rightarrow\\; \\gcd = 2^1 \\cdot 3^1 = 6, \\quad \\operatorname{lcm} = 2^2 \\cdot 3^2 = 36\\]\n\nA useful identity for any two positive integers: \\(\\gcd(a,b) \\times \\operatorname{lcm}(a,b) = a \\times b\\).\n\n**Perfect squares** have only **even** exponents in their prime factorization (e.g.\n\\(900 = 2^2 \\cdot 3^2 \\cdot 5^2\\)). To turn a number into a perfect square, supply whatever primes\nare needed to make every exponent even.\n\n**Quick divisibility rules.** Divisible by 3 if the digit sum is divisible by 3; by 9 if the digit\nsum is divisible by 9; by 4 if the last two digits form a multiple of 4; by 6 if it passes the 2\n**and** 3 tests.\n\n## Worked examples\n\n**LCM in disguise.** Two signals flash every 12 and 18 seconds and just flashed together. When do\nthey next coincide? That is \\(\\operatorname{lcm}(12,18) = 36\\) seconds — *not* their product.\n\n**Smallest multiplier for a perfect square.** What is the smallest \\(n\\) making \\(90n\\) a perfect\nsquare? \\(90 = 2 \\cdot 3^2 \\cdot 5\\). The 2 and the 5 have odd exponents, so \\(n = 2 \\cdot 5 = 10\\)\nand \\(90 \\cdot 10 = 900 = 30^2\\).\n\n## Common traps\n\n- **Confusing GCD with LCM.** GCD is the *small* shared part (lower exponents); LCM is the *big*\n covering number (higher exponents). \"How often do things line up again?\" is almost always LCM.\n- **Multiplying instead of taking the LCM.** \\(12 \\times 18 = 216\\) is a common multiple, but not\n the *least* one.\n- **Forgetting 1 and the number itself** when counting divisors — the \\((a+1)(b+1)\\) formula\n already includes them.\n\n## Key takeaways\n\n- Factor into primes first; divisor count, GCD, and LCM all read straight off the exponents.\n- Divisor count \\(= (a+1)(b+1)\\dots\\); GCD uses shared primes at the lower power; LCM uses all\n primes at the higher power.\n- \\(\\gcd \\times \\operatorname{lcm} = a \\times b\\), and perfect squares have all-even exponents." + }, + { + "id": "quant-number-properties-odds-evens-signs", + "section": "quant", + "topic": "number-properties", + "subtopic": "odds-evens-signs", + "title": "Odds, Evens & Sign Rules", + "tags": [ + "parity", + "odd", + "even", + "signs", + "positive", + "negative" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nMany \"which of the following **must** be true?\" questions never require you to find a value — only\nto track **parity** (odd/even) and **sign** (positive/negative). Memorize a handful of rules and\nthese become fast, reliable points.\n\n## Core concepts\n\n**Parity of sums.** odd \\(+\\) odd \\(=\\) even; even \\(+\\) even \\(=\\) even; odd \\(+\\) even \\(=\\) odd.\nA sum is **odd exactly when an odd number of its terms are odd.**\n\n**Parity of products.** A product is **odd only if every factor is odd.** If even one factor is\neven, the product is even. So \\(xy\\) odd \\(\\Rightarrow\\) both \\(x\\) and \\(y\\) are odd.\n\n**Consecutive integers.** Among any two consecutive integers, exactly one is even, so their product\nis always even. Among any three consecutive integers, the product is divisible by \\(3! = 6\\).\n\n**Sign rules for products.** A product is **negative when it has an odd number of negative\nfactors**, and **positive when it has an even number** (zero counts as even). So \\(xyz < 0\\) means\none or three of the factors are negative.\n\n**A \"must be true\" strategy.** When a question asks what *must* be true, try to build a\ncounterexample. If you can find even one case where a choice fails, eliminate it. What survives\nevery test is the answer.\n\n## Worked examples\n\n**Product forces parity.** If \\(xy\\) is odd, both \\(x\\) and \\(y\\) are odd, so \\(x + y\\) is even and\n\\(x^2 + y^2 = \\text{odd} + \\text{odd}\\) is even. Any choice claiming \\(x+y\\) is odd is out.\n\n**Chaining signs.** If \\(xyz < 0\\) and \\(xy > 0\\): \\(xy > 0\\) means \\(x\\) and \\(y\\) share a sign, so\ntheir product is positive; for the triple product to be negative, \\(z\\) must be **negative**. Note\nyou still can't pin down the signs of \\(x\\) and \\(y\\) individually.\n\n## Common traps\n\n- **Assuming \"integer\" means positive.** Integers include 0 and negatives. Test a negative and 0\n against every \"must be true\" choice.\n- **Forgetting 0.** Zero is even, is neither positive nor negative, and makes any product it\n touches equal 0.\n- **Reading \"could be true\" as \"must be true.\"** One counterexample kills a *must*-be-true choice,\n even if it's true most of the time.\n\n## Key takeaways\n\n- Product is odd only if **all** factors are odd; a sum is odd only with an **odd count** of odd terms.\n- Product sign flips with each negative factor: odd number of negatives → negative.\n- For \"must be true,\" hunt for a counterexample — and always test negatives and 0." + }, + { + "id": "quant-number-properties-remainders", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "title": "Remainders & Modular Thinking", + "tags": [ + "remainders", + "modular", + "divisibility", + "cycles" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nA remainder question is really a statement about form. \"\\(n\\) leaves remainder 4 when divided by 7\"\njust means \\(n = 7k + 4\\) for some integer \\(k\\). Write that form down and most remainder problems\nturn into ordinary algebra or a short list.\n\n## Core concepts\n\n**The division form.** \"\\(n\\) divided by \\(d\\) leaves remainder \\(r\\)\" means\n\n\\[n = dq + r, \\qquad 0 \\le r < d\\]\n\nThe remainder is always **less than the divisor** — a remainder of 7 (mod 7) is impossible; it is 0.\n\n**Remainders add and multiply.** To find the remainder of a sum or product, you may reduce each\npiece to its remainder first. If \\(n\\) leaves remainder 4 mod 7, then \\(3n\\) leaves the remainder of\n\\(3 \\times 4 = 12\\), which is \\(12 - 7 = 5\\).\n\n**Testing the smallest case.** When only the remainder is given, plug in the smallest value that\nfits (here \\(n = 4\\)) — the remainder of any sum/product won't depend on which valid \\(n\\) you pick.\n\n**Two conditions at once.** \"\\(n\\) leaves remainder 3 mod 5 **and** remainder 2 mod 3\": list each,\n\n\\[n \\equiv 3 \\pmod 5: \\; 3, 8, 13, 18, 23, \\dots \\qquad n \\equiv 2 \\pmod 3: \\; 2, 5, 8, 11, 14, 17, 20, 23, \\dots\\]\n\nThe first common value is 8, and solutions then repeat every \\(\\operatorname{lcm}(5,3) = 15\\):\n\\(8, 23, 38, \\dots\\)\n\n## Worked examples\n\n**Scaling a remainder.** \\(n\\) leaves remainder 4 mod 7. Remainder of \\(3n\\) mod 7? Reduce:\n\\(3 \\cdot 4 = 12 \\equiv 5\\). So the remainder is **5**.\n\n**Combining two moduli.** Smallest \\(n > 10\\) with remainder 3 mod 5 and remainder 2 mod 3? From the\nlists above the pattern is \\(8, 23, 38, \\dots\\); the first one over 10 is **23**.\n\n## Common traps\n\n- **Remainder \\(\\ge\\) divisor.** The remainder must be strictly smaller than what you divide by.\n Always reduce (e.g. \\(12 \\bmod 7 = 5\\)).\n- **Forgetting the small cases.** \\(n\\) could be the remainder itself (e.g. \\(n = 4\\)) — don't\n assume \\(n\\) is large.\n- **Checking only one condition.** With two divisors, verify a candidate against **both** before\n choosing it.\n\n## Key takeaways\n\n- Rewrite \"remainder \\(r\\) mod \\(d\\)\" as \\(n = dq + r\\) with \\(0 \\le r < d\\).\n- Reduce sums and products to their remainders, then reduce again if the result exceeds the divisor.\n- Two simultaneous conditions repeat every LCM of the divisors — list a few terms and find the overlap." + }, { "id": "quant-statistics-descriptive", "section": "quant",