From e587a6583df1ed67ebbdc79806d27782ac2394f8 Mon Sep 17 00:00:00 2001 From: Claude Date: Thu, 16 Jul 2026 09:09:26 +0000 Subject: [PATCH] Seed Number Properties content: +6 questions, +3 lessons MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Fills a real gap in the bank — Quant number-properties had questions but no lessons. Adds a coherent study unit covering three subtopics: - factors-multiples: prime factorization, divisor count, GCF/LCM, and the GCF*LCM = a*b identity (2 problem-solving questions) - odds-evens-signs: parity of sums/products, consecutive-integer trick, sign rules for products (2 questions) - remainders: the n = dq + r division equation and simultaneous conditions (2 questions) All questions solved fresh to confirm exactly one correct answer, with distractors mapped to specific common mistakes. Rebuilt content index (32->38 questions, 8->11 lessons). Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_015yeU7ESNtvyMQ8VhZ7U5sg --- ...ant-number-properties-factors-multiples.md | 63 +++++ ...uant-number-properties-odds-evens-signs.md | 63 +++++ .../quant-number-properties-remainders.md | 63 +++++ ...umber-properties-factors-multiples-0017.md | 46 ++++ ...umber-properties-factors-multiples-0018.md | 50 ++++ ...number-properties-odds-evens-signs-0019.md | 45 ++++ ...number-properties-odds-evens-signs-0020.md | 44 ++++ ...quant-number-properties-remainders-0021.md | 47 ++++ ...quant-number-properties-remainders-0022.md | 51 ++++ docs/data/content.json | 237 +++++++++++++++++- 10 files changed, 705 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..62f0052 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-factors-multiples.md @@ -0,0 +1,63 @@ +--- +id: quant-number-properties-factors-multiples +section: quant +topic: number-properties +subtopic: factors-multiples +title: "Factors, Multiples, GCF & LCM" +tags: [factors, multiples, gcf, lcm, prime-factorization] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Almost every number-properties question rests on one idea: **prime factorization**. Once you can +break an integer into its primes, counting its divisors, finding a greatest common factor (GCF), +or a least common multiple (LCM) all become mechanical. GMAT Focus rewards doing this quickly +instead of listing factors by hand. + +## Core concepts + +**Prime factorization.** Every integer greater than 1 is a unique product of primes. Write it in +exponent form: \(72 = 2^3 \times 3^2\), \(90 = 2 \times 3^2 \times 5\). + +**Counting divisors.** If \(N = p^a \times q^b \times \dots\), the number of positive divisors is +\[(a+1)(b+1)\dots\] +Each prime can appear in a divisor to a power from \(0\) up to its exponent — that's the "+1". + +**Greatest common factor (GCF).** For each prime the two numbers share, take the **lower** power. +\[18 = 2 \cdot 3^2, \quad 30 = 2 \cdot 3 \cdot 5 \;\Rightarrow\; \text{GCF} = 2 \cdot 3 = 6\] + +**Least common multiple (LCM).** For every prime that appears in either number, take the **higher** +power. +\[\text{LCM}(18, 30) = 2 \cdot 3^2 \cdot 5 = 90\] + +**The key identity.** For any two positive integers, +\[\text{GCF}(a,b) \times \text{LCM}(a,b) = a \times b\] +This lets you find one quantity when you know the other three. + +## Worked examples + +**Divisor count.** \(200 = 2^3 \cdot 5^2\), so it has \((3+1)(2+1) = 12\) positive divisors. + +**When do events coincide?** Two lights blink every 6 and every 8 seconds. They next blink +together after \(\text{LCM}(6,8)\) seconds. \(6 = 2 \cdot 3\), \(8 = 2^3\), so LCM \(= 2^3 \cdot 3 = 24\) +seconds. LCM answers "when do repeating cycles line up again." + +## Common traps + +- **Swapping GCF and LCM rules.** GCF takes the *lower* power of *shared* primes; LCM takes the + *higher* power of *all* primes. Mixing them up is the most common mistake. +- **Forgetting 1 and the number itself are divisors.** The \((a+1)(b+1)\) formula already counts them. +- **Using the product \(a \times b\) when the LCM is what's asked** (e.g. "when do the cycles meet?"). + The product is a common multiple, but rarely the *least* one. + +## Key takeaways + +- Factor into primes first — then divisor count, GCF, and LCM all follow mechanically. +- GCF = lowest powers of shared primes; LCM = highest powers of all primes. +- \(\text{GCF} \times \text{LCM} = a \times b\) — memorize it; it turns many problems into one division. +- LCM answers "when do repeating cycles coincide"; GCF answers "largest equal groups." 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..d4a3710 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-odds-evens-signs.md @@ -0,0 +1,63 @@ +--- +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 + +Parity (odd vs. even) and sign (positive vs. negative) questions look like arithmetic but are +really about **rules that always hold**. When a question asks what *must* be true, you're testing +a rule — not plugging in one lucky number. Knowing the handful of rules below lets you answer +instantly, and testing values lets you kill "must be true" traps. + +## Core concepts + +**Parity of sums and differences.** Same parity → even; different parity → odd. +\[\text{even} \pm \text{even} = \text{even}, \quad \text{odd} \pm \text{odd} = \text{even}, \quad \text{even} \pm \text{odd} = \text{odd}\] + +**Parity of products.** A product is **even if *any* factor is even**; it is odd only when **every** +factor is odd. +\[\text{even} \times \text{anything} = \text{even}, \quad \text{odd} \times \text{odd} = \text{odd}\] + +**A power keeps the base's parity.** \(n^k\) is odd exactly when \(n\) is odd. So \(n^2\) and \(n\) +always share parity — useful for "must be even" questions. + +**Consecutive integers.** Any two consecutive integers have opposite parity, so their **product is +always even**: \(n(n+1)\) is even for every integer \(n\). + +**Sign rules for products.** A product is positive when the number of negative factors is **even**, +negative when it's **odd**. For two numbers: \(xy > 0\) means *same sign*; \(xy < 0\) means *opposite +signs*. + +## Worked examples + +**Must-be-even.** Is \(n^2 + n\) always even? Factor: \(n^2 + n = n(n+1)\), a product of consecutive +integers — always even. Yes. + +**Reading two conditions together.** If \(xy > 0\) and \(x + y < 0\): same sign (from \(xy>0\)) plus a +negative sum forces **both** \(x\) and \(y\) negative. So \(x < 0\) must be true. + +## Common traps + +- **Assuming "integer" without being told.** Sign/parity rules apply to integers. If a variable + could be a fraction, "odd/even" doesn't apply. +- **Zero is even, and zero is neither positive nor negative.** Watch for it in "must be true" and + "could be" questions. +- **Confirming with one value instead of the rule.** For *must be true*, a single example can only + *disprove* a choice, never prove it. Find a counterexample to eliminate. + +## Key takeaways + +- Product is even if any factor is even; odd only if all factors are odd. +- \(n\) and \(n^2\) share parity; \(n(n+1)\) is always even. +- \(xy > 0\) → same sign; \(xy < 0\) → opposite signs; count negatives to get a product's sign. +- For "must be true," attack each choice with a counterexample; anything that survives is forced. 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..aad2ff3 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-remainders.md @@ -0,0 +1,63 @@ +--- +id: quant-number-properties-remainders +section: quant +topic: number-properties +subtopic: remainders +title: "Remainders & Modular Thinking" +tags: [remainders, modular-arithmetic, divisibility] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +A remainder question is really the statement "\(n\) is a multiple of \(d\), plus a little extra." +Writing that extra as an equation — \(n = dq + r\) — turns nearly every remainder problem into +straightforward algebra. The two reliable tools are the **division equation** and, when a problem +is small, **testing the smallest value** that fits. + +## Core concepts + +**The division equation.** When \(n\) divided by \(d\) leaves remainder \(r\), +\[n = dq + r, \quad 0 \le r < d\] +where \(q\) is the quotient. The remainder is always at least 0 and strictly less than the divisor. + +**Remainders combine.** To find the remainder of a sum, product, or multiple, you can work with the +remainders themselves. If \(n\) leaves remainder \(4\) on division by \(7\), then \(3n\) leaves the +same remainder as \(3 \times 4 = 12\), i.e. \(12 - 7 = 5\). Reduce at each step to keep numbers small. + +**Substitute the general form.** For "\(n\) has remainder \(4\) mod \(7\); find the remainder of +\(3n+2\)," write \(n = 7k + 4\): +\[3n + 2 = 3(7k+4) + 2 = 21k + 14 = 7(3k + 2)\] +That's a multiple of 7, so the remainder is \(0\). The \(21k\) term is always divisible by 7, so only +the constant part decides the remainder. + +**Two simultaneous conditions.** "Remainder 3 mod 5 and remainder 2 mod 4" — list one sequence and +scan for a value that fits the other. \(3, 8, 13, 18, \dots\) (mod 5); the first that is also +\(\equiv 2 \pmod 4\) is \(18\). + +## Worked examples + +**Constant-only shortcut.** \(n = 7k + 4\Rightarrow 3n+2 = 7(3k+2)\), remainder \(\mathbf{0}\). +Only the "+4 → +12 → +2 after adding 2 → then ×... " reasoning matters; drop the \(7k\). + +**Smallest common value.** Numbers with remainder 3 mod 5: \(3, 8, 13, 18\). Check each against +mod 4: \(18 \div 4 = 4\) r \(2\). So \(18\) is the smallest value satisfying both. + +## Common traps + +- **A remainder can't equal or exceed the divisor.** "Remainder 7 on division by 5" is impossible — + it would reduce to remainder 2. +- **Forgetting remainder 0 is allowed.** If the algebra gives a clean multiple, the answer is 0. +- **Testing only one value for "could be."** A single value proves *possible*; ruling something out + needs the general form \(n = dq + r\). + +## Key takeaways + +- Write \(n = dq + r\) with \(0 \le r < d\) — this equation solves most remainder problems. +- To combine remainders, reduce at each step; multiples of the divisor drop out. +- For simultaneous conditions, list one sequence and scan for the first term fitting the other. +- A remainder is always less than the divisor, and 0 is a valid remainder. 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..e6e0dcf --- /dev/null +++ b/content/questions/quant/quant-number-properties-factors-multiples-0017.md @@ -0,0 +1,46 @@ +--- +id: quant-number-properties-factors-multiples-0017 +section: quant +topic: number-properties +subtopic: factors-multiples +type: problem-solving +difficulty: medium +tags: [gcf, lcm, factors] +choices: + A: "15" + B: "30" + C: "45" + D: "54" + E: "60" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +The greatest common factor of two positive integers is 6 and their least common multiple is 90. +If one of the integers is 18, what is the other integer? + +## Explanation + +Use the identity that ties GCF and LCM to the two numbers: + +\[\text{GCF}(a,b) \times \text{LCM}(a,b) = a \times b\] + +Here \(6 \times 90 = 18 \times b\), so \(540 = 18b\) and \(b = 30\). + +Check it directly: \(18 = 2 \cdot 3^2\) and \(30 = 2 \cdot 3 \cdot 5\). The GCF takes the lower power +of shared primes: \(2 \cdot 3 = 6\) ✓. The LCM takes the higher power of all primes: +\(2 \cdot 3^2 \cdot 5 = 90\) ✓. + +**The trap:** choice **A (15)** is \(\text{LCM} \div \text{GCF} = 90 \div 6\), a tempting but +meaningless combination here. + +## Hints + +- There's a fixed relationship between two numbers, their GCF, and their LCM. +- \(\text{GCF} \times \text{LCM} = \text{the product of the two numbers}\). 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..6ff551d --- /dev/null +++ b/content/questions/quant/quant-number-properties-factors-multiples-0018.md @@ -0,0 +1,50 @@ +--- +id: quant-number-properties-factors-multiples-0018 +section: quant +topic: number-properties +subtopic: factors-multiples +type: problem-solving +difficulty: medium +tags: [lcm, multiples, word-problem] +choices: + A: "60" + B: "120" + C: "160" + D: "240" + E: "480" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +Three lighthouses flash at regular intervals of 8, 12, and 20 seconds, respectively. If all three +flash together at a certain instant, after how many seconds will they next all flash together? + +## Explanation + +Repeating cycles line up again after the **least common multiple** of their periods, so find +\(\text{LCM}(8, 12, 20)\). + +Factor each into primes: + +\[8 = 2^3, \quad 12 = 2^2 \cdot 3, \quad 20 = 2^2 \cdot 5\] + +The LCM takes the highest power of every prime that appears: + +\[\text{LCM} = 2^3 \cdot 3 \cdot 5 = 120\] + +So all three flash together again after **120 seconds**. + +**The trap:** choice **A (60)** is \(\text{LCM}(12, 20)\) — it ignores the 8-second light, which needs +the full \(2^3\). Multiplying all three periods (\(8 \cdot 12 \cdot 20 = 1920\)) gives a common +multiple, but not the *least* one. + +## Hints + +- "When do repeating cycles coincide again?" is always an LCM question. +- Factor each interval into primes and take the highest power of each prime. 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..f723c38 --- /dev/null +++ b/content/questions/quant/quant-number-properties-odds-evens-signs-0019.md @@ -0,0 +1,45 @@ +--- +id: quant-number-properties-odds-evens-signs-0019 +section: quant +topic: number-properties +subtopic: odds-evens-signs +type: problem-solving +difficulty: medium +tags: [parity, odd-even, must-be-true] +choices: + A: "\(2n + 1\)" + B: "\(3n\)" + C: "\(n^2 + n\)" + D: "\(n^2 + 1\)" + E: "\(n^3 - n + 1\)" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(n\) is an integer, which of the following must be even? + +## Explanation + +"Must be even" means even for **every** integer \(n\) — so a single odd result kills a choice. + +- **C: \(n^2 + n = n(n+1)\)** is the product of two *consecutive* integers. One of any two + consecutive integers is even, so the product is always even. **This is the answer.** + +Check the others fail for at least one \(n\): + +- **A: \(2n + 1\)** is even plus 1 → always **odd**. +- **B: \(3n\)** — at \(n = 1\) it's 3, odd. +- **D: \(n^2 + 1\)** — at \(n = 2\) it's 5, odd. +- **E: \(n^3 - n + 1 = (n-1)n(n+1) + 1\)** — the product of three consecutive integers is even, so + this is even + 1 → always **odd**. + +## Hints + +- "Must be even" fails the moment you find one odd result — try \(n = 1\) and \(n = 2\). +- A product of consecutive integers is always even. Can you rewrite a choice in that form? 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..020278f --- /dev/null +++ b/content/questions/quant/quant-number-properties-odds-evens-signs-0020.md @@ -0,0 +1,44 @@ +--- +id: quant-number-properties-odds-evens-signs-0020 +section: quant +topic: number-properties +subtopic: odds-evens-signs +type: problem-solving +difficulty: medium +tags: [signs, positive-negative, must-be-true] +choices: + A: "\(x > 0\)" + B: "\(y > 0\)" + C: "\(x < 0\)" + D: "\(x = y\)" + E: "\(xy < 0\)" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(xy > 0\) and \(x + y < 0\), which of the following must be true? + +## Explanation + +Read the two conditions together: + +- \(xy > 0\) means \(x\) and \(y\) have the **same sign** (both positive or both negative). +- \(x + y < 0\) means their sum is negative. + +Two positive numbers can't sum to something negative, so "both positive" is out. That leaves +**both negative**. Therefore \(x < 0\) (and \(y < 0\)) must be true — **choice C**. + +Why the others aren't forced: **A** and **B** claim a variable is positive — the opposite of what +we found. **D (\(x = y\))** isn't required (e.g. \(x = -1, y = -2\) works). **E (\(xy < 0\))** +directly contradicts the given \(xy > 0\). + +## Hints + +- \(xy > 0\) tells you the two variables share a sign. Which sign? +- If they had the same sign but summed to a negative, can they both be positive? 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..2db2197 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0021.md @@ -0,0 +1,47 @@ +--- +id: quant-number-properties-remainders-0021 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: medium +tags: [remainders, divisibility] +choices: + A: "0" + B: "1" + C: "2" + D: "5" + E: "6" +answer: A +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 +\(3n + 2\) is divided by 7? + +## Explanation + +Write \(n\) in the division form \(n = 7k + 4\), then substitute: + +\[3n + 2 = 3(7k + 4) + 2 = 21k + 12 + 2 = 21k + 14 = 7(3k + 2)\] + +That is a multiple of 7, so the remainder is **0**. + +The \(21k\) term is always divisible by 7, so only the constant part matters: +\(3 \times 4 + 2 = 14\), and \(14\) is itself a multiple of 7 — remainder 0. + +Sanity check with the smallest \(n\): \(n = 4 \Rightarrow 3(4) + 2 = 14\), and \(14 \div 7\) leaves 0. ✓ + +**The trap:** choice **D (5)** comes from forgetting the "\(+2\)" and taking the remainder of +\(3n = 12\) (i.e. \(12 - 7 = 5\)). + +## Hints + +- Substitute \(n = 7k + 4\) into \(3n + 2\) and simplify. +- The multiple-of-7 terms drop out; only the leftover constant sets the remainder. 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..3429086 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0022.md @@ -0,0 +1,51 @@ +--- +id: quant-number-properties-remainders-0022 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: medium +tags: [remainders, simultaneous-conditions] +choices: + A: "8" + B: "13" + C: "14" + D: "18" + E: "23" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +When a positive integer \(k\) is divided by 5, the remainder is 3, and when \(k\) is divided by 4, +the remainder is 2. What is the smallest possible value of \(k\)? + +## Explanation + +List the values that satisfy the first condition, then scan for one that also satisfies the second. + +Remainder 3 when divided by 5: \(k = 3, 8, 13, 18, 23, \dots\) + +Now test each against "remainder 2 when divided by 4": + +- \(3 \div 4\) → remainder 3. No. +- \(8 \div 4\) → remainder 0. No. +- \(13 \div 4\) → remainder 1. No. +- \(18 \div 4 = 4\) remainder 2. **Yes.** + +So the smallest value is **18**. Confirm both conditions: \(18 = 5(3) + 3\) ✓ and +\(18 = 4(4) + 2\) ✓. + +**The traps:** **A (8)** fits the mod-5 condition but is divisible by 4 (remainder 0, not 2). +**C (14)** fits the mod-4 condition (remainder 2) but \(14 \div 5\) leaves remainder 4, failing the +mod-5 condition. Each wrong choice satisfies at most one of the two requirements. + +## Hints + +- List every value with remainder 3 mod 5, then check each against the mod-4 condition. +- The answer must satisfy *both* conditions at once — don't stop at the first one. diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..24ee288 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 +418,184 @@ "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": [ + "gcf", + "lcm", + "factors" + ], + "choices": { + "A": "15", + "B": "30", + "C": "45", + "D": "54", + "E": "60" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "The greatest common factor of two positive integers is 6 and their least common multiple is 90.\nIf one of the integers is 18, what is the other integer?", + "explanation": "Use the identity that ties GCF and LCM to the two numbers:\n\n\\[\\text{GCF}(a,b) \\times \\text{LCM}(a,b) = a \\times b\\]\n\nHere \\(6 \\times 90 = 18 \\times b\\), so \\(540 = 18b\\) and \\(b = 30\\).\n\nCheck it directly: \\(18 = 2 \\cdot 3^2\\) and \\(30 = 2 \\cdot 3 \\cdot 5\\). The GCF takes the lower power\nof shared primes: \\(2 \\cdot 3 = 6\\) ✓. The LCM takes the higher power of all primes:\n\\(2 \\cdot 3^2 \\cdot 5 = 90\\) ✓.\n\n**The trap:** choice **A (15)** is \\(\\text{LCM} \\div \\text{GCF} = 90 \\div 6\\), a tempting but\nmeaningless combination here.", + "hints": [ + "There's a fixed relationship between two numbers, their GCF, and their LCM.", + "\\(\\text{GCF} \\times \\text{LCM} = \\text{the product of the two numbers}\\)." + ] + }, + { + "id": "quant-number-properties-factors-multiples-0018", + "section": "quant", + "topic": "number-properties", + "subtopic": "factors-multiples", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "lcm", + "multiples", + "word-problem" + ], + "choices": { + "A": "60", + "B": "120", + "C": "160", + "D": "240", + "E": "480" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "Three lighthouses flash at regular intervals of 8, 12, and 20 seconds, respectively. If all three\nflash together at a certain instant, after how many seconds will they next all flash together?", + "explanation": "Repeating cycles line up again after the **least common multiple** of their periods, so find\n\\(\\text{LCM}(8, 12, 20)\\).\n\nFactor each into primes:\n\n\\[8 = 2^3, \\quad 12 = 2^2 \\cdot 3, \\quad 20 = 2^2 \\cdot 5\\]\n\nThe LCM takes the highest power of every prime that appears:\n\n\\[\\text{LCM} = 2^3 \\cdot 3 \\cdot 5 = 120\\]\n\nSo all three flash together again after **120 seconds**.\n\n**The trap:** choice **A (60)** is \\(\\text{LCM}(12, 20)\\) — it ignores the 8-second light, which needs\nthe full \\(2^3\\). Multiplying all three periods (\\(8 \\cdot 12 \\cdot 20 = 1920\\)) gives a common\nmultiple, but not the *least* one.", + "hints": [ + "\"When do repeating cycles coincide again?\" is always an LCM question.", + "Factor each interval into primes and take the highest power of each prime." + ] + }, + { + "id": "quant-number-properties-odds-evens-signs-0019", + "section": "quant", + "topic": "number-properties", + "subtopic": "odds-evens-signs", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "parity", + "odd-even", + "must-be-true" + ], + "choices": { + "A": "\\(2n + 1\\)", + "B": "\\(3n\\)", + "C": "\\(n^2 + n\\)", + "D": "\\(n^2 + 1\\)", + "E": "\\(n^3 - n + 1\\)" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(n\\) is an integer, which of the following must be even?", + "explanation": "\"Must be even\" means even for **every** integer \\(n\\) — so a single odd result kills a choice.\n\n- **C: \\(n^2 + n = n(n+1)\\)** is the product of two *consecutive* integers. One of any two\n consecutive integers is even, so the product is always even. **This is the answer.**\n\nCheck the others fail for at least one \\(n\\):\n\n- **A: \\(2n + 1\\)** is even plus 1 → always **odd**.\n- **B: \\(3n\\)** — at \\(n = 1\\) it's 3, odd.\n- **D: \\(n^2 + 1\\)** — at \\(n = 2\\) it's 5, odd.\n- **E: \\(n^3 - n + 1 = (n-1)n(n+1) + 1\\)** — the product of three consecutive integers is even, so\n this is even + 1 → always **odd**.", + "hints": [ + "\"Must be even\" fails the moment you find one odd result — try \\(n = 1\\) and \\(n = 2\\).", + "A product of consecutive integers is always even. Can you rewrite a choice in that form?" + ] + }, + { + "id": "quant-number-properties-odds-evens-signs-0020", + "section": "quant", + "topic": "number-properties", + "subtopic": "odds-evens-signs", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "signs", + "positive-negative", + "must-be-true" + ], + "choices": { + "A": "\\(x > 0\\)", + "B": "\\(y > 0\\)", + "C": "\\(x < 0\\)", + "D": "\\(x = y\\)", + "E": "\\(xy < 0\\)" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(xy > 0\\) and \\(x + y < 0\\), which of the following must be true?", + "explanation": "Read the two conditions together:\n\n- \\(xy > 0\\) means \\(x\\) and \\(y\\) have the **same sign** (both positive or both negative).\n- \\(x + y < 0\\) means their sum is negative.\n\nTwo positive numbers can't sum to something negative, so \"both positive\" is out. That leaves\n**both negative**. Therefore \\(x < 0\\) (and \\(y < 0\\)) must be true — **choice C**.\n\nWhy the others aren't forced: **A** and **B** claim a variable is positive — the opposite of what\nwe found. **D (\\(x = y\\))** isn't required (e.g. \\(x = -1, y = -2\\) works). **E (\\(xy < 0\\))**\ndirectly contradicts the given \\(xy > 0\\).", + "hints": [ + "\\(xy > 0\\) tells you the two variables share a sign. Which sign?", + "If they had the same sign but summed to a negative, can they both be positive?" + ] + }, + { + "id": "quant-number-properties-remainders-0021", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "remainders", + "divisibility" + ], + "choices": { + "A": "0", + "B": "1", + "C": "2", + "D": "5", + "E": "6" + }, + "answer": "A", + "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 + 2\\) is divided by 7?", + "explanation": "Write \\(n\\) in the division form \\(n = 7k + 4\\), then substitute:\n\n\\[3n + 2 = 3(7k + 4) + 2 = 21k + 12 + 2 = 21k + 14 = 7(3k + 2)\\]\n\nThat is a multiple of 7, so the remainder is **0**.\n\nThe \\(21k\\) term is always divisible by 7, so only the constant part matters:\n\\(3 \\times 4 + 2 = 14\\), and \\(14\\) is itself a multiple of 7 — remainder 0.\n\nSanity check with the smallest \\(n\\): \\(n = 4 \\Rightarrow 3(4) + 2 = 14\\), and \\(14 \\div 7\\) leaves 0. ✓\n\n**The trap:** choice **D (5)** comes from forgetting the \"\\(+2\\)\" and taking the remainder of\n\\(3n = 12\\) (i.e. \\(12 - 7 = 5\\)).", + "hints": [ + "Substitute \\(n = 7k + 4\\) into \\(3n + 2\\) and simplify.", + "The multiple-of-7 terms drop out; only the leftover constant sets the remainder." + ] + }, + { + "id": "quant-number-properties-remainders-0022", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "remainders", + "simultaneous-conditions" + ], + "choices": { + "A": "8", + "B": "13", + "C": "14", + "D": "18", + "E": "23" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "When a positive integer \\(k\\) is divided by 5, the remainder is 3, and when \\(k\\) is divided by 4,\nthe remainder is 2. What is the smallest possible value of \\(k\\)?", + "explanation": "List the values that satisfy the first condition, then scan for one that also satisfies the second.\n\nRemainder 3 when divided by 5: \\(k = 3, 8, 13, 18, 23, \\dots\\)\n\nNow test each against \"remainder 2 when divided by 4\":\n\n- \\(3 \\div 4\\) → remainder 3. No.\n- \\(8 \\div 4\\) → remainder 0. No.\n- \\(13 \\div 4\\) → remainder 1. No.\n- \\(18 \\div 4 = 4\\) remainder 2. **Yes.**\n\nSo the smallest value is **18**. Confirm both conditions: \\(18 = 5(3) + 3\\) ✓ and\n\\(18 = 4(4) + 2\\) ✓.\n\n**The traps:** **A (8)** fits the mod-5 condition but is divisible by 4 (remainder 0, not 2).\n**C (14)** fits the mod-4 condition (remainder 2) but \\(14 \\div 5\\) leaves remainder 4, failing the\nmod-5 condition. Each wrong choice satisfies at most one of the two requirements.", + "hints": [ + "List every value with remainder 3 mod 5, then check each against the mod-4 condition.", + "The answer must satisfy *both* conditions at once — don't stop at the first one." + ] + }, { "id": "quant-statistics-descriptive-0009", "section": "quant", @@ -984,6 +1162,57 @@ "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, GCF & LCM", + "tags": [ + "factors", + "multiples", + "gcf", + "lcm", + "prime-factorization" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nAlmost every number-properties question rests on one idea: **prime factorization**. Once you can\nbreak an integer into its primes, counting its divisors, finding a greatest common factor (GCF),\nor a least common multiple (LCM) all become mechanical. GMAT Focus rewards doing this quickly\ninstead of listing factors by hand.\n\n## Core concepts\n\n**Prime factorization.** Every integer greater than 1 is a unique product of primes. Write it in\nexponent form: \\(72 = 2^3 \\times 3^2\\), \\(90 = 2 \\times 3^2 \\times 5\\).\n\n**Counting divisors.** If \\(N = p^a \\times q^b \\times \\dots\\), the number of positive divisors is\n\\[(a+1)(b+1)\\dots\\]\nEach prime can appear in a divisor to a power from \\(0\\) up to its exponent — that's the \"+1\".\n\n**Greatest common factor (GCF).** For each prime the two numbers share, take the **lower** power.\n\\[18 = 2 \\cdot 3^2, \\quad 30 = 2 \\cdot 3 \\cdot 5 \\;\\Rightarrow\\; \\text{GCF} = 2 \\cdot 3 = 6\\]\n\n**Least common multiple (LCM).** For every prime that appears in either number, take the **higher**\npower.\n\\[\\text{LCM}(18, 30) = 2 \\cdot 3^2 \\cdot 5 = 90\\]\n\n**The key identity.** For any two positive integers,\n\\[\\text{GCF}(a,b) \\times \\text{LCM}(a,b) = a \\times b\\]\nThis lets you find one quantity when you know the other three.\n\n## Worked examples\n\n**Divisor count.** \\(200 = 2^3 \\cdot 5^2\\), so it has \\((3+1)(2+1) = 12\\) positive divisors.\n\n**When do events coincide?** Two lights blink every 6 and every 8 seconds. They next blink\ntogether after \\(\\text{LCM}(6,8)\\) seconds. \\(6 = 2 \\cdot 3\\), \\(8 = 2^3\\), so LCM \\(= 2^3 \\cdot 3 = 24\\)\nseconds. LCM answers \"when do repeating cycles line up again.\"\n\n## Common traps\n\n- **Swapping GCF and LCM rules.** GCF takes the *lower* power of *shared* primes; LCM takes the\n *higher* power of *all* primes. Mixing them up is the most common mistake.\n- **Forgetting 1 and the number itself are divisors.** The \\((a+1)(b+1)\\) formula already counts them.\n- **Using the product \\(a \\times b\\) when the LCM is what's asked** (e.g. \"when do the cycles meet?\").\n The product is a common multiple, but rarely the *least* one.\n\n## Key takeaways\n\n- Factor into primes first — then divisor count, GCF, and LCM all follow mechanically.\n- GCF = lowest powers of shared primes; LCM = highest powers of all primes.\n- \\(\\text{GCF} \\times \\text{LCM} = a \\times b\\) — memorize it; it turns many problems into one division.\n- LCM answers \"when do repeating cycles coincide\"; GCF answers \"largest equal groups.\"" + }, + { + "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\nParity (odd vs. even) and sign (positive vs. negative) questions look like arithmetic but are\nreally about **rules that always hold**. When a question asks what *must* be true, you're testing\na rule — not plugging in one lucky number. Knowing the handful of rules below lets you answer\ninstantly, and testing values lets you kill \"must be true\" traps.\n\n## Core concepts\n\n**Parity of sums and differences.** Same parity → even; different parity → odd.\n\\[\\text{even} \\pm \\text{even} = \\text{even}, \\quad \\text{odd} \\pm \\text{odd} = \\text{even}, \\quad \\text{even} \\pm \\text{odd} = \\text{odd}\\]\n\n**Parity of products.** A product is **even if *any* factor is even**; it is odd only when **every**\nfactor is odd.\n\\[\\text{even} \\times \\text{anything} = \\text{even}, \\quad \\text{odd} \\times \\text{odd} = \\text{odd}\\]\n\n**A power keeps the base's parity.** \\(n^k\\) is odd exactly when \\(n\\) is odd. So \\(n^2\\) and \\(n\\)\nalways share parity — useful for \"must be even\" questions.\n\n**Consecutive integers.** Any two consecutive integers have opposite parity, so their **product is\nalways even**: \\(n(n+1)\\) is even for every integer \\(n\\).\n\n**Sign rules for products.** A product is positive when the number of negative factors is **even**,\nnegative when it's **odd**. For two numbers: \\(xy > 0\\) means *same sign*; \\(xy < 0\\) means *opposite\nsigns*.\n\n## Worked examples\n\n**Must-be-even.** Is \\(n^2 + n\\) always even? Factor: \\(n^2 + n = n(n+1)\\), a product of consecutive\nintegers — always even. Yes.\n\n**Reading two conditions together.** If \\(xy > 0\\) and \\(x + y < 0\\): same sign (from \\(xy>0\\)) plus a\nnegative sum forces **both** \\(x\\) and \\(y\\) negative. So \\(x < 0\\) must be true.\n\n## Common traps\n\n- **Assuming \"integer\" without being told.** Sign/parity rules apply to integers. If a variable\n could be a fraction, \"odd/even\" doesn't apply.\n- **Zero is even, and zero is neither positive nor negative.** Watch for it in \"must be true\" and\n \"could be\" questions.\n- **Confirming with one value instead of the rule.** For *must be true*, a single example can only\n *disprove* a choice, never prove it. Find a counterexample to eliminate.\n\n## Key takeaways\n\n- Product is even if any factor is even; odd only if all factors are odd.\n- \\(n\\) and \\(n^2\\) share parity; \\(n(n+1)\\) is always even.\n- \\(xy > 0\\) → same sign; \\(xy < 0\\) → opposite signs; count negatives to get a product's sign.\n- For \"must be true,\" attack each choice with a counterexample; anything that survives is forced." + }, + { + "id": "quant-number-properties-remainders", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "title": "Remainders & Modular Thinking", + "tags": [ + "remainders", + "modular-arithmetic", + "divisibility" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nA remainder question is really the statement \"\\(n\\) is a multiple of \\(d\\), plus a little extra.\"\nWriting that extra as an equation — \\(n = dq + r\\) — turns nearly every remainder problem into\nstraightforward algebra. The two reliable tools are the **division equation** and, when a problem\nis small, **testing the smallest value** that fits.\n\n## Core concepts\n\n**The division equation.** When \\(n\\) divided by \\(d\\) leaves remainder \\(r\\),\n\\[n = dq + r, \\quad 0 \\le r < d\\]\nwhere \\(q\\) is the quotient. The remainder is always at least 0 and strictly less than the divisor.\n\n**Remainders combine.** To find the remainder of a sum, product, or multiple, you can work with the\nremainders themselves. If \\(n\\) leaves remainder \\(4\\) on division by \\(7\\), then \\(3n\\) leaves the\nsame remainder as \\(3 \\times 4 = 12\\), i.e. \\(12 - 7 = 5\\). Reduce at each step to keep numbers small.\n\n**Substitute the general form.** For \"\\(n\\) has remainder \\(4\\) mod \\(7\\); find the remainder of\n\\(3n+2\\),\" write \\(n = 7k + 4\\):\n\\[3n + 2 = 3(7k+4) + 2 = 21k + 14 = 7(3k + 2)\\]\nThat's a multiple of 7, so the remainder is \\(0\\). The \\(21k\\) term is always divisible by 7, so only\nthe constant part decides the remainder.\n\n**Two simultaneous conditions.** \"Remainder 3 mod 5 and remainder 2 mod 4\" — list one sequence and\nscan for a value that fits the other. \\(3, 8, 13, 18, \\dots\\) (mod 5); the first that is also\n\\(\\equiv 2 \\pmod 4\\) is \\(18\\).\n\n## Worked examples\n\n**Constant-only shortcut.** \\(n = 7k + 4\\Rightarrow 3n+2 = 7(3k+2)\\), remainder \\(\\mathbf{0}\\).\nOnly the \"+4 → +12 → +2 after adding 2 → then ×... \" reasoning matters; drop the \\(7k\\).\n\n**Smallest common value.** Numbers with remainder 3 mod 5: \\(3, 8, 13, 18\\). Check each against\nmod 4: \\(18 \\div 4 = 4\\) r \\(2\\). So \\(18\\) is the smallest value satisfying both.\n\n## Common traps\n\n- **A remainder can't equal or exceed the divisor.** \"Remainder 7 on division by 5\" is impossible —\n it would reduce to remainder 2.\n- **Forgetting remainder 0 is allowed.** If the algebra gives a clean multiple, the answer is 0.\n- **Testing only one value for \"could be.\"** A single value proves *possible*; ruling something out\n needs the general form \\(n = dq + r\\).\n\n## Key takeaways\n\n- Write \\(n = dq + r\\) with \\(0 \\le r < d\\) — this equation solves most remainder problems.\n- To combine remainders, reduce at each step; multiples of the divisor drop out.\n- For simultaneous conditions, list one sequence and scan for the first term fitting the other.\n- A remainder is always less than the divisor, and 0 is a valid remainder." + }, { "id": "quant-statistics-descriptive", "section": "quant",