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..bf2d926 --- /dev/null +++ b/content/lessons/quant/quant-number-properties-factors-multiples.md @@ -0,0 +1,80 @@ +--- +id: quant-number-properties-factors-multiples +section: quant +topic: number-properties +subtopic: factors-multiples +title: "Factors, Multiples & Prime Factorization" +tags: [number-properties, factors, multiples, primes, gcd, lcm] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Almost every "clever" number-properties problem on the GMAT Focus Quant section becomes routine +once you write the number as a product of primes. **Prime factorization** is the master key: +counting divisors, finding a GCD or LCM, testing divisibility, and spotting perfect squares all +fall out of it directly. This lesson builds that one skill and shows the handful of formulas it +unlocks. + +## Core concepts + +**Factors vs. multiples.** A **factor** (divisor) of \(N\) divides \(N\) with no remainder; a +**multiple** of \(N\) is \(N\) times an integer. So the factors of 12 are \(\{1,2,3,4,6,12\}\), +while its multiples are \(12, 24, 36, \dots\) — a finite list versus an infinite one. + +**Prime factorization.** Every integer greater than 1 factors uniquely into primes: + +\[72 = 2^3 \times 3^2, \qquad 360 = 2^3 \times 3^2 \times 5\] + +**Counting divisors.** If \(N = p^a \times q^b \times \dots\), the number of positive divisors is + +\[(a+1)(b+1)\dots\] + +because each divisor picks an exponent for \(p\) from \(0\) to \(a\), independently for \(q\), and +so on. For \(72 = 2^3 \times 3^2\): \((3+1)(2+1) = 12\) divisors. + +**GCD and LCM from primes.** Line up the shared primes: + +- **GCD** — take the **lowest** power of each prime the numbers share. +- **LCM** — take the **highest** power of every prime that appears. + +For \(12 = 2^2 \times 3\) and \(18 = 2 \times 3^2\): \(\gcd = 2^1 \times 3^1 = 6\) and +\(\operatorname{lcm} = 2^2 \times 3^2 = 36\). + +**The product identity.** For any two positive integers, + +\[\gcd(a,b) \times \operatorname{lcm}(a,b) = a \times b\] + +This turns many two-number problems into one line of arithmetic. + +## Worked examples + +**Example 1 — counting perfect-square divisors.** How many divisors of \(N = 2^4 \times 3^2\) +are perfect squares? A divisor is a perfect square exactly when every prime exponent is **even**. +For \(2^4\) the choices are \(0, 2, 4\) (three options); for \(3^2\) they are \(0, 2\) (two +options). That gives \(3 \times 2 = 6\) perfect-square divisors. + +**Example 2 — the product identity.** Two positive integers have product 216 and GCD 6. Their +LCM is \(216 / 6 = 36\). No listing required. + +## Common traps + +- **1 is not prime.** It has exactly one divisor. The smallest prime is 2 (the only even prime). +- **Forgetting the "+1" in the divisor count.** The exponent can be \(0\) (the prime used zero + times), so a prime raised to \(a\) contributes \(a+1\) choices, not \(a\). +- **Swapping GCD and LCM.** GCD uses the *lowest* shared powers and is at most the smaller number; + LCM uses the *highest* powers and is at least the larger number. +- **The product identity is for two numbers only.** \(\gcd \times \operatorname{lcm} = a \times b\) + does not extend to three or more integers. + +## Key takeaways + +- Factor into primes first — divisor counts, GCD, LCM, and divisibility all follow mechanically. +- Divisor count of \(p^a q^b \dots\) is \((a+1)(b+1)\dots\); perfect-square divisors need all + exponents even. +- GCD = lowest shared prime powers; LCM = highest prime powers overall. +- For two numbers, \(\gcd \times \operatorname{lcm} = \) their product. diff --git a/content/questions/quant/quant-number-properties-factors-0001.md b/content/questions/quant/quant-number-properties-factors-0001.md new file mode 100644 index 0000000..6957473 --- /dev/null +++ b/content/questions/quant/quant-number-properties-factors-0001.md @@ -0,0 +1,42 @@ +--- +id: quant-number-properties-factors-0001 +section: quant +topic: number-properties +subtopic: factors-multiples +type: problem-solving +difficulty: easy +tags: [multiples, counting] +choices: + A: "15" + B: "16" + C: "17" + D: "18" + E: "20" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +How many multiples of 6 are there between 1 and 100, inclusive? + +## Explanation + +A multiple of 6 has the form \(6k\) for a positive integer \(k\). The largest one that does not +exceed 100 satisfies \(6k \le 100\), so \(k \le 16.67\), meaning \(k\) runs from 1 to 16. + +You can also just divide and take the whole-number part: + +\[\left\lfloor \frac{100}{6} \right\rfloor = 16\] + +Check the endpoints: \(6 \times 16 = 96 \le 100\) and \(6 \times 17 = 102 > 100\). So there are +**16** multiples of 6 in the range. + +## Hints + +- Every multiple of 6 looks like \(6 \times (\text{an integer})\). What is the biggest such number that stays at or below 100? +- Divide 100 by 6 and keep only the whole-number part. diff --git a/content/questions/quant/quant-number-properties-factors-0002.md b/content/questions/quant/quant-number-properties-factors-0002.md new file mode 100644 index 0000000..a1e1b66 --- /dev/null +++ b/content/questions/quant/quant-number-properties-factors-0002.md @@ -0,0 +1,45 @@ +--- +id: quant-number-properties-factors-0002 +section: quant +topic: number-properties +subtopic: factors-multiples +type: problem-solving +difficulty: medium +tags: [gcd, lcm, product-identity] +choices: + A: "6" + B: "18" + C: "36" + D: "72" + E: "216" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +The product of two positive integers is 216, and their greatest common divisor is 6. What is +their least common multiple? + +## Explanation + +For any two positive integers, the greatest common divisor and least common multiple are tied to +their product by the identity + +\[\gcd(a,b) \times \operatorname{lcm}(a,b) = a \times b\] + +Substituting the given values: + +\[6 \times \operatorname{lcm}(a,b) = 216 \implies \operatorname{lcm}(a,b) = \frac{216}{6} = 36\] + +So the least common multiple is **36**. (The two numbers are 12 and 18, whose product is 216, +GCD is 6, and LCM is 36 — consistent.) + +## Hints + +- There is a direct relationship between the GCD, the LCM, and the product of two numbers. +- \(\gcd \times \operatorname{lcm} = a \times b\). Solve for the one unknown. diff --git a/content/questions/quant/quant-number-properties-factors-0004.md b/content/questions/quant/quant-number-properties-factors-0004.md new file mode 100644 index 0000000..8802a71 --- /dev/null +++ b/content/questions/quant/quant-number-properties-factors-0004.md @@ -0,0 +1,51 @@ +--- +id: quant-number-properties-factors-0004 +section: quant +topic: number-properties +subtopic: factors-multiples +type: problem-solving +difficulty: hard +tags: [prime-factorization, perfect-squares, divisor-count] +choices: + A: "6" + B: "8" + C: "9" + D: "12" + E: "18" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(n = 2^4 \times 3^3 \times 5^2\), how many of the positive divisors of \(n\) are perfect +squares? + +## Explanation + +A divisor of \(n\) is formed by choosing an exponent for each prime, from \(0\) up to the exponent +in \(n\): + +\[d = 2^a \times 3^b \times 5^c, \quad 0 \le a \le 4,\; 0 \le b \le 3,\; 0 \le c \le 2\] + +That divisor is a **perfect square** exactly when every exponent is **even** (a perfect square has +even powers of all its primes). Count the even choices for each: + +- \(a \in \{0, 2, 4\}\) — **3** options +- \(b \in \{0, 2\}\) — **2** options +- \(c \in \{0, 2\}\) — **2** options + +Since the choices are independent, multiply: + +\[3 \times 2 \times 2 = 12\] + +So **12** of the divisors of \(n\) are perfect squares. + +## Hints + +- A perfect square has an even exponent on every prime in its factorization. +- For each prime, count only the even exponents available (including 0), then multiply the counts. diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0001.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0001.md new file mode 100644 index 0000000..ec8a7b9 --- /dev/null +++ b/content/questions/quant/quant-number-properties-odds-evens-signs-0001.md @@ -0,0 +1,44 @@ +--- +id: quant-number-properties-odds-evens-signs-0001 +section: quant +topic: number-properties +subtopic: odds-evens-signs +type: problem-solving +difficulty: easy +tags: [parity, odd-even, must-be] +choices: + A: "x + y" + B: "xy" + C: "x - y" + D: "x^2 - y" + E: "3x + y" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(x\) is an odd integer and \(y\) is an even integer, which of the following must be even? + +## Explanation + +Use the parity rules: odd + even = odd, odd − even = odd, odd × even = even, and odd × odd = odd. +Test each choice with \(x\) odd and \(y\) even. + +- **A.** \(x + y\) = odd + even = **odd**. +- **B.** \(xy\) = odd × even = **even**. ✓ +- **C.** \(x - y\) = odd − even = **odd**. +- **D.** \(x^2 - y\): \(x^2\) is odd × odd = odd, and odd − even = **odd**. +- **E.** \(3x + y\): \(3x\) is odd × odd = odd, and odd + even = **odd**. + +Only \(xy\) is guaranteed even (any product that includes an even factor is even). The answer is +**B**. + +## Hints + +- A product is even the moment any one of its factors is even. +- Odd ± even is always odd, and an odd number squared stays odd. diff --git a/content/questions/quant/quant-number-properties-remainders-0001.md b/content/questions/quant/quant-number-properties-remainders-0001.md new file mode 100644 index 0000000..be105b0 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0001.md @@ -0,0 +1,49 @@ +--- +id: quant-number-properties-remainders-0001 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: medium +tags: [remainders, division] +choices: + A: "1" + B: "2" + C: "3" + D: "4" + E: "5" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +When a positive integer \(n\) is divided by 7, the remainder is 4. What is the remainder when +\(3n + 5\) is divided by 7? + +## Explanation + +Write \(n\) in terms of its division by 7: + +\[n = 7k + 4\] + +for some non-negative integer \(k\). Then + +\[3n + 5 = 3(7k + 4) + 5 = 21k + 12 + 5 = 21k + 17\] + +The \(21k\) term is a multiple of 7, so it contributes remainder 0. That leaves the remainder of +17 upon division by 7: + +\[17 = 7 \times 2 + 3\] + +So the remainder is **3**. (Sanity check with \(n = 4\): \(3(4) + 5 = 17\), and \(17 \div 7\) +leaves 3.) + +## Hints + +- Replace \(n\) with \(7k + 4\) and expand — any multiple of 7 drops out of the remainder. +- Reduce the leftover constant modulo 7. diff --git a/content/questions/quant/quant-number-properties-remainders-0002.md b/content/questions/quant/quant-number-properties-remainders-0002.md new file mode 100644 index 0000000..e3d33d0 --- /dev/null +++ b/content/questions/quant/quant-number-properties-remainders-0002.md @@ -0,0 +1,47 @@ +--- +id: quant-number-properties-remainders-0002 +section: quant +topic: number-properties +subtopic: remainders +type: problem-solving +difficulty: hard +tags: [remainders, cyclicity, powers] +choices: + A: "0" + B: "1" + C: "2" + D: "3" + E: "4" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the remainder when \(7^{100}\) is divided by 5? + +## Explanation + +Only the remainder of the base matters, so reduce it first: \(7 \equiv 2 \pmod 5\). The problem +becomes the remainder of \(2^{100}\) upon division by 5. + +Look at the pattern of remainders of successive powers of 2 mod 5 — they **cycle**: + +\[2^1 \equiv 2,\quad 2^2 \equiv 4,\quad 2^3 \equiv 3,\quad 2^4 \equiv 1,\quad 2^5 \equiv 2,\ \dots\] + +The cycle has length 4: \((2, 4, 3, 1)\), repeating. The exponent that gives remainder 1 is any +multiple of 4. Since \(100 = 4 \times 25\) is a multiple of 4, it lands on the end of a full +cycle: + +\[2^{100} = (2^4)^{25} \equiv 1^{25} = 1 \pmod 5\] + +So the remainder is **1**. + +## Hints + +- First reduce the base modulo 5: \(7 \equiv 2\), so you only need powers of 2. +- Powers of 2 mod 5 repeat every 4 steps. Where does exponent 100 fall in that cycle? diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..a02383b 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": 6 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 32, - "lessons": 8 + "questions": 38, + "lessons": 9 }, "questions": [ { @@ -389,6 +389,65 @@ "Multiply P(first red) by P(second red given the first was red)." ] }, + { + "id": "quant-number-properties-factors-0001", + "section": "quant", + "topic": "number-properties", + "subtopic": "factors-multiples", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "multiples", + "counting" + ], + "choices": { + "A": "15", + "B": "16", + "C": "17", + "D": "18", + "E": "20" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "How many multiples of 6 are there between 1 and 100, inclusive?", + "explanation": "A multiple of 6 has the form \\(6k\\) for a positive integer \\(k\\). The largest one that does not\nexceed 100 satisfies \\(6k \\le 100\\), so \\(k \\le 16.67\\), meaning \\(k\\) runs from 1 to 16.\n\nYou can also just divide and take the whole-number part:\n\n\\[\\left\\lfloor \\frac{100}{6} \\right\\rfloor = 16\\]\n\nCheck the endpoints: \\(6 \\times 16 = 96 \\le 100\\) and \\(6 \\times 17 = 102 > 100\\). So there are\n**16** multiples of 6 in the range.", + "hints": [ + "Every multiple of 6 looks like \\(6 \\times (\\text{an integer})\\). What is the biggest such number that stays at or below 100?", + "Divide 100 by 6 and keep only the whole-number part." + ] + }, + { + "id": "quant-number-properties-factors-0002", + "section": "quant", + "topic": "number-properties", + "subtopic": "factors-multiples", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "gcd", + "lcm", + "product-identity" + ], + "choices": { + "A": "6", + "B": "18", + "C": "36", + "D": "72", + "E": "216" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "The product of two positive integers is 216, and their greatest common divisor is 6. What is\ntheir least common multiple?", + "explanation": "For any two positive integers, the greatest common divisor and least common multiple are tied to\ntheir product by the identity\n\n\\[\\gcd(a,b) \\times \\operatorname{lcm}(a,b) = a \\times b\\]\n\nSubstituting the given values:\n\n\\[6 \\times \\operatorname{lcm}(a,b) = 216 \\implies \\operatorname{lcm}(a,b) = \\frac{216}{6} = 36\\]\n\nSo the least common multiple is **36**. (The two numbers are 12 and 18, whose product is 216,\nGCD is 6, and LCM is 36 — consistent.)", + "hints": [ + "There is a direct relationship between the GCD, the LCM, and the product of two numbers.", + "\\(\\gcd \\times \\operatorname{lcm} = a \\times b\\). Solve for the one unknown." + ] + }, { "id": "quant-number-properties-factors-0003", "section": "quant", @@ -418,6 +477,125 @@ "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-0004", + "section": "quant", + "topic": "number-properties", + "subtopic": "factors-multiples", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "prime-factorization", + "perfect-squares", + "divisor-count" + ], + "choices": { + "A": "6", + "B": "8", + "C": "9", + "D": "12", + "E": "18" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(n = 2^4 \\times 3^3 \\times 5^2\\), how many of the positive divisors of \\(n\\) are perfect\nsquares?", + "explanation": "A divisor of \\(n\\) is formed by choosing an exponent for each prime, from \\(0\\) up to the exponent\nin \\(n\\):\n\n\\[d = 2^a \\times 3^b \\times 5^c, \\quad 0 \\le a \\le 4,\\; 0 \\le b \\le 3,\\; 0 \\le c \\le 2\\]\n\nThat divisor is a **perfect square** exactly when every exponent is **even** (a perfect square has\neven powers of all its primes). Count the even choices for each:\n\n- \\(a \\in \\{0, 2, 4\\}\\) — **3** options\n- \\(b \\in \\{0, 2\\}\\) — **2** options\n- \\(c \\in \\{0, 2\\}\\) — **2** options\n\nSince the choices are independent, multiply:\n\n\\[3 \\times 2 \\times 2 = 12\\]\n\nSo **12** of the divisors of \\(n\\) are perfect squares.", + "hints": [ + "A perfect square has an even exponent on every prime in its factorization.", + "For each prime, count only the even exponents available (including 0), then multiply the counts." + ] + }, + { + "id": "quant-number-properties-odds-evens-signs-0001", + "section": "quant", + "topic": "number-properties", + "subtopic": "odds-evens-signs", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "parity", + "odd-even", + "must-be" + ], + "choices": { + "A": "x + y", + "B": "xy", + "C": "x - y", + "D": "x^2 - y", + "E": "3x + y" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(x\\) is an odd integer and \\(y\\) is an even integer, which of the following must be even?", + "explanation": "Use the parity rules: odd + even = odd, odd − even = odd, odd × even = even, and odd × odd = odd.\nTest each choice with \\(x\\) odd and \\(y\\) even.\n\n- **A.** \\(x + y\\) = odd + even = **odd**.\n- **B.** \\(xy\\) = odd × even = **even**. ✓\n- **C.** \\(x - y\\) = odd − even = **odd**.\n- **D.** \\(x^2 - y\\): \\(x^2\\) is odd × odd = odd, and odd − even = **odd**.\n- **E.** \\(3x + y\\): \\(3x\\) is odd × odd = odd, and odd + even = **odd**.\n\nOnly \\(xy\\) is guaranteed even (any product that includes an even factor is even). The answer is\n**B**.", + "hints": [ + "A product is even the moment any one of its factors is even.", + "Odd ± even is always odd, and an odd number squared stays odd." + ] + }, + { + "id": "quant-number-properties-remainders-0001", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "remainders", + "division" + ], + "choices": { + "A": "1", + "B": "2", + "C": "3", + "D": "4", + "E": "5" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "When a positive integer \\(n\\) is divided by 7, the remainder is 4. What is the remainder when\n\\(3n + 5\\) is divided by 7?", + "explanation": "Write \\(n\\) in terms of its division by 7:\n\n\\[n = 7k + 4\\]\n\nfor some non-negative integer \\(k\\). Then\n\n\\[3n + 5 = 3(7k + 4) + 5 = 21k + 12 + 5 = 21k + 17\\]\n\nThe \\(21k\\) term is a multiple of 7, so it contributes remainder 0. That leaves the remainder of\n17 upon division by 7:\n\n\\[17 = 7 \\times 2 + 3\\]\n\nSo the remainder is **3**. (Sanity check with \\(n = 4\\): \\(3(4) + 5 = 17\\), and \\(17 \\div 7\\)\nleaves 3.)", + "hints": [ + "Replace \\(n\\) with \\(7k + 4\\) and expand — any multiple of 7 drops out of the remainder.", + "Reduce the leftover constant modulo 7." + ] + }, + { + "id": "quant-number-properties-remainders-0002", + "section": "quant", + "topic": "number-properties", + "subtopic": "remainders", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "remainders", + "cyclicity", + "powers" + ], + "choices": { + "A": "0", + "B": "1", + "C": "2", + "D": "3", + "E": "4" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the remainder when \\(7^{100}\\) is divided by 5?", + "explanation": "Only the remainder of the base matters, so reduce it first: \\(7 \\equiv 2 \\pmod 5\\). The problem\nbecomes the remainder of \\(2^{100}\\) upon division by 5.\n\nLook at the pattern of remainders of successive powers of 2 mod 5 — they **cycle**:\n\n\\[2^1 \\equiv 2,\\quad 2^2 \\equiv 4,\\quad 2^3 \\equiv 3,\\quad 2^4 \\equiv 1,\\quad 2^5 \\equiv 2,\\ \\dots\\]\n\nThe cycle has length 4: \\((2, 4, 3, 1)\\), repeating. The exponent that gives remainder 1 is any\nmultiple of 4. Since \\(100 = 4 \\times 25\\) is a multiple of 4, it lands on the end of a full\ncycle:\n\n\\[2^{100} = (2^4)^{25} \\equiv 1^{25} = 1 \\pmod 5\\]\n\nSo the remainder is **1**.", + "hints": [ + "First reduce the base modulo 5: \\(7 \\equiv 2\\), so you only need powers of 2.", + "Powers of 2 mod 5 repeat every 4 steps. Where does exponent 100 fall in that cycle?" + ] + }, { "id": "quant-statistics-descriptive-0009", "section": "quant", @@ -984,6 +1162,25 @@ "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 & Prime Factorization", + "tags": [ + "number-properties", + "factors", + "multiples", + "primes", + "gcd", + "lcm" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nAlmost every \"clever\" number-properties problem on the GMAT Focus Quant section becomes routine\nonce you write the number as a product of primes. **Prime factorization** is the master key:\ncounting divisors, finding a GCD or LCM, testing divisibility, and spotting perfect squares all\nfall out of it directly. This lesson builds that one skill and shows the handful of formulas it\nunlocks.\n\n## Core concepts\n\n**Factors vs. multiples.** A **factor** (divisor) of \\(N\\) divides \\(N\\) with no remainder; a\n**multiple** of \\(N\\) is \\(N\\) times an integer. So the factors of 12 are \\(\\{1,2,3,4,6,12\\}\\),\nwhile its multiples are \\(12, 24, 36, \\dots\\) — a finite list versus an infinite one.\n\n**Prime factorization.** Every integer greater than 1 factors uniquely into primes:\n\n\\[72 = 2^3 \\times 3^2, \\qquad 360 = 2^3 \\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\n\\[(a+1)(b+1)\\dots\\]\n\nbecause each divisor picks an exponent for \\(p\\) from \\(0\\) to \\(a\\), independently for \\(q\\), and\nso on. For \\(72 = 2^3 \\times 3^2\\): \\((3+1)(2+1) = 12\\) divisors.\n\n**GCD and LCM from primes.** Line up the shared primes:\n\n- **GCD** — take the **lowest** power of each prime the numbers share.\n- **LCM** — take the **highest** power of every prime that appears.\n\nFor \\(12 = 2^2 \\times 3\\) and \\(18 = 2 \\times 3^2\\): \\(\\gcd = 2^1 \\times 3^1 = 6\\) and\n\\(\\operatorname{lcm} = 2^2 \\times 3^2 = 36\\).\n\n**The product identity.** For any two positive integers,\n\n\\[\\gcd(a,b) \\times \\operatorname{lcm}(a,b) = a \\times b\\]\n\nThis turns many two-number problems into one line of arithmetic.\n\n## Worked examples\n\n**Example 1 — counting perfect-square divisors.** How many divisors of \\(N = 2^4 \\times 3^2\\)\nare perfect squares? A divisor is a perfect square exactly when every prime exponent is **even**.\nFor \\(2^4\\) the choices are \\(0, 2, 4\\) (three options); for \\(3^2\\) they are \\(0, 2\\) (two\noptions). That gives \\(3 \\times 2 = 6\\) perfect-square divisors.\n\n**Example 2 — the product identity.** Two positive integers have product 216 and GCD 6. Their\nLCM is \\(216 / 6 = 36\\). No listing required.\n\n## Common traps\n\n- **1 is not prime.** It has exactly one divisor. The smallest prime is 2 (the only even prime).\n- **Forgetting the \"+1\" in the divisor count.** The exponent can be \\(0\\) (the prime used zero\n times), so a prime raised to \\(a\\) contributes \\(a+1\\) choices, not \\(a\\).\n- **Swapping GCD and LCM.** GCD uses the *lowest* shared powers and is at most the smaller number;\n LCM uses the *highest* powers and is at least the larger number.\n- **The product identity is for two numbers only.** \\(\\gcd \\times \\operatorname{lcm} = a \\times b\\)\n does not extend to three or more integers.\n\n## Key takeaways\n\n- Factor into primes first — divisor counts, GCD, LCM, and divisibility all follow mechanically.\n- Divisor count of \\(p^a q^b \\dots\\) is \\((a+1)(b+1)\\dots\\); perfect-square divisors need all\n exponents even.\n- GCD = lowest shared prime powers; LCM = highest prime powers overall.\n- For two numbers, \\(\\gcd \\times \\operatorname{lcm} = \\) their product." + }, { "id": "quant-statistics-descriptive", "section": "quant",