Add Number Properties lesson + 6 practice questions

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.

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.

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.

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.

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.

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.

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?