Seed content: Number Properties batch (+5 questions, +1 lesson)

content/lessons/quant/quant-number-properties-remainders.md

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+---
+id: quant-number-properties-remainders
+section: quant
+topic: number-properties
+subtopic: remainders
+title: "Remainders & Modular Thinking"
+tags: [remainders, divisibility, modular-arithmetic]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+A remainder question is really a question about *structure*: every integer can be written as
+"a multiple of the divisor, plus a leftover." Once you write that down, most remainder problems
+turn into ordinary algebra. Focus Quant loves these because the slow way (plugging in numbers
+until something works) is beatable by a one-line setup.
+
+## Core concepts
+
+**The division identity.** When a positive integer \(n\) leaves remainder \(r\) on division by
+\(d\),
+
+\[n = dq + r, \qquad 0 \le r < d\]
+
+where \(q\) is the quotient. The remainder is always **non-negative and strictly less than the
+divisor** — a remainder of \(7\) on division by \(5\) is impossible; reduce it.
+
+**Remainders add and multiply.** You can do arithmetic on remainders directly and only reduce at
+the end. If \(n\) leaves remainder \(r\), then:
+
+- \(kn\) leaves the remainder of \(kr\) (then reduce mod \(d\)).
+- \(n + m\) leaves the remainder of \(r_n + r_m\) (then reduce).
+
+Example: \(n\) leaves remainder \(4\) on division by \(7\). Then \(3n\) leaves the remainder of
+\(3 \times 4 = 12\), and \(12 = 7 + 5\), so the remainder is \(5\) — no need to know \(n\) itself.
+
+**Two conditions at once.** "Remainder \(3\) mod \(5\) **and** remainder \(4\) mod \(7\)" pins the
+number down modulo \(5 \times 7 = 35\). List one condition (\(4, 11, 18, 25, 32, \dots\) for
+"\(\equiv 4 \bmod 7\)") and scan for the first that also satisfies the other. Here \(18\) works, so
+every such number is \(35k + 18\).
+
+**Remainder 0 means "divisible."** "Leaves no remainder" and "is a factor of" are the same
+statement — translate freely between them.
+
+## Worked examples
+
+**Multiplying a remainder.** \(n\) divided by \(6\) leaves remainder \(5\). Remainder of \(4n\)?
+\(4 \times 5 = 20 = 3\times 6 + 2\), so the remainder is \(2\).
+
+**Combining two moduli.** \(n \equiv 2 \pmod 4\) and \(n \equiv 3 \pmod 5\). Numbers that are
+\(3 \bmod 5\): \(3, 8, 13, 18, 23, \dots\); the first that is also \(2 \bmod 4\) is \(18\)
+(\(18 = 4\times4 + 2\)). So \(n = 20k + 18\).
+
+## Common traps
+
+- **Leaving an oversized remainder.** After multiplying or adding, always reduce so the remainder
+  is less than the divisor. \(12 \bmod 7\) is \(5\), not \(12\).
+- **Assuming a unique value.** Conditions like "remainder 3 mod 5" describe an infinite family
+  \(5k+3\); test the smallest valid case and, if the answer could change, a second case.
+- **Ignoring 0 as a possibility.** The smallest non-negative value satisfying "remainder \(r\)" can
+  be \(r\) itself (when the quotient is 0). Don't assume \(n > d\).
+
+## Key takeaways
+
+- Write \(n = dq + r\) with \(0 \le r < d\) — that single line unlocks most problems.
+- You can multiply or add remainders and reduce at the end; you rarely need the actual number.
+- Two modular conditions pin a value down modulo the product of the divisors — list and scan.
+- "Remainder 0," "divisible by," and "is a factor of" all say the same thing.

content/questions/quant/quant-number-properties-factors-0020.md

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+---
+id: quant-number-properties-factors-0020
+section: quant
+topic: number-properties
+subtopic: factors-multiples
+type: problem-solving
+difficulty: medium
+tags: [lcm, divisors, prime-factorization]
+choices:
+  A: "6"
+  B: "8"
+  C: "9"
+  D: "12"
+  E: "36"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Let \(N\) be the smallest positive integer that is divisible by both 12 and 18. How many positive
+divisors does \(N\) have?
+
+## Explanation
+
+The smallest positive integer divisible by both is the **least common multiple**. Use prime
+factorizations:
+
+\[12 = 2^2 \times 3, \qquad 18 = 2 \times 3^2.\]
+
+The LCM takes the **highest power of each prime** that appears:
+
+\[N = \operatorname{lcm}(12, 18) = 2^2 \times 3^2 = 36.\]
+
+Now count divisors. For \(N = p^a \times q^b\), the number of positive divisors is
+\((a+1)(b+1)\), because each divisor uses \(p\) to a power from \(0\) to \(a\) and \(q\) to a power
+from \(0\) to \(b\):
+
+\[(2+1)(2+1) = 3 \times 3 = 9.\]
+
+So \(N = 36\) has **9** divisors: 1, 2, 3, 4, 6, 9, 12, 18, 36.
+
+Watch two traps: taking the *lowest* powers gives the GCD (6), not the LCM; and answering **36** by
+reporting \(N\) itself instead of its divisor count.
+
+## Hints
+
+- "Smallest number divisible by both" is the LCM — take the highest power of each prime.
+- For \(N = p^a q^b\), the divisor count is \((a+1)(b+1)\), not \(N\) itself.

content/questions/quant/quant-number-properties-factors-0021.md

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+---
+id: quant-number-properties-factors-0021
+section: quant
+topic: number-properties
+subtopic: factors-multiples
+type: problem-solving
+difficulty: hard
+tags: [divisors, multiples, prime-factorization]
+choices:
+  A: "6"
+  B: "8"
+  C: "10"
+  D: "12"
+  E: "15"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Let \(n = 2^4 \times 3^2 \times 5\). How many of the positive divisors of \(n\) are multiples of
+10?
+
+## Explanation
+
+Every divisor of \(n\) has the form
+
+\[2^a \times 3^b \times 5^c, \qquad 0 \le a \le 4,\ 0 \le b \le 2,\ 0 \le c \le 1.\]
+
+A number is a **multiple of 10** exactly when it is divisible by both 2 and 5 — that is, when
+\(a \ge 1\) **and** \(c \ge 1\). Count the allowed choices for each exponent under those constraints:
+
+- \(a \in \{1, 2, 3, 4\}\): **4** choices (excluding \(a = 0\)).
+- \(b \in \{0, 1, 2\}\): **3** choices (no restriction — the factor of 3 doesn't affect being a multiple of 10).
+- \(c = 1\): **1** choice (must include the single 5).
+
+Multiplying the independent choices:
+
+\[4 \times 3 \times 1 = 12.\]
+
+So **12** of the divisors are multiples of 10.
+
+Common trap: requiring \(c \ge 1\) but forgetting to also force \(a \ge 1\) (giving \(5 \times 3 = 15\)),
+or dropping the free factor of 3 entirely (giving \(4 \times 1 = 4\)).
+
+## Hints
+
+- A divisor is a multiple of 10 exactly when it contains at least one 2 **and** at least one 5.
+- Count exponent choices separately: force \(a \ge 1\) and \(c \ge 1\), leave \(b\) unrestricted, then multiply.

content/questions/quant/quant-number-properties-odds-evens-signs-0019.md

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+---
+id: quant-number-properties-odds-evens-signs-0019
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: medium
+tags: [odd-even, parity]
+choices:
+  A: "a + b is odd"
+  B: "a - b is even"
+  C: "a·b + 1 is odd"
+  D: "a is even"
+  E: "2a + b is even"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(a\) and \(b\) are integers and the product \(ab\) is odd, which of the following **must** be
+true?
+
+- **A.** \(a + b\) is odd
+- **B.** \(a - b\) is even
+- **C.** \(ab + 1\) is odd
+- **D.** \(a\) is even
+- **E.** \(2a + b\) is even
+
+## Explanation
+
+A product of integers is odd only when **every factor is odd**. So \(ab\) odd forces both \(a\) and
+\(b\) to be odd. Test each choice under "both odd":
+
+- **A.** odd \(+\) odd \(=\) even, so \(a + b\) is even, not odd. False.
+- **B.** odd \(-\) odd \(=\) even. **Always true.** ✓
+- **C.** \(ab\) is odd, so \(ab + 1\) is even, not odd. False.
+- **D.** \(a\) is odd, not even. False.
+- **E.** \(2a\) is even and \(b\) is odd, so \(2a + b\) is even \(+\) odd \(=\) odd. False.
+
+Only **B** must hold. The key parity facts: odd \(\pm\) odd is even, and a product is odd exactly
+when all factors are odd.
+
+The tempting wrong answer is **A** — many students recall "odd + odd" but misremember it as odd.
+
+## Hints
+
+- For a product to be odd, what has to be true of *each* factor?
+- Once you know both are odd, use odd \(\pm\) odd \(=\) even.

content/questions/quant/quant-number-properties-remainders-0017.md

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+---
+id: quant-number-properties-remainders-0017
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: easy
+tags: [remainders, divisibility]
+choices:
+  A: "2"
+  B: "3"
+  C: "4"
+  D: "5"
+  E: "6"
+answer: D
+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
+\(3n\) is divided by 7?
+
+## Explanation
+
+Write \(n\) with the division identity: \(n = 7q + 4\) for some integer \(q\). Then
+
+\[3n = 3(7q + 4) = 21q + 12.\]
+
+The \(21q\) part is a multiple of 7 and contributes no remainder, so the remainder of \(3n\) is
+just the remainder of \(12\) on division by 7:
+
+\[12 = 7 + 5,\]
+
+which leaves **5**.
+
+The shortcut: you can multiply the remainder directly. \(3 \times 4 = 12\), then reduce mod 7 to
+get 5. You never need the actual value of \(n\).
+
+The trap answer is **4** — reusing the original remainder as if multiplying by 3 didn't change it.
+
+## Hints
+
+- Write \(n = 7q + 4\) and multiply through by 3.
+- Multiply the remainder (\(3 \times 4 = 12\)), then subtract 7 until it drops below 7.

content/questions/quant/quant-number-properties-remainders-0018.md

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+---
+id: quant-number-properties-remainders-0018
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: hard
+tags: [remainders, modular-arithmetic]
+choices:
+  A: "11"
+  B: "14"
+  C: "18"
+  D: "23"
+  E: "32"
+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 4 when
+divided by 7. What is the remainder when \(n\) is divided by 35?
+
+## Explanation
+
+Two modular conditions together pin \(n\) down modulo \(5 \times 7 = 35\), so there is a single
+remainder to find.
+
+List the numbers that leave remainder 4 on division by 7:
+
+\[4,\ 11,\ 18,\ 25,\ 32,\ \dots\]
+
+Now check which of these also leaves remainder 3 on division by 5 (i.e. ends in 3 or 8):
+
+- \(4 \to\) remainder 4. No.
+- \(11 \to\) remainder 1. No.
+- \(18 \to 18 = 3\times 5 + 3\), remainder 3. **Yes.**
+
+So \(n = 18\) works, and every valid \(n\) has the form \(35k + 18\). Dividing any of them by 35
+leaves remainder **18**.
+
+Check: \(18 \div 5 = 3\) remainder 3 ✓, and \(18 \div 7 = 2\) remainder 4 ✓.
+
+The distractors 11, 23, and 32 satisfy exactly *one* of the two conditions, which is why scanning
+both is essential.
+
+## Hints
+
+- Each condition alone allows infinitely many numbers; together they fix the remainder mod \(5\times7 = 35\).
+- List multiples-plus-4 of 7 and stop at the first one that also leaves remainder 3 mod 5.