Seed Number Properties content: +6 questions, +3 lessons

content/lessons/quant/quant-number-properties-factors-multiples.md

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+---
+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."

content/lessons/quant/quant-number-properties-odds-evens-signs.md

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+---
+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.

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, 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.

content/questions/quant/quant-number-properties-factors-multiples-0017.md

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+---
+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.
+- \(\text{GCF} \times \text{LCM} = \text{the product of the two numbers}\).

content/questions/quant/quant-number-properties-factors-multiples-0018.md

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+---
+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.

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: [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?

content/questions/quant/quant-number-properties-odds-evens-signs-0020.md

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+---
+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?