Day 3: Seed Number Properties topic (+3 lessons, +6 questions)

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, Primes & Divisibility"
+tags: [factors, multiples, primes, gcd, lcm, divisibility]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Almost every Number Properties question rests on one skill: **breaking an integer into its
+prime factors**. Once you have the prime factorization, divisor counts, greatest common divisors,
+least common multiples, and divisibility all follow mechanically. Get comfortable factoring fast
+and this whole topic gets easier.
+
+## Core concepts
+
+**Prime factorization.** Every integer greater than 1 is a unique product of primes. Split the
+number into any two factors and keep splitting until everything is prime.
+
+\[360 = 36 \times 10 = (2^2 \cdot 3^2)(2 \cdot 5) = 2^3 \cdot 3^2 \cdot 5\]
+
+**Counting divisors.** If \(N = p^a \cdot q^b \cdot r^c \dots\), the number of positive divisors is
+
+\[(a+1)(b+1)(c+1)\dots\]
+
+The "+1" is there because each divisor can use prime \(p\) to a power from \(0\) up to \(a\).
+For \(360 = 2^3 \cdot 3^2 \cdot 5^1\): \((3+1)(2+1)(1+1) = 24\) divisors.
+
+**GCD (greatest common divisor).** Take each prime the two numbers **share**, using the **lower**
+exponent. **LCM (least common multiple).** Take **every** prime that appears in either number,
+using the **higher** exponent.
+
+\[12 = 2^2 \cdot 3, \quad 18 = 2 \cdot 3^2 \;\Rightarrow\; \gcd = 2^1 \cdot 3^1 = 6, \quad \operatorname{lcm} = 2^2 \cdot 3^2 = 36\]
+
+A useful identity for any two positive integers: \(\gcd(a,b) \times \operatorname{lcm}(a,b) = a \times b\).
+
+**Perfect squares** have only **even** exponents in their prime factorization (e.g.
+\(900 = 2^2 \cdot 3^2 \cdot 5^2\)). To turn a number into a perfect square, supply whatever primes
+are needed to make every exponent even.
+
+**Quick divisibility rules.** Divisible by 3 if the digit sum is divisible by 3; by 9 if the digit
+sum is divisible by 9; by 4 if the last two digits form a multiple of 4; by 6 if it passes the 2
+**and** 3 tests.
+
+## Worked examples
+
+**LCM in disguise.** Two signals flash every 12 and 18 seconds and just flashed together. When do
+they next coincide? That is \(\operatorname{lcm}(12,18) = 36\) seconds — *not* their product.
+
+**Smallest multiplier for a perfect square.** What is the smallest \(n\) making \(90n\) a perfect
+square? \(90 = 2 \cdot 3^2 \cdot 5\). The 2 and the 5 have odd exponents, so \(n = 2 \cdot 5 = 10\)
+and \(90 \cdot 10 = 900 = 30^2\).
+
+## Common traps
+
+- **Confusing GCD with LCM.** GCD is the *small* shared part (lower exponents); LCM is the *big*
+  covering number (higher exponents). "How often do things line up again?" is almost always LCM.
+- **Multiplying instead of taking the LCM.** \(12 \times 18 = 216\) is a common multiple, but not
+  the *least* one.
+- **Forgetting 1 and the number itself** when counting divisors — the \((a+1)(b+1)\) formula
+  already includes them.
+
+## Key takeaways
+
+- Factor into primes first; divisor count, GCD, and LCM all read straight off the exponents.
+- Divisor count \(= (a+1)(b+1)\dots\); GCD uses shared primes at the lower power; LCM uses all
+  primes at the higher power.
+- \(\gcd \times \operatorname{lcm} = a \times b\), and perfect squares have all-even exponents.

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
+
+Many "which of the following **must** be true?" questions never require you to find a value — only
+to track **parity** (odd/even) and **sign** (positive/negative). Memorize a handful of rules and
+these become fast, reliable points.
+
+## Core concepts
+
+**Parity of sums.** odd \(+\) odd \(=\) even; even \(+\) even \(=\) even; odd \(+\) even \(=\) odd.
+A sum is **odd exactly when an odd number of its terms are odd.**
+
+**Parity of products.** A product is **odd only if every factor is odd.** If even one factor is
+even, the product is even. So \(xy\) odd \(\Rightarrow\) both \(x\) and \(y\) are odd.
+
+**Consecutive integers.** Among any two consecutive integers, exactly one is even, so their product
+is always even. Among any three consecutive integers, the product is divisible by \(3! = 6\).
+
+**Sign rules for products.** A product is **negative when it has an odd number of negative
+factors**, and **positive when it has an even number** (zero counts as even). So \(xyz < 0\) means
+one or three of the factors are negative.
+
+**A "must be true" strategy.** When a question asks what *must* be true, try to build a
+counterexample. If you can find even one case where a choice fails, eliminate it. What survives
+every test is the answer.
+
+## Worked examples
+
+**Product forces parity.** If \(xy\) is odd, both \(x\) and \(y\) are odd, so \(x + y\) is even and
+\(x^2 + y^2 = \text{odd} + \text{odd}\) is even. Any choice claiming \(x+y\) is odd is out.
+
+**Chaining signs.** If \(xyz < 0\) and \(xy > 0\): \(xy > 0\) means \(x\) and \(y\) share a sign, so
+their product is positive; for the triple product to be negative, \(z\) must be **negative**. Note
+you still can't pin down the signs of \(x\) and \(y\) individually.
+
+## Common traps
+
+- **Assuming "integer" means positive.** Integers include 0 and negatives. Test a negative and 0
+  against every "must be true" choice.
+- **Forgetting 0.** Zero is even, is neither positive nor negative, and makes any product it
+  touches equal 0.
+- **Reading "could be true" as "must be true."** One counterexample kills a *must*-be-true choice,
+  even if it's true most of the time.
+
+## Key takeaways
+
+- Product is odd only if **all** factors are odd; a sum is odd only with an **odd count** of odd terms.
+- Product sign flips with each negative factor: odd number of negatives → negative.
+- For "must be true," hunt for a counterexample — and always test negatives and 0.

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, divisibility, cycles]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+A remainder question is really a statement about form. "\(n\) leaves remainder 4 when divided by 7"
+just means \(n = 7k + 4\) for some integer \(k\). Write that form down and most remainder problems
+turn into ordinary algebra or a short list.
+
+## Core concepts
+
+**The division form.** "\(n\) divided by \(d\) leaves remainder \(r\)" means
+
+\[n = dq + r, \qquad 0 \le r < d\]
+
+The remainder is always **less than the divisor** — a remainder of 7 (mod 7) is impossible; it is 0.
+
+**Remainders add and multiply.** To find the remainder of a sum or product, you may reduce each
+piece to its remainder first. If \(n\) leaves remainder 4 mod 7, then \(3n\) leaves the remainder of
+\(3 \times 4 = 12\), which is \(12 - 7 = 5\).
+
+**Testing the smallest case.** When only the remainder is given, plug in the smallest value that
+fits (here \(n = 4\)) — the remainder of any sum/product won't depend on which valid \(n\) you pick.
+
+**Two conditions at once.** "\(n\) leaves remainder 3 mod 5 **and** remainder 2 mod 3": list each,
+
+\[n \equiv 3 \pmod 5: \; 3, 8, 13, 18, 23, \dots \qquad n \equiv 2 \pmod 3: \; 2, 5, 8, 11, 14, 17, 20, 23, \dots\]
+
+The first common value is 8, and solutions then repeat every \(\operatorname{lcm}(5,3) = 15\):
+\(8, 23, 38, \dots\)
+
+## Worked examples
+
+**Scaling a remainder.** \(n\) leaves remainder 4 mod 7. Remainder of \(3n\) mod 7? Reduce:
+\(3 \cdot 4 = 12 \equiv 5\). So the remainder is **5**.
+
+**Combining two moduli.** Smallest \(n > 10\) with remainder 3 mod 5 and remainder 2 mod 3? From the
+lists above the pattern is \(8, 23, 38, \dots\); the first one over 10 is **23**.
+
+## Common traps
+
+- **Remainder \(\ge\) divisor.** The remainder must be strictly smaller than what you divide by.
+  Always reduce (e.g. \(12 \bmod 7 = 5\)).
+- **Forgetting the small cases.** \(n\) could be the remainder itself (e.g. \(n = 4\)) — don't
+  assume \(n\) is large.
+- **Checking only one condition.** With two divisors, verify a candidate against **both** before
+  choosing it.
+
+## Key takeaways
+
+- Rewrite "remainder \(r\) mod \(d\)" as \(n = dq + r\) with \(0 \le r < d\).
+- Reduce sums and products to their remainders, then reduce again if the result exceeds the divisor.
+- Two simultaneous conditions repeat every LCM of the divisors — list a few terms and find the overlap.

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: [lcm, cycles, multiples]
+choices:
+  A: "6"
+  B: "36"
+  C: "54"
+  D: "72"
+  E: "216"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A red beacon flashes every 12 seconds and a green beacon flashes every 18 seconds. If both beacons
+flash at the same instant right now, how many seconds from now will they next flash at the same
+instant?
+
+## Explanation
+
+"Flash together again" means the first time that is a multiple of **both** 12 and 18 — the **least
+common multiple**.
+
+Factor each interval into primes:
+
+\[12 = 2^2 \cdot 3, \qquad 18 = 2 \cdot 3^2\]
+
+The LCM takes each prime at its **highest** power:
+
+\[\operatorname{lcm}(12, 18) = 2^2 \cdot 3^2 = 36\]
+
+So the beacons next coincide after **36** seconds.
+
+Watch the two classic traps: \(6\) is the *greatest common divisor* (lower powers), not the LCM,
+and \(12 \times 18 = 216\) is *a* common multiple but not the *least* one.
+
+## Hints
+
+- "Line up again" points to a common multiple — and you want the *least* one.
+- Factor both numbers into primes; the LCM uses the higher power of each prime, the GCD the lower.

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: hard
+tags: [prime-factorization, perfect-square]
+choices:
+  A: "2"
+  B: "5"
+  C: "10"
+  D: "15"
+  E: "90"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the smallest positive integer \(n\) such that \(90n\) is the square of an integer?
+
+## Explanation
+
+A perfect square has **only even exponents** in its prime factorization. Factor 90:
+
+\[90 = 2 \cdot 3^2 \cdot 5\]
+
+The exponent of 3 is already even (2), but the exponents of 2 and 5 are each **odd** (1). To make
+every exponent even, \(n\) must supply one more 2 and one more 5:
+
+\[n = 2 \cdot 5 = 10\]
+
+Check: \(90 \cdot 10 = 900 = 2^2 \cdot 3^2 \cdot 5^2 = 30^2\). ✓
+
+Smaller candidates fail because they fix only one offending prime: \(n = 2\) leaves the 5 odd, and
+\(n = 5\) leaves the 2 odd. The answer is **10**.
+
+## Hints
+
+- Factor 90 into primes and look at which exponents are odd.
+- A perfect square needs every prime exponent to be even — multiply in exactly the primes required
+  to fix the odd ones.

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, must-be-true]
+choices:
+  A: "\(x + y\) is odd"
+  B: "\(x + y\) is even"
+  C: "\(x\) is even"
+  D: "\(x - y\) is odd"
+  E: "\(x^2 + y^2\) is odd"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(x\) and \(y\) are integers and the product \(xy\) is odd, which of the following must be true?
+
+## Explanation
+
+A product is odd **only when every factor is odd**. So \(xy\) odd forces **both** \(x\) and \(y\) to
+be odd. Now test each choice with "odd, odd":
+
+- **A.** odd \(+\) odd \(=\) even, so \(x + y\) is *not* odd. Eliminate.
+- **B.** odd \(+\) odd \(=\) **even**. This must be true. ✓
+- **C.** \(x\) is odd, not even. Eliminate.
+- **D.** odd \(-\) odd \(=\) even, so \(x - y\) is not odd. Eliminate.
+- **E.** \(x^2\) and \(y^2\) are both odd, and odd \(+\) odd \(=\) even, so \(x^2 + y^2\) is even, not
+  odd. Eliminate.
+
+Only **B** holds. The answer is **\(x + y\) is even**.
+
+## Hints
+
+- What does an *odd product* tell you about each of the two factors?
+- Once you know the parity of \(x\) and \(y\), just apply odd/even sum rules to each choice.

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, must-be-true]
+choices:
+  A: "\(x < 0\)"
+  B: "\(y > 0\)"
+  C: "\(z < 0\)"
+  D: "\(xz > 0\)"
+  E: "\(yz > 0\)"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(x\), \(y\), and \(z\) are nonzero numbers with \(xyz < 0\) and \(xy > 0\), which of the
+following must be true?
+
+## Explanation
+
+Work with the sign of each product.
+
+Since \(xy > 0\), \(x\) and \(y\) have the **same sign**, so their product is positive. Then
+
+\[xyz = (xy) \cdot z < 0 \quad\text{with}\quad xy > 0 \;\Rightarrow\; z < 0.\]
+
+So \(z\) must be negative — choice **C**.
+
+The other choices aren't forced: \(x\) and \(y\) could **both** be positive or **both** be negative
+(either keeps \(xy > 0\)), so A and B fail. And because \(x\)'s sign is unknown while \(z < 0\), the
+product \(xz\) could be positive or negative, so D fails; the same reasoning kills E. The answer is
+**\(z < 0\)**.
+
+## Hints
+
+- Treat \(xy\) as a single quantity — its sign is given directly.
+- A product is negative when it has an odd number of negative factors; use that to pin down \(z\).