Add Quant number-properties: odds/evens/signs lesson + 5 questions

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 & Signs"
+tags: [parity, odd-even, signs, number-properties]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Many GMAT Focus Quant questions never ask you to compute a value — they ask what **must** be
+true about whether a number is odd or even, or positive or negative. These **parity** and **sign**
+questions look abstract, but they collapse to a handful of fixed rules. Once you know the rules you
+can answer in seconds, and when you don't, you can always fall back on **plugging in small numbers**.
+
+## Core concepts
+
+**Parity of sums and differences.** Addition and subtraction have the *same* parity behavior:
+
+\[\text{even} \pm \text{even} = \text{even}, \quad \text{odd} \pm \text{odd} = \text{even}, \quad \text{even} \pm \text{odd} = \text{odd}\]
+
+The takeaway: \(a + b\) and \(a - b\) always have the **same** parity. A sum (or difference) is odd
+**only** when exactly one of the two numbers is odd.
+
+**Parity of products.** A product is even if **any** factor is even; it is odd **only** when **every**
+factor is odd:
+
+\[\text{even} \times \text{anything} = \text{even}, \qquad \text{odd} \times \text{odd} = \text{odd}\]
+
+**Two consecutive integers.** \(k\) and \(k+1\) are always one odd and one even, so their product
+\(k(k+1)\) is **always even**. This is why expressions like \(k^2 + k = k(k+1)\) are guaranteed even.
+
+**Signs of products and quotients.** For multiplication and division, only the *count* of negative
+factors matters:
+
+- An **even** number of negative factors → **positive** result.
+- An **odd** number of negative factors → **negative** result.
+
+So a product of three numbers is negative exactly when one or three of them are negative — never
+when exactly two are.
+
+**Zero is even, and it has no sign.** \(0\) is an even integer. It is neither positive nor negative,
+and any product containing a \(0\) is \(0\) — a case that quietly breaks "must be positive/negative"
+claims, so watch for whether the problem says the numbers are *nonzero*.
+
+## Worked examples
+
+**Parity of an expression.** Is \(3p + 5q\) odd or even? The coefficients don't change parity:
+\(3p\) has the parity of \(p\), and \(5q\) has the parity of \(q\). So \(3p + 5q\) has the parity of
+\(p + q\) — it's even exactly when \(p\) and \(q\) share the same parity.
+
+**Sign of a quotient.** If \(a < 0\), \(b > 0\), \(c < 0\), then \(\dfrac{ab}{c}\) has two negative
+inputs (\(a\) and \(c\)) — an even count — so the result is **positive**.
+
+## Common traps
+
+- **Assuming subtraction behaves differently from addition.** \(x - y\) has the *same* parity as
+  \(x + y\). If \(x + y\) is odd, so is \(x - y\).
+- **Forgetting the odd-count rule for signs.** "Exactly two negatives" gives a **positive** product,
+  not a negative one.
+- **Ignoring zero.** If the problem doesn't say "nonzero," a variable could be \(0\) — which is even
+  and sign-less, and often kills a "must be positive" answer.
+- **Testing only one case.** When you plug in numbers to eliminate a "must be true" choice, try both
+  parities and both signs; one clean counterexample is enough to reject a choice.
+
+## Key takeaways
+
+- \(a + b\) and \(a - b\) always share parity; a sum/difference is odd only when exactly one term is odd.
+- A product is odd only if every factor is odd; a single even factor makes it even.
+- Consecutive integers \(k(k+1)\) are always even.
+- Sign of a product/quotient depends only on the parity of the *count* of negatives: even count → positive, odd count → negative.
+- Zero is even and sign-less — check whether variables are allowed to be zero.

content/questions/quant/quant-number-properties-odds-evens-signs-0017.md

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+---
+id: quant-number-properties-odds-evens-signs-0017
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: easy
+tags: [parity, odd-even, must-be-true]
+choices:
+  A: "\(x - y\)"
+  B: "\(xy\)"
+  C: "\(x^2 + y^2\)"
+  D: "\(x + 2y\)"
+  E: "\(2x + y\)"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(x\) and \(y\) are integers and \(x + y\) is odd, which of the following must be even?
+
+## Explanation
+
+If \(x + y\) is odd, then exactly one of \(x\) and \(y\) is odd and the other is even.
+
+- **A. \(x - y\):** subtraction has the same parity as addition, so \(x - y\) is **odd**, not even.
+  This is the classic trap.
+- **B. \(xy\):** the product of an even number and an odd number is always **even**. This must be even. ✓
+- **C. \(x^2 + y^2\):** a square keeps the parity of its base, so \(x^2 + y^2\) has the same parity as
+  \(x + y\), which is **odd**.
+- **D. \(x + 2y\):** \(2y\) is even, so this has the parity of \(x\). Since \(x\) could be the odd one
+  or the even one, this isn't guaranteed either way.
+- **E. \(2x + y\):** similarly has the parity of \(y\), which isn't fixed.
+
+Only **B** is guaranteed even.
+
+## Hints
+
+- If \(x + y\) is odd, one of the two is even and the other is odd — write that down first.
+- A product is even whenever *any* factor is even.

content/questions/quant/quant-number-properties-odds-evens-signs-0018.md

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+---
+id: quant-number-properties-odds-evens-signs-0018
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: easy
+tags: [signs, must-be-true, inequalities]
+choices:
+  A: "\(m > 0\) and \(n > 0\)"
+  B: "\(m < 0\) and \(n < 0\)"
+  C: "\(m\) and \(n\) have opposite signs"
+  D: "\(m = -n\)"
+  E: "exactly one of \(m, n\) is zero"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(mn > 0\) and \(m + n < 0\), which of the following must be true?
+
+## Explanation
+
+Because \(mn > 0\), the product is positive, so \(m\) and \(n\) have the **same** sign (and neither is
+zero — a zero would make the product \(0\), not positive).
+
+Same sign leaves two possibilities: both positive or both negative. But \(m + n < 0\) is negative,
+and two positive numbers can't add to something negative. So both must be **negative**.
+
+- **A** is the opposite conclusion.
+- **C** would make the product negative, contradicting \(mn > 0\).
+- **D** (\(m = -n\)) would give \(m + n = 0\), not less than \(0\).
+- **E** would make \(mn = 0\), not positive.
+
+The answer is **B**.
+
+## Hints
+
+- A positive product means the two numbers share a sign.
+- Can two positive numbers sum to a negative? Use that to pick between the two same-sign cases.

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, consecutive-integers, must-be-true]
+choices:
+  A: "\(k^2 + 1\)"
+  B: "\(k^2 + k\)"
+  C: "\(2k + 1\)"
+  D: "\(k^2 + k + 1\)"
+  E: "\(3k\)"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(k\) is a positive integer, which of the following is always even?
+
+## Explanation
+
+Test each choice against both parities of \(k\), or use parity rules directly.
+
+- **A. \(k^2 + 1\):** if \(k\) is even, \(k^2\) is even and \(k^2 + 1\) is **odd**. Not always even.
+- **B. \(k^2 + k = k(k+1)\):** this is the product of two **consecutive** integers, one of which is
+  always even. So \(k(k+1)\) is **always even**. ✓
+- **C. \(2k + 1\):** \(2k\) is even, so \(2k + 1\) is always **odd**.
+- **D. \(k^2 + k + 1 = k(k+1) + 1\):** an even number plus \(1\) is always **odd**.
+- **E. \(3k\):** has the parity of \(k\); if \(k\) is odd (e.g. \(k = 1\)), \(3k = 3\) is **odd**.
+
+Only **B** is guaranteed even.
+
+## Hints
+
+- Try \(k = 1\) and \(k = 2\) in each choice and keep only the ones that stay even.
+- Factor \(k^2 + k\) — what do you know about the product of two consecutive integers?

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: hard
+tags: [signs, product, cannot-be-true]
+choices:
+  A: "All three integers are negative."
+  B: "Exactly one of the integers is negative."
+  C: "Exactly two of the integers are negative."
+  D: "One of the integers equals \(-1\)."
+  E: "The sum of the three integers is positive."
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+The product of three integers is negative. Which of the following CANNOT be true?
+
+## Explanation
+
+A product is negative only when it is nonzero and the **count of negative factors is odd** — so with
+three integers, either **one** or **all three** are negative. (None can be zero, or the product would
+be \(0\).)
+
+- **A. All three negative:** three negatives is an odd count → negative product. **Possible.**
+- **B. Exactly one negative:** one negative, two positive → negative product. **Possible.**
+- **C. Exactly two negative:** two negatives is an *even* count, so those two multiply to a positive,
+  and the third (nonzero) factor keeps it positive. The product would be **positive**, never negative.
+  This **cannot** be true. ✓
+- **D. One equals \(-1\):** e.g. \(-1, 2, 3\) gives \(-6\). **Possible.**
+- **E. Sum is positive:** e.g. \(-1, 5, 5\) gives product \(-25\) with sum \(9 > 0\). **Possible.**
+
+The answer is **C**.
+
+## Hints
+
+- A negative product needs an *odd* number of negative factors.
+- Multiply out "exactly two negatives" with a small example — what sign do you always get?

content/questions/quant/quant-number-properties-odds-evens-signs-0021.md

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+---
+id: quant-number-properties-odds-evens-signs-0021
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: medium
+tags: [parity, coefficients, must-be-true]
+choices:
+  A: "\(pq\)"
+  B: "\(p + q\)"
+  C: "\(2p + q\)"
+  D: "\(p + 1\)"
+  E: "\(pq + 1\)"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(p\) and \(q\) are integers and \(3p + 5q\) is even, which of the following must be even?
+
+## Explanation
+
+An odd coefficient doesn't change parity: \(3p\) has the parity of \(p\), and \(5q\) has the parity of
+\(q\). So \(3p + 5q\) has the parity of \(p + q\). For that to be even, \(p\) and \(q\) must have the
+**same parity** (both odd or both even).
+
+- **A. \(pq\):** both even → even, but both odd → odd. Not guaranteed.
+- **B. \(p + q\):** same parity means their sum is **even** every time. This must be even. ✓
+- **C. \(2p + q\):** \(2p\) is even, so this has the parity of \(q\), which could be odd or even.
+- **D. \(p + 1\):** flips the parity of \(p\); could be either.
+- **E. \(pq + 1\):** since \(pq\) isn't fixed (see A), neither is \(pq + 1\).
+
+The answer is **B**.
+
+## Hints
+
+- Drop the coefficients: \(3p + 5q\) has the same parity as \(p + q\).
+- If \(p + q\) is even, what does that tell you about whether \(p\) and \(q\) match in parity?