Problem 17 · 1986 AJHSME
Hard
Number Theory
factor-outparity-of-product
Let o be an odd whole number and let n be any whole number. Which of the following statements about the whole number (o² + no) is always true?
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Answer: E — it is odd only if n is even.
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Hint 1 of 3
Don't grind through cases — both terms share a common factor of o. Pulling it out turns the whole thing into a *product* of two pieces, and a product's odd/even-ness is easy to read off.
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Hint 2 of 3
o² + no = o(o + n). A product of whole numbers is odd only when *both* parts are odd. The first part o is already odd, so everything hinges on whether o + n is odd or even.
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Hint 3 of 3
o + n is odd ⟺ you added an odd and an even. Since o is odd, that needs n even.
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Approach: factor out o, then read the parity of the product
- Factor the shared o: o² + no = o(o + n). Now it's one odd number (o) times another number (o + n), and a product is odd only when both factors are odd — otherwise an even factor drags the whole thing even.
- The first factor o is odd no matter what. So the product is odd exactly when o + n is also odd. Adding to the odd o, you get an odd total only by adding an *even* n (odd + even = odd; odd + odd = even).
- Therefore the expression is odd only if n is even — answer E.
- Sanity check with numbers: o = 3, n = 2 → 9 + 6 = 15 (odd, n even ✓); o = 3, n = 1 → 9 + 3 = 12 (even, n odd ✓). The 'odd happens only with even n' pattern holds.
- Why factoring first helps: turning a sum into a product lets you judge odd/even instantly, since one even factor makes the product even — a tactic worth reaching for whenever a common factor is in sight.
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