Problem 14 · 2015 AMC 8
Medium
Number Theory
divisibilityalgebra-from-pattern
Which of the following integers cannot be written as the sum of four consecutive odd integers?
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Answer: D — 100.
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Hint 1 of 2
Don't test each answer by hunting for four odds — first ask what every such sum has in common. Add a general block of four consecutive odd integers and a hidden divisibility rule pops out.
Still stuck? Show hint 2 →
Hint 2 of 2
Writing them as n, n+2, n+4, n+6, the sum is 4n + 12 = 4(n + 3). Since n is odd, n+3 is even, so there's a second factor of 2 — the sum is always a multiple of 8. Now just find the choice that isn't.
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Approach: find the hidden invariant: the sum is always a multiple of 8
- Let the four be n, n+2, n+4, n+6. Sum = 4n + 12 = 4(n + 3).
- n is odd, so n + 3 is even — that even factor donates another 2, making the sum divisible by 4 × 2 = 8. So every valid sum is a multiple of 8.
- Scan the choices for the odd one out: 16, 40, 72, 200 are all multiples of 8, but 100 = 8·12 + 4 is not.
- Only 100 can't be made.
- Why this transfers: on 'which one can't be written as …' problems, derive the form's invariant (here, divisibility) and let it disqualify the choices — far faster than constructing examples.
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