AMC 8 · 2015 · #14

Grade 4 algebranumber-theory
divisibility-rulesparitymultiples convert-to-algebracasework ↑ Prerequisites: multi-digit-arithmeticdivisibility-rules
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Problem
Among the five numbers 16, 40, 72, 100, and 200, find the one that can NOT be written as the sum of four consecutive odd integers (like 1+3+5+7 or 7+9+11+13).

Pick an answer.

(A)
16
(B)
40
(C)
72
(D)
100
(E)
200

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

We do not yet know which sums are possible, so start by listing a few real sums of four consecutive odd integers (Tool #9): 1+3+5+7, 3+5+7+9, 5+7+9+11, and so on. Then look at the resulting sums for a pattern (Tool #5) — every sum jumps by 8, so every possible sum is a multiple of 8. Finally test the five answer choices against that pattern (Tool #3, Eliminate) — only one choice fails the divisibility-by-8 test.

1STEP 1

Start easy: add the four smallest consecutive odd integers, then shift up, listing the totals 16, 24, 32, 40.

1+3+5+7 = 16, 3+5+7+9 = 24, 5+7+9+11 = 32, 7+9+11+13 = 40
2STEP 2

Each total is 8 more than the last, so every possible sum is a multiple of 8.

24 - 16 = 8, 32 - 24 = 8, 40 - 32 = 8 → sums = 16, 24, 32, 40, 48, … = 8 × (whole number)
3STEP 3

Test each choice: the one that is not a multiple of 8 is the impossible sum.

16 = 8 × 2, 40 = 8 × 5, 72 = 8 × 9, 100 = 8 × 12 + 4 (not a multiple of 8), 200 = 8 × 25
4STEP 4

Only 100 is not a multiple of 8, so it is the number that cannot be written this way — the answer is (D).

100 → (D)
Answer
100
Double-check by trying to actually build 100 as a sum of four consecutive odd integers. The middle of those four integers would be near 100 ÷ 4 = 25, so try 23 + 25 + 27 + 29 = 104 (too big) and 21 + 23 + 25 + 27 = 96 (too small). The sum jumps straight from 96 to 104 — skipping 100 — which confirms 100 is impossible. Each of the other four choices can be hit: 16 = 1+3+5+7, 40 = 7+9+11+13, 72 = 15+17+19+21, 200 = 47+49+51+53.
💡Key takeaway

Adding four small odd numbers a few times and spotting the jump-by-8 pattern is a Grade 4 skill — no algebra needed to crack this AMC 8 problem!