AMC 10 · 2012 · #5

Grade 2 logicnumber-theory
parityoptimizationlogical-deduction extremal-constructionidentify-subproblems ↑ Prerequisites: parity
📏 Short solution 💡 2 insights
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Problem
Six whole numbers arrive two at a time with three running totals reported. Find the fewest even ones possible.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
How to solve
Strategy Extreme Principle

The question asks for a minimum, so Tool #14 (Extreme Principle) sets the shape of the work: find a floor no set can go below, then produce one set that sits exactly on the floor. Tool #7 (Identify Subproblems) makes the floor findable — the running totals cut the six integers into three independent pairs, each with its own sum. Tool #16 (Change Focus) drops the actual sizes and keeps only even-or-odd, which is all the count depends on. Tool #6 (Guess and Check) then builds a concrete list of six integers that hits every stated total while carrying as few even numbers as the floor allows.

1STEP 1

Cut the six into three pairs

The running totals split the six into three pairs.

26, 41-26=15, 57-41=16
2STEP 2

Watch parity, not size

Only each pair's parity matters.

odd+odd=even, even+even=even, even+odd=odd
3STEP 3

Prove zero evens is impossible

One odd pair sum forces at least one even.

26, 16 even→those pairs may be odd+odd; 15 odd→at least 1 even
4STEP 4

Build a list that reaches one

A real list reaches 1, choice (A).

15+11=26; 26+8+7=41; 41+9+7=57; evens={8}
Answer
1
Check the witness against the original wording rather than against the pairs. The six numbers 15, 11, 8, 7, 9, 7 give 15+11=26 for the first two, then 26+8+7=41 once two more are added, then 41+9+7=57 once the last two are added, so all three stated totals are exact. Counting even numbers in that list gives one, the 8. Step 3 already showed zero is impossible, so the minimum is pinned between 1 and 1. Both halves of the minimum claim hold, so (A) is consistent.
💡Key takeaway

Two odd numbers always add up to an even number, so any pair that must add to an odd total is forced to hide exactly one even number inside it.

  • Cut the six into three pairs
  • Watch parity, not size
  • Prove zero evens is impossible
  • Build a list that reaches one