AMC 10 · 2012 · #9

Grade 2 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
parity extremal-construction ↑ Prerequisites: parity
📏 Medium solution 💡 2 insights
📘 View easy version →
Problem
Six integers are added two at a time. The first two sum to 26. Adding the next two makes the running total 41. Adding the last two makes the total 57. Find the fewest of the six integers that can be odd.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

The question asks for a minimum, which is the direct signal for Tool #14 (Extreme Principle): push the six numbers to the boundary where odds are as scarce as possible. Tool #11 (Work Backwards) recovers each pair's own sum from the three running totals by subtraction. Tool #7 (Identify Subproblems) then treats the three pairs independently, because each pair's sum by itself decides the fewest odds that pair can hold.

1STEP 1

Recover each pair's sum

Subtract consecutive running totals to recover the three pair sums: 26, 15, 16.

pair₁ = 26, pair₂ = 41 - 26 = 15, pair₃ = 57 - 41 = 16
2STEP 2

Parity of a pair sum

Two integers sum to an even number when they share parity, and to an odd number only when exactly one of them is odd.

even sum → 0 or 2 odds, odd sum → exactly 1 odd
3STEP 3

Push each pair to fewest odds

Give each pair the fewest odds its sum allows: even sums 26 and 16 need 0 odds, odd sum 15 forces exactly 1.

26 : 0, 15 : 1, 16 : 0
4STEP 4

Add the minimums

Add the per-pair minimums: 0 + 1 + 0 = 1. The build (12, 14), (7, 8), (8, 8) hits 26, 41, 57 with 7 the lone odd.

0 + 1 + 0 = 1 → (A)
Answer
1
Only the pair whose sum is odd (15) is forced to hold an odd number, and one odd there is unavoidable because two evens or two odds always make an even sum. The other two pairs (sums 26 and 16) are cleanly all-even, contributing zero. The explicit build (12, 14), (7, 8), (8, 8) realizes exactly one odd while matching the totals 26, 41, 57, so 1 is achievable and cannot be beaten — answer (A).
💡Key takeaway

Two numbers only make an odd total when exactly one of them is odd, so this AMC 10 question is really a Grade 2 odd/even puzzle: only the pair that jumps by an odd amount is forced to carry a single odd number.

  • Recover each pair's sum
  • Parity of a pair sum
  • Push each pair to fewest odds
  • Add the minimums

More like this

Same archetype — closest grade level first.