Printable

2012 AMC 10B Problem 9

2012 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Number Properties · difficulty 2 of 5 · about 2 min · position P1-10

Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?

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

Answer

Key insight

An even sum needs no odd addends but an odd sum needs one; the middle pair adds to 15, so exactly one odd integer is forced.

Solution

Separate the three pairs by subtracting consecutive totals: the first pair sums to $26$ , the second pair to $41 - 26 = 15$ , and the third pair to $57 - 41 = 16$ .

Two integers with an even sum can both be even, so the first and third pairs need no odd integers at all (for instance $13 + 13$ and $8 + 8$ ). Two integers with an odd sum must have opposite parity, so the second pair contains exactly one odd integer (for instance $7 + 8$ ).

Hence at least one odd integer is required, and one is enough.

The answer is $\boxed{\textbf{(A)}\ 1}$ .

Why this works

Parity of a sum depends only on how many odd terms it has: an odd total forces an odd number of odd terms, an even total forces an even number (possibly zero). "Minimum" questions then reduce to choosing zero odd terms wherever allowed and one wherever an odd total demands it. Look at each stage's increment, not the running totals.

The trap

Answering 0 by overlooking that the second pair sums to the odd number 15, or answering 3 by assuming every pair needs an odd number.

Common mistakes

Answering 0 by overlooking that the second pair sums to the odd number 15, or answering 3 by assuming every pair needs an odd number.
Reading $41$ and $57$ as odd totals and concluding two odd integers are needed, forgetting that $57 - 41 = 16$ is even.

Techniques

Use an invariant, parity, or coloring argument