2012 AMC 10B Problem 9

P1-10~2 minPrint

Number Properties Use an invariant, parity, or coloring argument

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?

Show solution without answering

Key insight

Split the totals into three pairs that add to 26, 15 and 16. Only the pair adding to 15, an odd number, is forced to contain an odd integer.

Solution

1. The problem gives running totals. Subtract to find what each new pair adds:
- first pair: $26$ ;
- second pair: $41 - 26 = 15$ ;
- third pair: $57 - 41 = 16$ .
2. Recall the rules for adding two integers: even + even = even, odd + odd = even, even + odd = odd.
3. The first pair adds to $26$ , which is even. Both numbers can be even, for example $12 + 14$ . So this pair needs no odd numbers.
4. The third pair adds to $16$ , also even. Take $8 + 8$ . No odd numbers needed.
5. The second pair adds to $15$ , which is odd. Two even numbers always add to an even number, so this pair must contain an odd number. One is enough, for example $7 + 8$ .
6. So at least one odd integer is needed, and the example $12, 14, 7, 8, 8, 8$ has exactly one.

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

Why this works

Whether a sum is odd or even depends only on how many odd numbers are in it: an odd sum needs an odd count of odd numbers, and an even sum can have none. For "fewest odd numbers" questions, use zero odd numbers wherever the sum is even and one wherever it is odd. Look at what each new pair adds, not at the running totals.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(A)}\ 1}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 20: Special Functions (§20.3)