2011 AMC 10B Problem 21

Problem 21 of 25HarderAlgebra

Brian writes down four integers

w > x > y > z

whose sum is

44.

The pairwise positive differences of these numbers are

1,

3,

4,

5,

6,

and

9.

What is the sum of the possible values for

w?

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Solution

The largest difference is

9

, so

w-z=9

. For either middle number

n

, the two differences

w-n

and

n-z

must add to

9

. The available pairs of differences that add to

9

are

3+6

and

4+5

, and the remaining difference between the two middle numbers is

1

. One possible set is

\{w,w-5,w-6,w-9\}

, giving

4w-20=44

and

w=16

. The other is

\{w,w-3,w-4,w-9\}

, giving

4w-16=44

and

w=15

. The sum of the possible values of

w

16+15=31

. Thus, B is the correct answer.