2011 AMC 12B Problem 13

Problem 13 of 25IntermediateAlgebra

Brian writes down four integers

w \gt x \gt y \gt 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

w-z=9.

For either interior number

n,

we have

9=(w-n)+(n-z).

The only pairs among the listed differences that sum to

9

are

3+6

and

4+5,

so the remaining difference must be

x-y=1.

The second largest difference

6

is either

w-y

x-z.

w-y=6,

the numbers are

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

4w-20=44

and

w=16.

x-z=6,

the numbers are

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

4w-16=44

and

w=15.

The possible values are

16

and

15,

which sum to

31.

Thus, the correct answer is B.