2011 AMC 10B Problem 21

Question

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?

16

31

48

62

93

The largest difference must be

w-z=9

The remaining paired differences must split

9

as

3+6

or

4+5

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

is

16+15=31

.

Thus, B is the correct answer.

Problem 21 in other years