2011 AMC 12B Problem 13

Question

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?

16

31

48

62

93

The largest difference is

w-z=9,

and the smallest positive difference must be

x-y=1

The second largest difference

6

is either

w-y

or

x-z

; handle each case

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

or

x-z.

If

w-y=6,

the numbers are

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

so

4w-20=44

and

w=16.

If

x-z=6,

the numbers are

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

so

4w-16=44

and

w=15.

The possible values are

16

and

15,

which sum to

31.

Thus, the correct answer is B.

Problem 13 in other years