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 $w?$

$16$

$31$

$48$

$62$

$93$

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 $w$ is $16+15=31$ .

Thus, B is the correct answer.

2011 AMC 10B Problem 21

Solution:

Problem 21 in Other Years