2011 AMC 12B Problem 13

Below is the professionally curated solution for Problem 13 of the 2011 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.

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Concepts:system of equations casework

Difficulty rating: 1610

13.

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 $w?$

$16$

$31$

$48$

$62$

$93$

Solution:

The largest difference is $w-z=9.$ Writing $9=(w-x)+(x-z)$ style splits, the interior differences pair as $3+6$ and $4+5,$ which forces the smallest difference $x-y=1.$

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.

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