2011 AMC 10B Problem 21

Attempt Problem 21 of the 2011 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

Brian writes down four integers w>x>y>zw > x > y > z whose sum is 44.44. The pairwise positive differences of these numbers are 1,3,4,5,6,1, 3, 4, 5, 6, and 9.9. What is the sum of the possible values for w?w?

1616

3131

4848

6262

9393

Answer: B
Concepts:system of equationscasework
Difficulty rating: 1990
Solution:

The largest difference is 99, so wz=9w-z=9. For either middle number nn, the two differences wnw-n and nzn-z must add to 99.

The available pairs of differences that add to 99 are 3+63+6 and 4+54+5, and the remaining difference between the two middle numbers is 11.

One possible set is {w,w5,w6,w9}\{w,w-5,w-6,w-9\}, giving 4w20=444w-20=44 and w=16w=16. The other is {w,w3,w4,w9}\{w,w-3,w-4,w-9\}, giving 4w16=444w-16=44 and w=15w=15.

The sum of the possible values of ww is 16+15=3116+15=31.

Thus, B is the correct answer.

← Problem 20#20
Full Exam

Problem 21 in Other Years