2014 AMC 12B Problem 18

Attempt Problem 18 of the 2014 AMC 12B 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 2014 AMC 12B solutions, or check the answer key.

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18.

The numbers 1,2,3,4,51, 2, 3, 4, 5 are to be arranged in a circle. An arrangement is bad if it is not true that for every nn from 11 to 1515 one can find a subset of the numbers that appear consecutively on the circle that sum to n.n. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?

11

22

33

44

55

Answer: B
Concepts:circular arrangementscomplementary countingcasework
Difficulty rating: 2150
Solution:

Any single number covers sums 11 through 5.5. If a consecutive block sums to n,n, the remaining numbers form a consecutive block summing to 15n,15 - n, so sums 1010 through 1414 are automatically covered as well. Thus an arrangement is bad only if it fails to produce 66 or 7.7.

If 66 cannot be formed, rotate and reflect so the order is 1bc5e.1bc5e. The pair {b,c}\{b,c\} cannot be {2,3}\{2,3\} or {2,4},\{2,4\}, so e=2;e=2; avoiding the block 213213 then forces 14352.14352. If 77 cannot be formed, write the order as 2bc5e.2bc5e. Now {b,c}\{b,c\} cannot be {3,4}\{3,4\} or {1,4},\{1,4\}, so e=4,e=4, and avoiding 421421 forces 23154.23154.

These are the only two bad arrangements up to rotation and reflection.

Thus, the correct answer is B.

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