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2014 AMC 12B Problem 18

Problem 18 of 25IntermediateCounting & Probability

The numbers 1,1, 2,2, 3,3, 4,4, 55 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?

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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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Concepts: circular arrangements · complementary counting · casework

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