Problem:
The numbers are to be arranged in a circle. An arrangement is bad if it is not true that for every from 1 to 15 one can find a subset of the numbers that appear consecutively on the circle that sum to . 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:
The circular arrangement 14352 is bad because the sum 6 cannot be achieved with consecutive numbers, and the circular arrangement 23154 is bad because the sum 7 cannot be so achieved. It remains to show that these are the only bad arrangements. Given a circular arrangement, sums 1 through 5 can be achieved with a single number, and if the sum can be achieved, then the sum can be achieved using the complementary subset. Therefore an arrangement is not bad as long as sums 6 and 7 can be achieved. Suppose 6 cannot be achieved. Then 1 and 5 cannot be adjacent, so by a suitable rotation and/or reflection, the arrangement is . Furthermore, cannot equal because ; similarly cannot equal . It follows that , which then forces the arrangement to be 14352 in order to avoid consecutive 213. This arrangement is bad. Next suppose that 7 cannot be achieved. Then 2 and 5 cannot be adjacent, so again without loss of generality the arrangement is . Reasoning as before, cannot equal or , so , and then and , to avoid consecutive 421 ; therefore the arrangement is 23154 , which is also bad. Thus there are only bad arrangements up to rotation and reflection.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions