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 on this page are the property of the MAA's American Mathematics Competitions