Show that there must be at least 90 ways to choose six integers from 1 to 15 so that all choices have the same sum.

Answer 1

There are 6C15 = 5005 ways to choose 6 numbers

The minimum sum is 1+2+2+4+5+6 = 21
the maximum sum is 10+11+12+13+14+15 = 75
Thus there are only 55 different sums possible

5005/55 = 91

So, at least one of the sums must appear at least 91 times.

Answer 2

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