How many positive integers smaller than 400 can you get as a sum of eleven consecutive positive integers?

Solution

The sum of eleven consecutive integers starting at $$n$$ is $$11(n+5)$$, which must be less than 400 and at least $$11 \times 6=66$$ (for $$n=1$$). This gives $$n$$ ranging from 1 to 31, so there are 31 possible sums.

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