2015/16 AHSMC Part 1 Problem 7

The sum of $k$ positive consecutive integers is $$a + (a+1) + \cdots + (a+k-1) = ka + \frac{k(k-1)}{2} = \frac{k(2a+k-1)}{2},$$ and thus $k(2a+k-1) = 110 = 2 \cdot 5 \cdot 11$. The solutions $(k,a)$ are $(2,27)$, $(5,9)$, and $(10,1)$ for which $$55 = 27+28 = 9+10+11+12+13 = 1+2+\cdots+10.$$ If one considers the set of whole numbers $W = \{1,2,3,\ldots\}$ then there are three sets of consecutive whole numbers having the sum $55$. However, if $W = \{0,1,2,3,\ldots\}$ then also $$55 = 0+1+2+\cdots+10,$$ and we find four sets with the required property.

The answer is (B) or (C).