Berkeley Math Circle Monthly Contest 5

Solution:

No. Let $n, n+1, \ldots, n+99$ be any 100 consecutive positive integers. Then
$$n + (n+1) + (n+2) + \cdots + (n+99) = 100n + (1+2+\cdots+99).$$
However,
$$1+2+\cdots+99 = (1+99) + (2+98) + (3+97) + \cdots + (49+51) + 50 = 49 \cdot 100 + 50 = 50(2 \cdot 49 + 1) = 50 \cdot 99.$$
Thus
$$n + (n+1) + \cdots + (n+99) = 100n + 50 \cdot 99 = 50(2n + 99)$$
and this is not prime.