Putnam 1940 
Topic: Algebra Number Theory

Problem: Given that $f(x)$ is a polynomial with integral coefficients, and there exists an integer $k$ such that none of the integers $f(1), f(2), \ldots, f(k)$ is divisible by $k$, then find the number of integral roots of $f(x)$.

Solution: Suppose $f$ has an integral root $r$. Then $f(x) = (x - r)g(x)$ where $g(x)$ is also a polynomial with integral coefficients. Then there are integers $p$ and $q$ such that $r = p + kq$ and $1 \leq p \leq k$. But 
\[ f(p) = (p - r)g(p) = -kqg(p), \]
and hence $f(p)$ is divisible by $k$ contrary to the hypothesis. This contradiction shows that $f(x)$ has no integral root making the answer \boxed{0}.