1984 AIME Problems/Problem 12

Problem

A function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$ ?

If $f(2+x)=f(2-x)$ , then substituting $t=2+x$ gives $f(t)=f(4-t)$ . Similarly, $f(t)=f(14-t)$ . In particular,

Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$ ) are also roots. To see that these may be the only integer roots, observe that the function $f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}$ satisfies the conditions and has no other roots.

In the interval $-1000\leq x\leq 1000$ , there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \pmod{10}$ , therefore the minimum number of roots is $\boxed{401}$ .