Question

A function $f$ is defined for all real numbers and satisfies $\begin{aligned} f(2+x)&=f(2-x),\\ f(7+x)&=f(7-x) \end{aligned}$ for all real $x.$ If $x=0$ is a root of $f(x)=0,$ what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x\leq1000?$

Answer

Answer: 401

Concepts:functional equation reflection (geometry)counting integers in a range

Difficulty rating: 2110

Small Hint:

Interpret the two identities as reflection symmetries about $2$ and $7$

Big Hint:

Composing the two reflections produces a translation by $10$

Solution:

The identities make the root set invariant under reflection about $2$ and about $7.$ Their composition is translation by $10.$ Starting from the root $0,$ these operations force every number in $\begin{gathered} \{10k:k\in\mathbb Z\}\\ {}\cup\{4+10k:k\in\mathbb Z\} \end{gathered}$ to be a root. In the interval, the first set contributes $201$ roots and the second contributes $200,$ for a total of $401.$

This bound is attainable: define $f$ to be $0$ on this invariant set and $1$ elsewhere. It has both required reflection symmetries. Hence the least possible number is $401.$

1984 AIME Problem 12

Answer: 401

Difficulty rating: 2110

Small Hint:

Big Hint:

Solution:

Problem 12 in Other Years