Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an odd function, that is, a function that satisfies $-f(x)=f(-x)$ for all $x \in \mathbb{R}$. Suppose that $f(x+5)=f(x)$ for all $x \in \mathbb{R}$ and that $f(1 / 3)=1$. Determine the value of the sum:

$$
f(16 / 3)+f(29 / 3)+f(12)+f(-7)
$$

#
Solution

We have

$$
\begin{aligned}
f\left(\frac{1}{3}\right) & =1 \\
f\left(\frac{1}{3}+5\right) & =1 \\
f\left(\frac{16}{3}\right) & =1
\end{aligned}
$$

Since $f$ is an odd function, i.e., $f(-x)=-f(x)$, it follows that

$$
\begin{aligned}
f\left(-\frac{1}{3}\right) & =-1 \\
f\left(-\frac{1}{3}+5+5\right) & =-1 \\
f\left(\frac{29}{3}\right) & =-1
\end{aligned}
$$

and

$$
\begin{aligned}
f(-7) & =f(-12+5) \\
& =f(-12) \\
& =-f(12)
\end{aligned}
$$

Finally,

$$
\begin{aligned}
f(16 / 3)+f(29 / 3)+f(12)+f(-7) & = \\
1+(-1)+f(12)-f(12) & =0
\end{aligned}
$$