# 5. Place the signs $\times$ and : instead of the squares in the expression

$$
1 \square 3 \square 3^{2} \square 3^{4} \square 3^{8} \square 3^{16} \square 3^{32} \square 3^{64}=3^{99}
$$

such that the value of the expression becomes $3^{99}$.
Answer: $1 \times 3: 3^{2}: 3^{4}: 3^{8} \times 3^{16} \times 3^{32} \times 3^{64}=3^{99}$. Solution. The problem reduces to choosing between the signs (plus) and (minus) in the identity $0 \pm 1 \pm 2 \pm 4 \pm 8 \pm 16 \pm 32 \pm 64=99$. This can be done simply by trial and error. However, we can also denote the sum of numbers with a plus sign as $x$, and the sum of the absolute values of numbers "with a minus" as $y$. Then, since the sum $1+2+3+\ldots+64=127$, we have the system $\left\{\begin{array}{l}x+y=127, \\ x-y=99\end{array}\right.$ from which we find $x=113, y=14$. Since $y=14=2+4+8$, the minus sign will stand before $2,4,8$.