1. If the equality holds

$$
\sqrt{17^{2}+17^{2}+17^{2}+\cdots+17^{2}+17^{2}+17^{2}}=17^{2}+17^{2}+17^{2}
$$

how many times does $17^{2}$ appear as an addend under the root?
1. Let $17^{2}$ appear under the root $k$ times. We can write

$3 \cdot 17^{2}=17^{2}+17^{2}+17^{2}=\sqrt{17^{2}+17^{2}+17^{2}+\cdots+17^{2}+17^{2}+17^{2}}=\sqrt{k \cdot 17^{2}}=17 \sqrt{k}$, so it follows that $k=(3 \cdot 17)^{2}=51^{2}=2601$.