1. Let $n$ be a two-digit number such that the square of the sum of digits of $n$ is equal to the sum of digits of $n^{2}$. Find the sum of all possible values of $n$.
(1 mark)
設 $n$ 為兩位數, 而 $n$ 的數字之和的平方等於 $n^{2}$ 的數字之和 。求 $n$ 所有可能值之和。
1. 170
1. Since $n$ is a two-digit number, we have $n^{2} \leq 99^{2}=9801<9999$ and so the sum of digits of $n^{2}$ is less than $9 \times 4=36$. Since the sum of digits of $n^{2}$ is equal to the square of the sum of digits of $n$, the sum of digits of $n$ is less than $\sqrt{36}=6$. It remains to search through all the two-digit numbers whose sum of digits not greater than 5 . There are 15 such numbers, namely, 10, 11, $12,13,14,20,21,22,23,30,31,32,40,41,50$. By checking these numbers one by one, we know that the answer is $10+11+12+13+20+21+22+30+31=170$.