Highlander

[Sk_Anonymous]

Given 1962 -digit number. It is divisible by 9. Let x be the sum of its
digits. Let the sum of the digits of x be y. Let the sum of the digits
of y be z. Find z.

[Gi81]

In general, given \( n \in \mathbb{N} \), if \( d_n \in \mathbb{N}\) is the sum of the digits of \( n \), then \( \displaystyle 9 \ \mid \ n \ \Leftrightarrow \ 9 \mid d_n \)

So \(\displaystyle x,y,z \) are all divisible by \(\displaystyle 9 \).
Moreover, \(\displaystyle 1\leqslant x \leqslant 9 \cdot 1962 < 10 \cdot 2000 = 20000 \), so \(\displaystyle x \) is a \(m\)-digits number, with \(\displaystyle 1\leqslant m\leqslant 5 \).
Then \(\displaystyle y \leqslant 5\cdot 9=45 \Rightarrow y \in \{ 9,18,27,36,45 \} \), so \(\displaystyle z=9 \).

[Sk_Anonymous]

well done!