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Problem 879

AIME late

Number theory Difficulty 5.7 Prove it

396. Does there exist a natural number $n$ with the following property: the sum of the digits of the number $n$ (in decimal notation) is 1000, and the sum of the digits of the number $n^{2}$ is $1000^{2}$ ?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

396. Answer: it exists. Let the $n-k$ -digit number, written using $m$ ones and some number of zeros, be such that $S\left(n^{2}\right)=m^{2}$ . Then for the number $n_{1}=10^{k+1} n+1$ , where $k$ is the number of digits in the representation of $n, \quad S\left(n_{1}\right)=m+1, S\left(n_{1}^{2}\right)=(m+1)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.