Competition problem — Number theory

Library / /158 of 520

Number theory Difficulty 6.3 National olympiad Prove it

$1 \cdot 1$ Let $S(n)$ denote the sum of all digits of the natural number $n$ .
（1）Does there exist a natural number $n$ such that $n+S(n)=1980$ ?
（2）Prove: Among any two consecutive natural numbers, one can be expressed in the form $n+S(n)$ , where $n$ is some natural number.

Solution

[Solution] (1) $1962+S(1962)=1980$ .
(2) We denote $S(n)+n$ as $S_{n}$ . If the last digit of the number $n$ is 9, then $S_{n+1}2$ . Select the largest $N$ such that $S_{N}<m$ , then $S_{N+1} \geqslant m$ . Clearly, the last digit of $N$ is not 9. Therefore, $S_{N+1}=m$ or $S_{N+1}=m+1$ . It is also clear that $S_{1}=2$ , thus the proposition is proved.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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