6. Find all natural numbers that are 36 times the sum of their digits. ANSWER: $324 ; 648$.
Solution. Let $S(x)$ denote the sum of the digits of the number $x$. Then the equation $x=36 \cdot S(x)$ has no solutions for $n \geq 5$, since $x \geq 10^{n-1}$, while $36 \cdot S(x) \leq 36 \cdot 9 \cdot n = 324 n$. For $36(a+b+c+d) \Leftrightarrow 964 a + 64 b > 26 \cdot 9 + 35 \cdot 9 \geq 26 c + 35 d$. For $n=3: a \cdot 100 + b \cdot 10 + c = 36(a + b + c) \Leftrightarrow 64 a = 26 b + 35 c$. After this, we perform a check. As a result, the numbers 324 and 648 are obtained.

$$
\text{For } n=2: a \cdot 10 + b = 36(a + b) \Leftrightarrow 26 a = 35 b - \text{no solutions}
$$

Lomonosov Moscow State University

## School Olympiad "Conquer Sparrow Hills" in Mathematics

Final stage tasks for the 2015/2016 academic year for 7-8 grades