5. 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 $36 \cdot S(x) \leq 36 \cdot 9 \cdot n=324 n36(a+b+c+d) \Leftrightarrow 964 a+64 b>26 \cdot 9+35 \cdot 9 \geq 26 c+35 d$.

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.

For $n=2: a \cdot 10+b=36(a+b) \Leftrightarrow 26 a=35 b-$ there are no solutions.