Problem 1429

1. Denote by $s(N)$ the sum of the digits of $N$. It is a well-known result that $s(N) \equiv N \pmod{9}$. This means that the sum of the digits of a number $N$ is congruent to $N$ modulo 9.
2. We need to prove that for every natural number $k$, there exists an infinite set of natural numbers $t$ such that the decimal notation of $t$ does not contain zeroes and the sums of the digits of the numbers $t$ and $kt$ are equal. In other words, we need $s(t) = s(kt)$.
3. Using the property $s(N) \equiv N \pmod{9}$, we have:
$$s(t) \equiv t \pmod{9} \quad \text{and} \quad s(kt) \equiv kt \pmod{9}$$
Therefore, we need:
$$t \equiv kt \pmod{9}$$
4. This simplifies to:
$$t(1 - k) \equiv 0 \pmod{9}$$
This means that $t$ must be a multiple of $\frac{9}{\gcd(k, 9)}$.
5. We consider two cases based on the value of $\gcd(k, 9)$:
- Case 1: $\gcd(k, 9) = 1$. In this case, $\frac{9}{\gcd(k, 9)} = 9$. Therefore, $t$ must be a multiple of 9. We can choose $t$ to be any number of the form $9m$ where $m$ is a natural number and the decimal notation of $t$ does not contain zeroes. For example, $t = 9, 18, 27, \ldots$.
- Case 2: $\gcd(k, 9) > 1$. The greatest common divisor can be either 3 or 9.
- If $\gcd(k, 9) = 3$, then $\frac{9}{\gcd(k, 9)} = 3$. Therefore, $t$ must be a multiple of 3. We can choose $t$ to be any number of the form $3m$ where $m$ is a natural number and the decimal notation of $t$ does not contain zeroes. For example, $t = 3, 6, 9, 12, 15, \ldots$.
- If $\gcd(k, 9) = 9$, then $\frac{9}{\gcd(k, 9)} = 1$. This is trivial, and every $t$ works. We can choose $t$ to be any natural number whose decimal notation does not contain zeroes.

6. In all cases, we can find an infinite set of natural numbers $t$ such that the decimal notation of $t$ does not contain zeroes and $s(t) = s(kt)$.

$\blacksquare$