Tokoreve C.

Does there exist a six-digit number $A$ such that among the numbers $A, 2A, \ldots, 500000A$ there is not a single number ending in six identical digits?
Let $A$ be coprime with 10. Then the numbers $A, 2A, \ldots, 10^{6}A$ give all possible remainders modulo $10^{6}$ exactly once (see problem 6 0733). Therefore, it is sufficient to find a number $A$ such that the remainders 111111, 222222, ..., 999999 appear in the last nine digits, that is, among the numbers from $\left(10^{6}-9\right)A$ to $\left(10^{6}-1\right)A$.

Take $A=888889 \equiv -111111 \left(\bmod 10^{6}\right)$. Then $\left(10^{6}-m\right)A \equiv -mA \equiv 111111m \left(\bmod 10^{6}\right)$.

## Answer

There exists.