[ $\underline{\text { equations in integers }}]$ [GCD and LCM. Mutual simplicity ]

The last digit in the notation of a natural number is 2016 times less than the number itself. Find all such numbers.

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Let $x$ be the last digit of the number. We can reason in different ways.

The first way. The number 2016x should end with the digit x. Therefore, $x$ is an even digit, and $x \neq$ 0. By checking, we find that the values of $x$, equal to 2, 4, 6, and 8, satisfy the condition.

The second way. The desired number has the form $10a + x$, where a is some natural number. Then $10a + x = 2016x$, from which $2a = 403x$. Since the numbers 2 and 403 are coprime, $a$ is divisible by 403, and $x$ is even. Substituting the values of $x$, equal to $2, 4, 6$, and 8, into the obtained equation, we find the corresponding values of $a$ and get the answer.

## Answer

$4032, 8064, 12096, 16128$.