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

The first method. 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.

We can also perform a complete enumeration of all possible values of $x$: check all digits from 0 to 9.

The second method. Let the required numbers have the form $a x = 10 a + x$, where $a$ is some natural number. Then $10 a + x = 2016 x \Leftrightarrow 2 a = 403 x$. Since 2 and 403 are coprime numbers, $a$ is divisible by 403, and $x$ is divisible by 2. By substituting, for example, 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.

Grading criteria.

"+" - a complete and justified solution is provided (by any method, regardless of its rationality)

“ $\pm$ " - only the completely correct answer is provided

"Ғ" - a correct reasoning is provided, but one of the numbers is missing in the answer

“-” - no more than two numbers are correctly provided in the answer

“-” - the problem is not solved or solved incorrectly
1. The last digit in the notation of a natural number is 2016 times smaller than the number itself. Find all such numbers. Answer: $4032,8064,12096,16128$.