Auto: Shapovesov A.B. On the board, four three-digit numbers are written, which sum up to 2012. Only two different digits were used to write all of them.

Provide an example of such numbers.

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The first method. $2012=503+503+503+503$. By replacing zero with five, we increase the sum by 200; to compensate, we replace one of the fives in the hundreds place with a three: $2012=353+553+553+553$.

The second method. Let's try to find two numbers, each written with two digits, whose sum is $2012: 2=$ 1006. A carry must occur from the tens place, so the sum of the digits in the hundreds place without this carry must be 9. This means that both unknown digits are present in this place, and their sum is 9. On the other hand, the sum of the digits in the units place ends in 6. Therefore, either this sum is 6 and is obtained as $3+3$ with the second digit being 6 (but then we cannot get a sum of 10 in the tens place), or it is 16 and is obtained as 8 + 8 with the second digit being 1. From this, we get the representations $1006=118$ + $888=188$ + 818; $2012=118+888+118+888=188+818+188+818=118+888+188+818$.

## Answer

$2012=353+553+553+553=118+118+888+888=118+188+818+888=188+188+818+818$.

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