[ [divisibility rules for 3 and 9]

Igor wrote one non-zero digit on each of three cards. Katya formed all possible three-digit numbers from them. Can the sum of these numbers be equal to 2018?

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Let the digits on the cards be $a, b$, and $c$. If they are all distinct, then Katya formed six numbers. Since each digit appears in the hundreds, tens, and units places twice, their sum is $200(a+b+c)+20(a+b+c)+2(a+b+c)=222(a+b+c)$. The equation $222(a+b+c)=2018$ cannot hold because 222 is divisible by 3, while 2018 is not.

If any two digits are the same, for example, $b=c$, then Katya formed three numbers. Their sum:

![](https://cdn.mathpix.com/cropped/2024_05_06_79aa1f75de69bad6c7b2g-25.jpg?height=60&width=1950&top_left_y=1456&top_left_x=17)
is a number that is definitely less than 2018. Thus, in these cases, the sum cannot be equal to 2018 either.

## Answer

it cannot.