2. For which four-digit number $\overline{a b c d}$ does $\overline{a b c d}+\overline{a b c}+\overline{a b}+a=2023$ hold?
2. For which four-digit number $\overline{a b c d}$ is $\overline{a b c d}+\overline{a b c}+\overline{a b}+a=2023$ ?

## Solution.

$\overline{a b c d}+\overline{a b c}+\overline{a b}+a=2023$

Expand the left side of the equation in decimal form.

$(1000 a+100 b+10 c+d)+(100 a+10 b+c)+(10 a+b)+a=2023$

$1 \text{ POINT}$

We have:

$1111 a+111 b+11 c+d=2023$, where $a, b, d, c \in\{0,1,2,3,4,5,6,7,8,9\}, a \neq 0 . \quad 1$ POINT

The digit $a$ cannot be greater than 1 because otherwise the number on the left side would be greater than 2023. Therefore, $a=1$.

1 POINT

Then we have:

$1111+111 b+11 c+d=2023$

$111 b+11 c+d=912$

We conclude that $b<9$ because otherwise the number on the left side would be greater than 999.

$1 \text{ POINT}$

If $b=8$, then:

$888+11 c+d=912$

$11 c+d=24$

$1 \text{ POINT}$

If $b=7$, then:

$777+11 c+d=912$

$11 c+d=135$

This is impossible, as the largest digit is 9, so $11 c+d$ is at most 108. Therefore, $b$ cannot be 7 or less than 7.

This means that $b=8$.

Then $11 c+d=24$, so $c \leq 2$.

$1 \text{ POINT}$

If $c=2$, then:

$22+d=24$

$d=24-22$

$d=2$

$1 \text{ POINT}$

If $c=1$, then:

$11+d=24$

$d=24-11$

$d=13$

This is impossible, as the largest digit is 9, so $11+d$ is at most 20. Therefore, $d$ cannot be 1 or less than 1.

Finally, $a=1, b=8, c=2$ and $d=2$ so the desired number is 1822.