3. The sum of three five-digit numbers of the form $\overline{\text { mat31 }}, \overline{\text { mat41 }}, \overline{\text { mat51 }}$ is 202023. What is $m+a+t$?
First solution.

|  | $m$ | $a$ | $t$ | 3 | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
|  | $m$ | $a$ | $t$ | 4 | 1 |
| + | $m$ | $a$ | $t$ | 5 | 1 |
| 2 | 0 | 2 | 0 | 2 | 3 |

$1+1+1=3$

$3+4+5=12$, we write 2 and carry 1 to the left column, resulting in

$3 \cdot t+1$ which can be 10 or 20 or 30 (no more because $t$ is a digit).

$1 \text{ POINT}$

If $3 \cdot t+1=30$ then $3 \cdot t=29$, which is not possible because 29 cannot be divided by 3 without a remainder.

If $3 \cdot t+1=20$ then $3 \cdot t=19$, which is not possible because 19 cannot be divided by 3 without a remainder.

If $3 \cdot t+1=10$ then $3 \cdot t=9$, from which it follows that $t=3$.

$1 \text{ POINT}$

From the sum $t+t+t+1=10$, we write 0 and carry 1 to the left column, resulting in $3 \cdot a+1$ which can be 2 (not possible because it is too small) or 12 or 22 or 32 (no more because $a$ is a digit).

$1 \text{ POINT}$

If $3 \cdot a+1=12$ then $3 \cdot a=11$, which is not possible because 11 cannot be divided by 3 without a remainder.

If $3 \cdot a+1=32$ then $3 \cdot a=31$, which is not possible because 31 cannot be divided by 3 without a remainder.

If $3 \cdot a+1=22$ then $3 \cdot a=21$, so $a=7$.

$1 \text{ POINT}$

From the sum $a+a+a+1=22$, we write 2 and carry 2 to the left column, resulting in

$m+m+m+2=20$, from which it follows that $3 \cdot m=18$, or $m=6$.

$1 \text{ POINT}$

Then $m+a+t=6+7+3=16$.

1 POINT

TOTAL 6 POINTS

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