Problem 5.6. On the board, there is one three-digit number and two two-digit numbers. The sum of the numbers that have a seven in their notation is 208. The sum of the numbers that have a three in their notation is 76. Find the sum of all three numbers.
Answer: 247.

Solution. Let the three-digit number be $A$, the two-digit numbers be $-B$ and $C$. Among the numbers whose sum is 76, there cannot be a three-digit number. The sum cannot consist of a single number either, because otherwise that number would be 76, but it does not contain a three. Therefore, 76 is the sum of two two-digit numbers, each of which contains a three:

$$
B+C=76
$$

Among the numbers whose sum is 208, there must be a three-digit number (since the sum of two-digit numbers is only 76). It cannot be the only one, because the number 208 does not contain a seven. Therefore, there is at least one two-digit number in this sum - but not both at the same time (otherwise the three-digit number would be $208-76=132$, but it does not contain a seven).

Without loss of generality, let's assume that the two-digit number in this sum is $B$. Then

$$
A+B=208
$$

and both of these numbers contain a seven.

Therefore, the number $B$ contains both a three and a seven, i.e., $B=37$ or $B=73$.

If $B=73$, then $C=76-B=3$ - not a two-digit number. Therefore, $B=37, C=76-37=39$, $A=208-37=171$.

The sum of all the numbers is $171+37+39=247$.