3. On the board, three two-digit numbers are written, one of which starts with 5, the second with 6, and the third with 7. The teacher asked three students to each choose any two of these numbers and add them. The first student got 147, and the answers of the second and third students are different three-digit numbers starting with 12. What can the number starting with 7 be? If there are multiple answers, list them all.
Answer: Only 78.

Solution: Let the number starting with 7 be denoted as $a$, the number starting with 6 as $b$, and the number starting with 5 as $c$. The sum $a+b \geqslant 70+60=130$, so it must be equal to 147. The maximum sum of numbers starting with 7 and 6 is $69+79=148$. The number 147 is only 1 less than 148, so the numbers $a$ and $b$ are 79 and 68 or 78 and 69.

We will show that $a$ is not equal to 79. Suppose $a=79$. The sum $a+c \leqslant 129$, so $c \leqslant 129-79=50$. But then the sum $b+c \leqslant 69+50=119$ cannot start with 12. Therefore, $a$ is not equal to 79.

Thus, $a=78$ and $b=69$. If we take $c=51$, we get a set that fits the condition.

Comment: The problem did not require this, but let's prove that nothing other than $c=51$ works. Note that $c$ is not equal to 50, because otherwise the sum $b+c \leqslant 69+50=119$ cannot start with 12. Also, $c<52$ because otherwise the sum $a+c \geqslant 78+52=130$ cannot start with 12. Therefore, $c=51$.