15. (6 points) Given a three-digit number $\mathrm{abc}$, and $a(b+c)=33, b(a+c)=40$, then this three-digit number is $\qquad$
【Solution】Solution: $a \times(b+c)=33=3 \times 11$, there is only one possibility, $a=3$;
$$
b(a+c)-a(b+c)=40-33=7,
$$

which means $c \times(b-a)=7$, and $7=1 \times 7$, so $c \times(b-a)=1 \times 7$, there is only one possibility, $c=7$;
so $3 \times(b+7)=33$
$$
\begin{aligned}
b+7 & =11 \\
b & =4
\end{aligned}
$$

Therefore, the three-digit number is 347.
The answer is: 347.