3-5. The factorial of a number $n$ is defined as the product of all integers from 1 to $n$ inclusive. Find all three-digit numbers that are equal to the sum of the factorials of their digits.
Solve 5. Answer: 145. Let $N=100x+10y+z$ be the desired number, for which $N=x!+y!+z!$. The number $7!=5040$ is four-digit, so no digit of the number $N$ exceeds 6. Therefore, the number $N$ is less than 700. But then no digit of the number $N$ exceeds 5, since $6!=720$. The inequality $3 \cdot 4!=72<100$ shows that at least one digit of the number $N$ is 5. At the same time, $x \neq 5$, since $3 \cdot 5!=360<500$. This inequality also shows that $x \leqslant 3$. Moreover, $x \leqslant 2$, since $3!+2 \cdot 5!=246<300$. The number 255 does not satisfy the condition of the problem, and if only one digit of the desired number is 5, then $x \leqslant 1$, since $2!+5!+4!=146<200$. Since $1!+5!+4!=145<150$, we get $y \leqslant 4$. Therefore, $z=5$. Considering that $x=1$ and $0 \leqslant y \leqslant 4$, we find the unique solution $N=145$.

## Second Round

## 7 - 8 grades