If $n$ is a positive integer, we denote by $n$ ! the product of all integers from 1 to $n$. For example, 5 ! = $1 \times 2 \times 3 \times 4 \times 5=120$ and 13 ! $=$ $1 \times 2 \times 3 \times 4 \times 5 \times \cdots \times 12 \times 13$. By convention, we write $0!=1!=1$. Find three integers $a, b$ and $c$ between 0 and 9, that are distinct and such that the three-digit number $a b c$ is equal to $a!+b!+c$!.
We want $a b c=a!+b!+c!$ with digits $0 \leq a, b, c \leq 9$. Since $0!=1!=1, 2!=2, 3!=6$, and $4!=24$, one of the digits $a, b$, or $c$ must be greater than 4, because $0!+1!+2!+3!+4!=34$ has only two digits. If one of the digits $a, b$, or $c$ is greater than or equal to 6, we will have $a b c=a!+b!+c!>6!=720$, which implies that one of the digits $a, b$, or $c$ is at least equal to 7; but then $a b c=a!+b!+c!>7!=5040$ has at least four digits, which is impossible.

Thus, one of $a, b$, and $c$ is equal to 5 and the others are less than 5. The smallest possible number is $5!+1!+0!=120+1+1=122$ and the largest possible number is $5!+3!+4!=120+6+24=150$. Therefore, the digit $a$ in the hundreds place is 1. If the digit $b$ in the tens place is 5, then $c \leq 4$ and

$$
1!+5!+c!=1+120+c!=121+c! \leq 121+4!=121+24=145 \neq 15c
$$

If the digit $b$ in the tens place is 0, 2, or 3, then $b!$ is equal to 1, 2, or 6, and since $c$ must be 5, we have that $1!+b!+5!=1+b!+120=121+b!$ is equal to 122, 123, or 127, all different from $1b5$. The only remaining option is $b=4$ and $c=5$. In this case, indeed $1!+4!+5!=1+24+120=145$, as we wanted. The three integers are $a=1, b=4$, and $c=5$.