Which are the three-digit numbers that are equal to the sum of the factorials of their digits?
I. solution. The digits of the three-digit number fall between 0 and 9, and 0 cannot be in the hundreds place. Let's write down the factorial of the first few numbers:

$$
0!=1, \quad 1! \text { by convention } \quad 1!=1, \quad 2!=2, \quad 3!=6, \quad 4!=24, \quad 5!=120, \quad 6!=720, \quad 7!>1000
$$

Since $7!>1000$, the digits cannot include $7, 8, 9$.

Since $6!=720$, there cannot be a 6 among the digits, because then at least a 7 would have to be in the hundreds place, which we have excluded.

If there were no 5 among the digits, then according to the above, the sum of the three digits could be at most $3 \cdot 24=72$, which is not a three-digit number.

The number cannot consist of three 5s, because $555 \neq 5!+5!+5!=360$.

If two 5s were among the digits, then the sum of the factorials would be at least 241 and at most 264, so the third digit would have to be a 2 in the hundreds place, but $255 \neq 2!+5!+5!=242$.

If only one 5 is among the digits, then a 1 must stand in the hundreds place ( $120+24+244$ !, that is, $m=5$. Then $A=255 \neq f(A)=2!+2 \cdot 5!=242$.

If $a=1$, then $A=100+10 b+c=1+120+m$ !, and thus

$$
5!>100>10 b+c=21+m!>m!
$$

so $m \leq 4$. This means that

$$
10 b+c \leq 21+4!=45
$$

that is, $b \leq 4$. Since $\max (b, c)=5>b$, from this $c=5$ and $b=m \leq 4$. Now from (2)

$$
10 b+5=21+b! \quad \text { and thus } \quad 10 b-b!=16
$$

For the remaining possible values of $b=1,2,3,4$, this is satisfied in one case, when $b=4$. Thus, there is only one three-digit number $A$ for which $A=f(A)$, which is $145=1!+4!+5!$.