21.15. Find all three-digit numbers that are equal to the sum of the factorials of their digits.
21.15. Answer: 145. Let \( N = 100x + 10y + z \) be the desired number, for which \( N = x! + y! + z! \). The number \( 7! = 5040 \) is a four-digit number, so no digit of \( N \) exceeds 6. Therefore, the number \( N \) is less than 700. But then no digit of \( N \) exceeds 5, since \( 6! = 720 \). The inequality \( 3 \cdot 4! = 72 < 100 \) shows that at least one digit of \( N \) is equal to 5. At the same time, \( x \neq 5 \), since \( 3 \cdot 5! = 360 < 500 \). The equality \( 3 \cdot 5! = 360 \) also shows that \( x \leq 3 \). Moreover, \( x \leq 2 \), 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 equal to 5, then \( x \leq 1 \), since \( 2! + 5! + 4! = 146 < 200 \). Since \( 1! + 5! + 4! = 145 < 150 \), we get \( y \leq 4 \). Therefore, \( z = 5 \). Considering that \( x = 1 \) and \( 0 \leq y \leq 4 \), we find the unique solution \( N = 145 \).