If $n$ is a natural number, we denote by $n!$ the product of all integers from 1 to $n$. For example: $5! = 1 \times 2 \times 3 \times 4 \times 5$ and 13! $=$ $1 \times 2 \times 3 \times 4 \times 5 \times \ldots \times 12 \times 13$. By convention, $0! = 1$. Find three different integers $a, b$, and $c$, between 0 and 9 such that the three-digit number $abc$ is equal to $a! + b! + c!$.
Firstly, note that since the number has 3 digits, the largest digit must be less than or equal to 6, as $7!>1000$. Since the number must have 3 digits and $4!=1 \times 2 \times 3 \times 4=24$, one of the digits must be 5 or 6, but $6!=720$ would imply that the sum has a digit greater than or equal to 7, so the largest digit is 5. On the other hand, $5!=120$ and $5!+4!+3!=120+24+6=150$, so the sum of the factorials is between 100 and 150, thus the hundreds digit is 1. Finally, since $5!+1!+4!=145$, 145 is a solution.