The number that equals the sum of the factorials of its digits

Question

Find a three-digit number that equals the sum of the factorials of its digits. For example, 123 would qualify if 1! + 2! + 3! were 123 (it is 9, so it does not). Then explain why such numbers can only have a handful of digits, however far you search.

Accepted Answer

145 = 1! + 4! + 5! = 1 + 24 + 120. Narrowing the search. Digits 7, 8 and 9 have factorials above 999, so a three-digit solution cannot contain them. A 6 contributes 720, leaving at most 279 for the other two digits' factorials, so those digits are at most 5, and the number would be at least 600: but 720 + 120 + 120 = 960 is the most you can make with a 6 and two 5s, and the combinations 6, 5, 5 and 6, 5, x and 6, x, y with small x, y all fail to match (for example 655 would need 720 + 120 + 120 = 960, not 655). So there is no 6. With digits at most 5, the largest possible sum is 3 × 120 = 360…

The number that equals the sum of the factorials of its digits

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