Find the sum of the values of A, B and C if ABC=A!+B!+C! where ABC is a three…

A number is called a factorion when it equals the sum of the factorials of its own digits. For a three-digit number with hundreds digit A, tens digit B and units digit C, this means 100A + 10B + C = A! + B! + C!. Because factorial values grow extremely fast, only digits 0 to 6 can ever appear in such an equation — 7! = 5040 alone already exceeds every three-digit number, so 7, 8 and 9 are ruled out immediately.

1. Bound the digits: a three-digit number is at most 999, and since 6! = 720 fits under that limit but 7! = 5040 does not, every digit here must be 0, 1, 2, 3, 4, 5, or 6.

2. Rule out 6 with a positional check: if A = 6, then ABC is below 700, yet the factorial sum already includes 6! = 720 — too big, a contradiction. If 6 appears as B or C instead (with A at most 5), then ABC is below 600, but the factorial sum still includes 720 from that one digit — again too big. So no digit can be 6, leaving 0 through 5 for every digit.

3. Show a digit must be 5: without any digit equal to 5, every digit is at most 4, and the largest possible factorial sum then is 4! + 4! + 4! = 72, which is too small to equal any three-digit number (the smallest is 100). So at least one of A, B, C must be 5, contributing 120 to the sum.

Since one digit must be 5, exhaustively check every possible pair of partner digits (X, Y), each from 0 to 5, against the actual three-digit number their combined digit set could form:

Only the pair (1, 4) succeeds — confirming this is the sole digit combination among every possibility that satisfies the equation.

Cross-check: substituting back, 1! + 4! + 5! = 1 + 24 + 120 = 145, which matches the digits 1, 4, 5 forming the number 145 exactly — confirming the digit set is genuine and not a coincidence of digit order.

Sum of the digits: 1 + 4 + 5 = 10.