Let be a 3 digit number, with , , and being distinct nonzero single digits, such that , where is the factorial notation
Find . (I.e, if you think the answer is that , you would enter your answer as .)
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None of the digits can be > 5 because 7 ! = 5 0 4 0 > 9 9 9 and 6 ! = 7 2 0 > 7 0 0 . Thus, our digits are bound to [ 1 , 5 ] . Observe that 4 ! + 3 ! + 2 ! = 3 2 < 1 0 0 , so one of the digits must be 5 . Also note that 1 ! + 2 ! + 3 ! + 4 ! + 5 ! = 1 5 3 , meaning that 1 0 0 ≤ A B C ≤ 1 5 3 , so A must equal 1 . Because A , B , C are distinct digits, our remaining digit is bound to [ 2 , 4 ] , making it easier to check for it. Let's examine the remaining cases: 1 B 5 or 1 5 C .
Case 1: A B C = 1 5 C
1 ! + 5 ! + C ! 1 2 1 + C ! ⟹ C ! − C = 1 5 C = 1 5 0 + C = 2 9
There exist no solutions C ∈ [ 2 , 4 ] for the above equation, so we must have A B C = 1 B 5 .
Case 2: A B C = 1 B 5
1 + B ! + 5 ! 1 2 1 + B ! ⟹ 1 0 B − B ! ⟹ B = 1 B 5 = 1 0 0 + 1 0 B + 5 = 1 6 = 4
We have A B C = 1 4 5 , and verify that, indeed, 1 ! + 4 ! + 5 ! = 1 4 5 .