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ABC = A! + B! + C!

Cannot have 6, 7, 8, 9 as digits

At least one digit to be equal to 5

Exactly one digit is equal to 5

 Apr 26, 2019
 #1
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I don't think there is a solution for this using the "factorial function"!. However, there is a solution using the "Gamma function". It gives the following solution:

 

A*B*C =A! + B! + C!, solve for A, B, C

where: A ≈ 4.39056.... and B = -53/10 ≈ -5.30000       and C = -17/10 ≈ -1.70000

Note: This is courtesy of: Mathematica 11 Home Edition.

 Apr 26, 2019
 #2
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Assuming that ABC is a three digit number (rather than the product A*B*C), 

145 = 1! + 4! + 5! = 1 + 24 + 120 = 145.

 Apr 27, 2019
 #3
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Very clever, Tiggsy!.

 Apr 27, 2019

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