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Solution for "That's something interesting!"

https://brilliant.org/problems/thats-something-interesting/?group=j0gXDzTOQ64o

Given: a!b! = a!+b! a>0; b>0; Solution: Factorial definition (http://en.wikipedia.org/wiki/Factorial) is a!=1234...a Let's assume a>b, so b!=a!(a+1)(a+2)(a+3)...b or b!=a!x, where x>1 Next a!b! = a!+b! => \({ (a!) }^{ 2 }\)x = (a!)(1+x) => x=\(\frac { 1 }{ a!-1 }\) .
x will never be greater than 1. Therefore, a=b => \({ (a!) }^{ 2 }\) = 2
(a!) => a! = 2 => a = 2, b = 2, a+b = 4

Note by Alex Gawkins
1 year, 11 months ago

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FYI You can enter your solution directly on the problem. After you solved it, and clicked on reveal solutions, you will see:

Calvin Lin Staff · 1 year, 11 months ago

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