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Factorials everywhere (Posted on 2004-08-16) Difficulty: 3 of 5
The equation a!b!=c! is trivial if we allow a or b to be 0 or 1.

However, adding that 1<a<b<c, also allows trivial solutions: which? What condition should be added to disallow such solutions?

Finally, adding that condition, can you find any solution to the problem?

See The Solution Submitted by Federico Kereki    
Rating: 2.6000 (5 votes)

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Solution Solution / Thoughts | Comment 1 of 6

There are an infinite number of trivial solutions. If we allow b to equal a! - 1, then c is equal to a!. For example, if a=3 and b=5, then since a!=6 and c!=5!*6=b!*a!, the solution is trivial. These trivial solutions get incalculable very quickly.

I don't believe there are any quick solutions if you eliminate that condition.


  Posted by Eric on 2004-08-16 09:21:19
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