Multiplication of Factorials

  • Thread starter msudidi
  • Start date
  • #1
2
0
Given that a, b, and c are positive integers solve the following equation.

a!b! = a! + b! + c^2

anyone?
 

Answers and Replies

  • #2
773
0
I found the answer through brute force: a=2, b=3, c=2.

Not sure if there is a more elegant solution though.
 
  • #3
2
0
vorde, thanks for trying, im getting the same answer too:smile:

but i'm seeking for method to solve it: how to relate multiplication of 2 factorials and their sums? if we can, then c wouldn't be a problem.

there should be a way to solve it:rolleyes:
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
956
"Brute force" is a method! Please clarify what you are looking for.
 
  • #5
611
23
Given that a, b, and c are positive integers solve the following equation.

a!b! = a! + b! + c^2

anyone?
Doesn't it work for all positive integers c such that [itex]c = \sqrt{a!b! -a! -b!}[/itex]? :biggrin:

Spit-balling here, we have [itex]a!b! = a! + b! + c^2[/itex]? Doesn't that imply that [itex]\displaystyle a! = 1 + \frac{a! + c^2}{b!} = 1 + \frac{a(a-1)(a-2)(a-3)...}{b(b-1)(b-2)...} + \frac{c^2}{b!} = 1 + \prod_{k = (b+1)}^a k + \frac{c^2}{b!}[/itex]. Don't know where I'm going with that...
 

Related Threads on Multiplication of Factorials

Replies
6
Views
1K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
1
Views
13K
  • Last Post
2
Replies
27
Views
38K
  • Last Post
Replies
2
Views
9K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
13K
  • Last Post
Replies
2
Views
490
  • Last Post
Replies
2
Views
2K
Top