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Non-negative Integers Crossed Expression Product puzzle (Posted on 2023-09-07) Difficulty: 2 of 5
Determine all possible pairs (x,y) of non-negative integers that satisfy this equation:
(x!+1)(y!+1)= (x+y)!

  Submitted by K Sengupta    
Rating: 5.0000 (1 votes)
Solution: (Hide)
(x,y) = (1,2) and, (2,1)

EXPLANATION:

Assume both x>=2 and y>=2. Then the left side of the equation will be odd and the right side will be even. This is a contradiction, so at least one of x and y is less than 2.

Without loss of generality, assume x<2. Then for either x=0 or x=1 we have x!=1 and the equation reduces to 2*(y!+1) = (y+1)! Now assume y>=3. Then working mod 3 the left side reduces to 2 and the right side reduces to 0. This is a contradiction. Therefore, y is less than 3.

This leaves exactly two cases to check, y=0, y=1 making y!=1
or y=2 making y!=2.
By direct evaluation only the last case works. Then (x,y)=(1,2) is a solution. Because x and y are symmetric then their reversals also work and (x,y)= (2,1) is also a solution.

Consequently, the possible pairs (x,y) of non-negative integers that satisfy the given equation are (1,2), and (2,1).

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: Another solutionLarry2023-09-07 12:40:32
SolutionAnother solutionBrian Smith2023-09-07 10:51:07
SolutionAnalytic solutionLarry2023-09-07 08:49:14
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