perplexus.info :: Just Math : Some factorial and some power

Some factorial and some power (Posted on 2024-03-26)

Find all possible non-negative integer solution (x,y) of the following equation

x! + 2^y = (x+1)!

No Solution Yet Submitted by Danish Ahmed Khan

Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Another solution

| Comment 2 of 5 |

Assume x>=3. Then take each side mod 3.

x! and (x+1)! are congruent to 0 mod 3 while 2^y is congruent to 1 or 2 mod 3.

Then the equation would reduce to 1=0 mod 3 or 2=0 mod 3.

These are false so there are no solutions with x>=3.

Then x=0, 1, or 2.

If x=0 then 1+2^y=1, no solution for y.

If x=1 then 1+2^y=2, then 2^y=1 which makes y=0.

If x=2 then 2+2^y=6, then 2^y=4 which makes y=2.

The set of all non-negative integer solutions (x,y) of the given equation is (x,y)=(1,0) or (2,2).

Posted by Brian Smith on 2024-03-26 22:56:26

Please log in:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info