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.)

Puzzle Answer

| Comment 1 of 5

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

Thus, each of x and x! must be a power of 2.

However, for x>=3, it follows that x! cannot be a power of 2.

For x= 0, we have: 2^y = 0*0! = 0*1 =0, which does not yield a solution.

For x=1, we have: 2^y = 1!*1 = 1, so that: y= 0

For x=2, we have: 2^y = 2!*2 = 2*2 =4, so that: y =2

Therefore, (x,y) = (1,0), (2,2) are the only valid solutions to the given equation.

Posted by K Sengupta on 2024-03-26 08:02:46

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