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Factorials and 2 powers (Posted on 2021-08-13) Difficulty: 3 of 5
Find all pairs (k,n) of positive integers such that

k!=(2n-1)(2n-2)(2n-4)...(2n-2n-1)

No Solution Yet Submitted by Danish Ahmed Khan    
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Some Thoughts Poor attempt at an explanation | Comment 2 of 5 |
Tracking the number of factors of 2 and 3 on both sides of the equation:

On the RHS k! has approximately k factors of 2 and k/2 factors of 3.  So a given factorial has about twice as many factors of 2 as 3.
(Both are roughly linear)

On the LHS, the 2's grow very fast.  There are n(n-1)/2 factors of 2.
Quadratic growth
The 3's grow linearly again, this time to about 3k/4.

When n is small, the difference is small.  n doesn't need to be very big to make the difference significant.

For example when n=6, there are 15 factors of 2 and only 4 factors of 3.
k! has 15 factors of 2 when k=16, but k! has 4 factors of 3 when k=9,10, or 11.


  

  Posted by Jer on 2021-08-13 15:12:34
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