(In reply to
Poor attempt at an explanation by Jer)
I think you had it with the 2s:
On the RHS k! has approximately k factors of 2
On the LHS, the 2's grow very fast. There are n(n-1)/2 factors of 2.
Note that here we are comparing like with like; number of factors of 2, versus number of factors of 2.
Thus for some n we need n>=n(n-1)/2, true only if n={0,1,2,3}.
If n>3, then RHS will always have more factors of 2 than LHS.
So the solutions given by Charlie, plus k=0, are exhaustive.
Edited on August 14, 2021, 10:06 pm
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Posted by broll
on 2021-08-14 22:03:28 |
re: Poor attempt at an explanation
