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[SOLVED] find n and k in n!+8=2^k

kaliprasad

Well-known member
Mar 31, 2013
1,295
Find all pairs of integers n and k such that $n!+8 = 2^k$
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Find integers n and k such that $n!+8 = 2^k$
From inspection I can find $n=4,\,k=5$.
As a bonus I can see that $n=5,\,k=7$ works as well.
I kind of suspect that you intended to find all such integers, did you? Ah well, that was not the question.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,697
From inspection I can find $n=4,\,k=5$.
As a bonus I can see that $n=5,\,k=7$ works as well.
I kind of suspect that you intended to find all such integers, did you? Ah well, that was not the question.
I really don't like that ISPOILER effect! :sick:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736

kaliprasad

Well-known member
Mar 31, 2013
1,295
From inspection I can find $n=4,\,k=5$.
As a bonus I can see that $n=5,\,k=7$ works as well.
I kind of suspect that you intended to find all such integers, did you? Ah well, that was not the question.
I have edited the question to make it more clear

your solution is correct. please mention the steps
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,697
If $k>3$ then $2^k-8 = 8(2^{k-3} - 1)$, which is an odd multiple of $8$ and therefore not divisible by $16$. But if $n>5$ then $n!$ is divisible by $2*4*6$, which is a multiple of $16$. Those two facts ensure that there are no more solutions of $n! + 8 = 2^k$ apart from the two already mentioned.