# Find n and k in n!+8=2^k

• MHB
Gold Member
MHB
Find all pairs of integers n and k such that $n!+8 = 2^k$

Last edited:

Homework Helper
MHB
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.

Gold Member
MHB
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:

Homework Helper
MHB
I really don't like that ISPOILER effect! :sick:
It's new in Xenforo. I felt I just had to try it out. 😢

Gold Member
MHB
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

Gold Member
MHB
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.

Arlynnnn
Where's the solutions?

Gold Member
MHB
Where's the solutions?
If you click on the blurred sections in the above comments, they will become clear and reveal the solutions.

Arlynnnn