MHB What are the values of n and k in n+8=2^k when n!+8 = 2^k?

[kaliprasad]

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

 

[I like Serena]

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]

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:

 

[I like Serena]

I really don't like that ISPOILER effect! :sick:

It's new in Xenforo. I felt I just had to try it out.

 

[kaliprasad]

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

Spoiler

your solution is correct. please mention the steps

 

[Opalg]

Spoiler

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?

 

[Opalg]

Where's the solutions?

If you click on the blurred sections in the above comments, they will become clear and reveal the solutions.

 

[Arlynnnn]

your solution is correct. please mention the steps