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

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The discussion focuses on finding integer pairs (n, k) that satisfy the equation n! + 8 = 2^k. The solutions reveal that for n = 0, 1, 2, and 3, the corresponding values of k are 3, 4, 5, and 6 respectively. For n ≥ 4, n! becomes even, making n! + 8 odd, which cannot equal 2^k. Thus, the only valid pairs are (0, 3), (1, 4), (2, 5), and (3, 6).

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Find all pairs of integers n and k such that $n!+8 = 2^k$
 
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kaliprasad said:
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.
 
Klaas van Aarsen said:
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:
 
Opalg said:
I really don't like that ISPOILER effect! :sick:
It's new in Xenforo. I felt I just had to try it out. 😢
 
Klaas van Aarsen said:
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
 
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.
 
Where's the solutions?
 
Arlynnnn said:
Where's the solutions?
If you click on the blurred sections in the above comments, they will become clear and reveal the solutions.
 
your solution is correct. please mention the steps
 

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