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

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Discussion Overview

The discussion revolves around finding integer pairs (n, k) that satisfy the equation n! + 8 = 2^k. Participants explore the implications of this equation, potentially delving into factorial properties and powers of two.

Discussion Character

  • Exploratory, Homework-related

Main Points Raised

  • Some participants seek to find all pairs of integers n and k that satisfy the equation n! + 8 = 2^k.
  • Others express frustration with the forum's ISPOILER feature, indicating a desire for clearer communication of solutions.
  • A participant mentions editing the question for clarity, suggesting that the original phrasing may have caused confusion.
  • There are repeated inquiries about the location of solutions, indicating a lack of visibility or accessibility of the answers within the thread.
  • One participant confirms that a proposed solution is correct but requests a detailed explanation of the steps involved.

Areas of Agreement / Disagreement

The discussion appears to lack consensus, with multiple participants expressing confusion and seeking clarity on the solutions, while others focus on the mathematical problem itself.

Contextual Notes

Some participants have noted issues with the presentation of solutions due to the ISPOILER feature, which may hinder the flow of information. There is also an indication that the question may have been unclear prior to edits.

kaliprasad
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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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