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

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In summary, n and k are variables in the equation n+8=2^k which can be solved using algebraic methods to understand the relationship between the numbers. It can also be applied in various fields to calculate exponential growth or decay and can be represented in other ways such as n=2^k-8.
  • #1
kaliprasad
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Find all pairs of integers n and k such that $n!+8 = 2^k$
 
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  • #2
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.
 
  • #3
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:
 
  • #4
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. 😢
 
  • #5
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
 
  • #6
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.
 
  • #7
Where's the solutions?
 
  • #8
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.
 
  • #9
your solution is correct. please mention the steps
 

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

1. What does "n" and "k" represent in the equation n+8=2^k?

In this equation, "n" represents an unknown number and "k" represents the exponent of the power of 2.

2. How can I solve for the values of n and k in this equation?

To solve for the values of n and k, you can use algebraic manipulation and substitution. First, subtract 8 from both sides of the equation to get n=2^k-8. Then, you can substitute this value of n into the second equation, n!+8=2^k, to get (2^k-8)!+8=2^k. From here, you can use mathematical techniques such as factoring, expanding, and simplifying to solve for the values of n and k.

3. Can there be multiple solutions for n and k in this equation?

Yes, there can be multiple solutions for n and k in this equation. This is because there are many different combinations of numbers that can be raised to a power of 2 and then have 8 added to them to equal another power of 2. For example, if n=4 and k=4, the equation becomes 4+8=2^4 and 4!+8=2^4, but if n=3 and k=5, the equation also holds true as 3+8=2^5 and 3!+8=2^5.

4. Is there a specific method or formula for solving this equation?

There is not a specific method or formula for solving this equation. It requires a combination of algebraic manipulation and mathematical techniques such as factoring and simplifying to solve for the values of n and k.

5. What are some real-world applications of this equation?

This equation can be used in various fields of science and mathematics, such as computer science, cryptography, and number theory. It can also be used to solve problems involving combinations and permutations, as well as in the study of prime numbers and their properties.

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