+0  
 
0
58
9
avatar

Find the greatest common factor of all nubmers of the form 2^n*3^(2n) - 1, where n is an integer greater than 1.

 Dec 24, 2019
 #1
avatar
0

Find the greatest common factor of all nubmers of the form 2^n*3^(2n) - 1, where n is an integer greater than 1.

 

The greatest GCF = 2^4  = 16

 Dec 24, 2019
 #2
avatar+106887 
+1

Find the greatest common factor of all nubmers of the form 2^n*3^(2n) - 1, where n is an integer greater than 1.

 

\(2^n*3^{2n} - 1=2^n*9^n-1=18^n-1\) 

 

I think the greatest common factor is probably 1.  

I don't know how to show or prove it though. 

 Dec 25, 2019
 #3
avatar
+1

Hi Melody: The way you understand it, 2^n*3^(2n) - 1, it gives the following 10 terms:

 

GCD = (17, 323, 5831, 104975, 1889567, 34012223, 612220031, 11019960575, 198359290367, 3570467226623) = 17

 Dec 25, 2019
 #4
avatar+106887 
+1

Thanks guest, I'll take another  look.

 

 

2^n*3^(2n) - 1

\(2^n*3^{2n} - 1=18^n-1 \)

 

n 2 3 4 5 6
f(n) 323 5831 104975    
factor(f(n)) 17* 19 17*7^3 17* ...    

 

ok, your answer could well be correct, it certainly works for the first 3 terms,

This is not a proof that it will always be so.  Do you know how to prove it?

Melody  Dec 25, 2019
 #6
avatar
+1

Hi Melody: I'm not sure if I know how to prove it formally, but notice the following:

 

1 - Your result of 18^n - 1 = 17 when n = 1

2 - The result of 18^2 - 1 = 17 x 19

3 - The result of 18^3 - 1 = 17 x 7^3

.

.

.

10 - The result of 18^10 - 1 =17 x  11 x 19........etc.

Because you are subtracting 1 from 18 =17, no matter what value n takes, the result will ALWAYS have a 17 as a prime factor. I don't know if this makes sense mathematically, but it seems clear to me that you will always have 17 as factor for any value of n.

Guest Dec 25, 2019
 #7
avatar+106887 
+1

Hi Guest,

using your logic

 

f(n)=15^n-1      would always have a facter of  14 . is this true?

 

if n =1 it is true

if n=2 then f(2)=224   yes, that is divisable by 14

if n=3 then f(3)=3374  yes that is divisable by 14

 

I am surprised, I did not expect that to work.

I am still not completely convinced that this will always be the case but maybe it is.

 

I am intrigued.

Can anyone prove that this will always be the case?

Melody  Dec 25, 2019
 #8
avatar+76 
+2

Hi Melody

 

Consider it from the point of view of the equation

\(\displaystyle x^{n}-1=(x-1)k(x). \)

All that you are looking for is that \(\displaystyle (x-1) \text{ is a factor of }x^{n}-1.\)  

Tiggsy  Dec 25, 2019
 #9
avatar+106887 
0

Of course, thanks Tiggsy.  

 

Much appreciated.

Melody  Dec 26, 2019

17 Online Users

avatar