Find the greatest common factor of all nubmers of the form 2^n*3^(2n) - 1, where n is an integer greater than 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.
The greatest GCF = 2^4 = 16
+118609 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.
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
+118609 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?
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.
+118609 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?
