OK this is the final one. Question is:
What is the greatest prime factor of (4^17) - (2^28)?
Answer is, apparently, 7.
Again there is a VERY long way of working this out but there must be a short cut. There's obviously some relation between 4 and 2 when powered up (?!) but I can't get all the way to the answer.
Multo thanks.
2007-12-30 23:41:23 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Hmmm...
Well... 4^17 is also equal to 2^34...
So, you end up with 2^34 - 2^28...
Factor out 2^28 from both...
2^28 ( 2^6 - 1 ) since you can divide 2^28 out of 2^34.
You can ignore the 2^28 cause the GCF of 2^28 will automatically be 2.
2^6 - 1 = 63
The GCF of 63 (that's prime), is 7. :)
2007-12-30 23:55:23 · answer #1 · answered by "Speedy" 4 · 1⤊ 1⤋
First, note that 4^17 = (2²)^17 = 2^34.
So N= 4^17 - 2^28 = 2^34 - 2^28.
Now 2^28 is a common factor of both terms,
so N = 2^28(2^6-1) = 2^28*63 = 2^28*3² * 7,
so your answer is 7.
2007-12-31 12:23:00 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
greatest prime factor of 4^17 -- 2^28 = 2^28(2^6 -- 1)
= 2^28(63) = 7*9*2^28
greatest prime factor is 7
2007-12-31 07:52:15 · answer #3 · answered by sv 7 · 1⤊ 0⤋
4^17 = 2^34 !
So the question is 2^34 - 2^28
So then you have 2^28 * (2^6 - 1)
2^6 - 1 = 63
Greatest primary of 63 is 7 .
2007-12-31 07:50:52 · answer #4 · answered by Anonymous · 2⤊ 0⤋
4^16=2^34
therefore 2^34-2^28=2^28(2^6-1)=2^28*63
since 63=3^2*7
therefore the largest prime is 7
2007-12-31 07:48:55 · answer #5 · answered by someone else 7 · 0⤊ 3⤋
2^34-2^28
=2^28(2^6-1)
=2^28 ( 63)
primes are 2, 3, 7
2007-12-31 07:48:36 · answer #6 · answered by iyiogrenci 6 · 1⤊ 1⤋
4^17 = (2^2)^17 = 2^(17*2) = 2^34
Now,
2^34 - 2^28 = 2^28(2^6 - 1) = 2^28(64-1) = 63 * 2^28
= 7 * 9 * 2^28
= 7 * 3 * 3 * 2^28
Therefore, 7 is the greatest prime number here. The other remaining numbers are 3 and 2 which are lesser than 7.
Hope this helps
2007-12-31 07:48:10 · answer #7 · answered by {flick} 3 · 2⤊ 1⤋