What is the greatest prime factor of 4^142^28?
2
3
5
7
11
GREATEST PRIME FACTOR
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Is the question correct?. From how I read the question when you simplify the equation it comes out to be zero. 4^14  2^28 => 2^28 2^28 => 0.

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oh shoot i'm sorry, i typed it wrong. heres the right question...
What is the greatest prime factor of 4^172^28?
2
3
5
7
11
What is the greatest prime factor of 4^172^28?
2
3
5
7
11

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You can simplify the above equation as follows
4^17  2 ^ 28
(2^2)^17  2 ^ 28
2^34  2^28
this can be written as
2^28(2^6  1)
2^28(641)
2^28 * 63 => 2^28*7*3*3
The greatest prime number for the above would be 7. So the answer should be D.
4^17  2 ^ 28
(2^2)^17  2 ^ 28
2^34  2^28
this can be written as
2^28(2^6  1)
2^28(641)
2^28 * 63 => 2^28*7*3*3
The greatest prime number for the above would be 7. So the answer should be D.
For some reason I am missing how you went from step 3 to step 4 here. How does 2^34  2^28 become 2^28(2^6  1)?ramyaravindran wrote:You can simplify the above equation as follows
4^17  2 ^ 28
(2^2)^17  2 ^ 28
2^34  2^28
this can be written as
2^28(2^6  1)
2^28(641)
2^28 * 63 => 2^28*7*3*3
The greatest prime number for the above would be 7. So the answer should be D.
Thank you!
2^34=2^28 * 2^6uncbeers wrote: For some reason I am missing how you went from step 3 to step 4 here. How does 2^34  2^28 become 2^28(2^6  1)?
Thank you!
a^(b+c)= a^b * a^c
So we have
A= 2^34  2^28
A= 2^28 * 2^6  2^28
A= 2^28 ( 2^6 1)
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4^17 = (2^2)^17 = 2^34
2^34  2^28 = 2^28*(2^6  1) = 2^28*63 = 2^28 * 7 * 3^2
so 7
2^34  2^28 = 2^28*(2^6  1) = 2^28*63 = 2^28 * 7 * 3^2
so 7
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Ahhh.... I finally figured it out after writing it down on paper. You turned 2^34 into two number 2^6 and 2^28. Now the two 2^28 cancel each out out leaving 2^1 * 2^6
The rest is just simplifying.
Thanks!
The rest is just simplifying.
Thanks!