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I have a hard time ignoring that 2^28 and trusting that breaking down 63 into its prime factors will give me the right answer.

Is there a hard and fast rule for why you can forget about 2^28 and trust that the 7 derived from 63 is correct?
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rockzom
I have a hard time ignoring that 2^28 and trusting that breaking down 63 into its prime factors will give me the right answer.

Is there a hard and fast rule for why you can forget about 2^28 and trust that the 7 derived from 63 is correct?

\(4^{17}-2^{28}\) equals to \(2^{28}*3^2*7\), which means that the prime factors of this number are 2, 3, and 7, so the greatest prime factor is 7 (2^28=2*2*...*2, so this expression has only one prime: 2).
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I see. So essentially any time that you have an expression where all of the bases are prime, you can assume that the highest base would be the greatest prime factor?

For example, the expression \(7^7 * 13^3 * 17\) would have 17 as the greatest prime factor. Correct?
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rockzom
I see. So essentially any time that you have an expression where all of the bases are prime, you can assume that the highest base would be the greatest prime factor?

For example, the expression \(7^7 * 13^3 * 17\) would have 17 as the greatest prime factor. Correct?

How else? Exponentiation does not "produce" primes: if p is a prime number then p^12 or p^10000 will still have only one prime - p.
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4^17 - 2^28
= 2^34 - 2^28
= 2^28 * (2^6 - 1)
= (2*2*2*...) * (63)
= (2*2*2*...) * (3 * 3 * 7)

Greatest prime factor is 7
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Let's simplify (4)^17-2^28
={(2)^2}^17-(2)^28
=(2)^34-(2)^28
=(2)^28{(2)^6-1}
=(2)^28{64-1}
=(2)^28{63}
=(2)^28{3x3x7}

clearly 7
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What is the greatest prime factor of \(4^17 - 2^28\)?

(A) 2
(B) 3
(C) 5
(D) 7
(E) 11

4^17 can be written as 2^34.
Hence we have to find out the greatest prime factor of 2^34-2^28.
Take 2^28 as common.
2^28(2^6-1)
It will become, 2^28 * (64-1)
2^28 * 63
Greatest prime factor of 2^28=2
Greatest prime factor of 63 is 7.
2^28 * 7*9
Therefore the answer is 7.
Hence D
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thevenus

=(2)^34-(2)^28
=(2)^28{(2)^6-1}

I don't get the transition here... what's the process involved that these two equal one another??
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dbiersdo
thevenus

=(2)^34-(2)^28
=(2)^28{(2)^6-1}

I don't get the transition here... what's the process involved that these two equal one another??

\(4^{17}-2^{28}=2^{34}-2^{28}\) --> factor out 2^28: \(2^{34}-2^{28}=2^{28}(2^6-1)=2^{28}*63=2^{28}*3^2*7\) --> the greatest prime factor is 7.

Hope it's clear.
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Flexxice
What is the greatest prime factor of 4^17 - 2^28?

A) 2

B) 3

C) 5

D) 7

E) 11


First of all 4^17 = (2^2)^17 = 2^34

So, 4^17 - 2^28 = 2^34 - 2^28
= 2^28(2^6 - 1) [factored out 2^28]
= 2^28(2^3 + 1)(2^3 - 1) [factored again]
= 2^28(8 + 1)(8 - 1) [evaluated]
= 2^28(9)(7) [evaluated]
= 2^28(3)(3)(7)

So, as you can see, the greatest prime factor is 7

Answer = D

Cheers,
Brent
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Flexxice
What is the greatest prime factor of 4^17 - 2^28?

A) 2

B) 3

C) 5

D) 7

E) 11


First of all 4^17 = (2^2)^17 = 2^34

So, 4^17 - 2^28 = 2^34 - 2^28
= 2^28(2^6 - 1) [factored out 2^28]
= 2^28(2^3 + 1)(2^3 - 1) [factored again]
= 2^28(8 + 1)(8 - 1) [evaluated]
= 2^28(9)(7) [evaluated]
= 2^28(3)(3)(7)

So, as you can see, the greatest prime factor is 7

Answer = D

Cheers,
Brent


Thank you! I tried it in another way and I do not get why I got another answer than you did.

4^17 - 2^28 = 4^17 - 4^14 = 4^3 = 2^6

Because of that, I chose answer A.
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Flexxice
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Flexxice
What is the greatest prime factor of 4^17 - 2^28?

A) 2

B) 3

C) 5

D) 7

E) 11


First of all 4^17 = (2^2)^17 = 2^34

So, 4^17 - 2^28 = 2^34 - 2^28
= 2^28(2^6 - 1) [factored out 2^28]
= 2^28(2^3 + 1)(2^3 - 1) [factored again]
= 2^28(8 + 1)(8 - 1) [evaluated]
= 2^28(9)(7) [evaluated]
= 2^28(3)(3)(7)

So, as you can see, the greatest prime factor is 7

Answer = D

Cheers,
Brent


Thank you! I tried it in another way and I do not get why I got another answer than you did.

4^17 - 2^28 = 4^17 - 4^14 = 4^3 = 2^6

Because of that, I chose answer A.

Hi,
you are wrong in the highlighted portion--
\(\frac{4^{17}}{4^{14}}= 4^{17-14} =4^3\)
\(4^{17} - 4^{14} = 4^{14}(4^3-1) = 4^{14} *(64-1) = 4^{14}*63= 4^{14}*7*9..\)
so 7 is the answer..
You can not subtract the way you have done, it can be done ONLY if two terms are being divided as shown above..

a simpler examplewill be
\(2^3 -2^1\).. as per you it will be\(2^2=4.\).
BUT \(2^3-2^1=8-2=6..\)
hope it helps
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Flexxice


Thank you! I tried it in another way and I do not get why I got another answer than you did.

4^17 - 2^28 = 4^17 - 4^14 = 4^3 = 2^6

Because of that, I chose answer A.

We cannot simply subtract the terms.

4^17 - 4^14 = 4^14 (4^3 - 1) = 4^14*(64 - 1) = 4^14 * 63
Here we have taken 4^14 common from both the terms and written the remainder inside the brackets.
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Bunuel
student26
What is the greatest prime factor of 4^17 - 2^28?

A. 2
B. 3
C. 5
D. 7
E. 11

\(4^{17}-2^{28}=2^{34}-2^{28}=2^{28}(2^6-1)=2^{28}*63=2^{28}*3^2*7\) --> greatest prime factor is 7.

Answer: D.


Why is it (2^6-1)? :oops:
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Aline2289
Bunuel
student26
What is the greatest prime factor of 4^17 - 2^28?

A. 2
B. 3
C. 5
D. 7
E. 11

\(4^{17}-2^{28}=2^{34}-2^{28}=2^{28}(2^6-1)=2^{28}*63=2^{28}*3^2*7\) --> greatest prime factor is 7.

Answer: D.


Why is it (2^6-1)? :oops:

Added extra step:

\(4^{17}-2^{28}=2^{34}-2^{28}=2^{28}*2^6-2^{28}=2^{28}(2^6-1)=2^{28}*63=2^{28}*3^2*7\) --> greatest prime factor is 7.

So, we are factoring out \(2^{28}\) from \(2^{34}-2^{28}\).

Hope it's clear.
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student26
What is the greatest prime factor of 4^17 - 2^28?

A. 2
B. 3
C. 5
D. 7
E. 11

We need to determine the greatest prime factor of 4^17 – 2^28. We can start by breaking 4^17 into prime factors.

4^17 = (2^2)^17 = 2^34

Now our equation is as follows:

2^34 – 2^28

Note that the common factor in each term is 2^28; thus, the expression can be simplified as follows:

2^28(2^6 – 1)

2^28(64 – 1)

2^28(63)

2^28 x 9 x 7

2^28 x 3^2 x 7

We see that the greatest prime factor must be 7.

Answer: D
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A video explanation can be found here:
https://www.youtube.com/watch?v=gCL6DmaIqK4

The first step is noting that we want to work from a common base, i.e. instead of 4 and 2, we want 2^2 and 2. Now we have

(2^2)^17 - 2^28

What’s tested are your knowledge of exponent rules, factoring, and prime numbers. We have

2^34 – 2^28, which is

2^28(2^6 – 1) [note that 2^6 can be viewed more simply as 2^3*2^3, or 8*8], which is

2^28(64 – 1), which is

2^28 (63), which is

2^28 (3^2)(7)

Prime factors are 2, 3 and 7. Greatest prime factor is 7
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