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I have seen the solution to this problem.

What is the greatest prime factor of $ \ 4^{17} - 2^{28} \ $?

Answer: 7

$$ 4^{17}-2^{28} \ = \ 2^{34}-2^{28} \ = \ 2^{28} \ (2^6-1) \ = \ 2^{28} \ \cdot \ 63 \ = \ 2^{28}\ \cdot \ 3^2 \ \cdot \ 7 $$

I understand the problem up to this part: $2^{28} \ (2^6-1) \ $ . Why did this occur? I can't explain it.

Thanks!

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3 Answers 3

$2^{28}$ is a common factor between the two terms. Thus, it can be factored out.

You can factor anything out incidentally: if you factored out a $3$, you'd get

$$2^{34} - 2^{28} = 3 \left( \frac{2^{34}}{3} - \frac{2^{28}}{3} \right) $$

of course, for the purposes of working on this question, factoring out a $3$ is not very useful. But since $2$ divides both terms $28$ times, factoring $2^{28}$ might be a simplification, and does turn out to be quite useful here.

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$$4^{17} = (2^2)^{17} = 2^{2 \cdot 17} = 2^{34} = 2^{28 + 6} = 2^{28} \cdot 2^6$$

Now recognize the common factor of $2^{28}$ and factor it out.

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You are asking how you went from $2^{34} - 2^{28} = 2^{28}(2^6-1)$? You are factoring out $2^{28}$ from both of the numbers. $2^{34} - 2^{28}=(2^{34} - 2^{28})*\frac{2^{28}}{2^{28}}=\frac{2^{34} - 2^{28}}{2^{28}}*2^{28}=(2^6-1)*2^{28}$

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