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2^100 - 2^96=2^96 ( 2^4-1)=2^96*3*5

so greatest prime factor =5
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Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ?

Thanks!
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Bunuel
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Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ?

Thanks!

Sure.

What is the greatest prime factor of 2^100 - 2^96?

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

\(2^{100} - 2^{96}=2^4*2^{96}-2^{96}\).

Now, factor out 2^{96}: \(2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5\). The greatest prime factor is 5.

Answer: C.

Hope it's clear.

@bunuel..2^96-2^96=2^1 according to index laws..in our factorised equation 2^4*2^96-2^96 the answer is 16*2=32..can we do prime factorisation on 32 where by k/2 +k/3 +k/5--->32/2+32/3+32/5?would we just take the 5 as our largest factor?but we know 5 is not a factor of 32..where am i missing the details please tell me..thanks in advance :)

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chiccufrazer1
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Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ?

Thanks!

Sure.

What is the greatest prime factor of 2^100 - 2^96?

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

\(2^{100} - 2^{96}=2^4*2^{96}-2^{96}\).

Now, factor out 2^{96}: \(2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5\). The greatest prime factor is 5.

Answer: C.

Hope it's clear.

@bunuel..2^96-2^96=2^1 according to index laws..in our factorised equation 2^4*2^96-2^96 the answer is 16*2=32..can we do prime factorisation on 32 where by k/2 +k/3 +k/5--->32/2+32/3+32/5?would we just take the 5 as our largest factor?but we know 5 is not a factor of 32..where am i missing the details please tell me..thanks in advance :)

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Not sure I understand your post.

First of all, 2^96-2^96=0 not 2.

Next, \(2^4*2^{96}-2^{96}\) equals to \(2^{96}*3*5\) not 32.
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­Hi
Can we use the concept of cyclicity in this? 

2^100 will have units digit as 4
and so will 2^96
Subtracting, 4-4 = 0
So 0 is the units place of the difference 
Thus, out ouf all the options, 5 fits the best 

Please let me know­
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