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# How many positive two-digit integers are factors of 2^24 - 1?

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Math Expert
Joined: 02 Sep 2009
Posts: 55609
How many positive two-digit integers are factors of 2^24 - 1?  [#permalink]

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17 Apr 2019, 23:42
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Difficulty:

95% (hard)

Question Stats:

17% (02:15) correct 83% (02:25) wrong based on 29 sessions

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How many positive two-digit integers are factors of $$2^{24}-1$$?

(A) 4
(B) 8
(C) 10
(D) 12
(E) 14

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Math Expert
Joined: 02 Aug 2009
Posts: 7735
How many positive two-digit integers are factors of 2^24 - 1?  [#permalink]

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20 Apr 2019, 19:27
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1
Bunuel wrote:
How many positive two-digit integers are factors of $$2^{24}-1$$?

(A) 4
(B) 8
(C) 10
(D) 12
(E) 14

$$2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^{12}+1)=63*65*(64*64+1)=3*3*7*5*13*4097=3*3*5*7*13*17*241$$

We are looking for two digits factors..
So let us start from the biggest..
1) 241..None
2) 17...17, 17*3, 17*5.....three
3) 13.....13, 13*3, 13*5, 13*7.....four
4) 7.......7*3, 7*5, 7*3*3, ......three
5) 5.....5*3, 5*3*3.....two

Total 3+4+3+2=12

D
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Re: How many positive two-digit integers are factors of 2^24 - 1?  [#permalink]

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20 Apr 2019, 21:10
1
(a^24-1)= (a^12-1)(a^12+1)
=(a^6-1)(a^6+1)(a^4+1)(a^8+1-a^4)
(2^24-1) = (2^6-1)(2^6+1)(2^4+1)(2^8+1-2^4)
=63*65*17*241=(3^2)*5*7*13*17*241
Now 2 digits factors of (2^24-1)=3*5, 3*7, 3*13, 3*17, (3^2)*5, (3^2)*7, 5*7, 5*13, 5*17, 7*13, 13*1, 17*1
Hence 12 values are possible.
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Re: How many positive two-digit integers are factors of 2^24 - 1?  [#permalink]

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21 Apr 2019, 03:39
1
How to tackle a question like this in actual GMAT test? I started on the right track but got stuck because i didn't know the square of 64. I went ahead anyway and calculated the square and ended up with 4097 whose factors i needed to find to arrive at the solution, but at this point of time it was already way past the 3 minutes. I would have surely guessed and jumped to the next question in actual GMAT condition.
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Re: How many positive two-digit integers are factors of 2^24 - 1?  [#permalink]

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21 Apr 2019, 04:07
1
you could further factorize $$(2^{12})+1= [(2^4)^3]+1^3$$ by using formula $$a^3 + b^3=(a+b) (a^2 +b^2-ab)$$
In this way you neither have to calculate 64^2 nor have to factorize 4097.( both are time consuming)

prashant212 wrote:
How to tackle a question like this in actual GMAT test? I started on the right track but got stuck because i didn't know the square of 64. I went ahead anyway and calculated the square and ended up with 4097 whose factors i needed to find to arrive at the solution, but at this point of time it was already way past the 3 minutes. I would have surely guessed and jumped to the next question in actual GMAT condition.
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Posts: 7735
Re: How many positive two-digit integers are factors of 2^24 - 1?  [#permalink]

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21 Apr 2019, 04:35
prashant212 wrote:
How to tackle a question like this in actual GMAT test? I started on the right track but got stuck because i didn't know the square of 64. I went ahead anyway and calculated the square and ended up with 4097 whose factors i needed to find to arrive at the solution, but at this point of time it was already way past the 3 minutes. I would have surely guessed and jumped to the next question in actual GMAT condition.

Hi,

Neither do you require to know the factors of 4097 nor hidden formulas or a^3+b^3 etc.
You will never see such a question in GMAT.
Yes you should know how to tackle 2-digit positive factors from all factors.
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Re: How many positive two-digit integers are factors of 2^24 - 1?   [#permalink] 21 Apr 2019, 04:35
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