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# If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by

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Intern
Joined: 06 Jan 2013
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If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]  05 May 2013, 10:38
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51% (03:36) correct 49% (02:29) wrong based on 79 sessions
If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240
B. 345
C. 270
D. 120
E. None of these
[Reveal] Spoiler: OA

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Last edited by Bunuel on 05 May 2013, 21:56, edited 1 time in total.
RENAMED THE TOPIC.
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Re: GMAT quant topic [#permalink]  05 May 2013, 11:58
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Numbers divisible by 50 not divisible by 100 would be divisible by $$2^1$$ and $$5^2$$, but power of 2 cannot be more than 1 as anything including $$2^2$$ and $$5^2$$ would be divisible by 100. So only 1 power of 2 allowed(2^1) and 5 powers of 5($$5^2$$,$$5^3$$,$$5^4$$,$$5^5$$,$$5^6$$)
=1*6*5*9
(1 powers of 2)*(6 powers of 3)*(5 powers of 5)*(9 powers of 7)
=270
Intern
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Re: GMAT quant topic [#permalink]  05 May 2013, 11:59
Another method can be
factors of 50 - factors of 100
1890-1620=270
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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]  06 May 2013, 09:12
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Expert's post
GMATtracted wrote:
If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240
B. 345
C. 270
D. 120
E. None of these

To be divisible by 50, the factor must have $$2*5^2$$ and to be not divisible by 100, it must NOT have $$2^2*5^2$$. Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2)

The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 270
1 - because 2 can have a power in only one way
6 - because 3 can have 6 different powers (0/1/2/3/4/5)
5 - because 5 can have 5 different powers (2/3/4/5/6)
9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8)

If you are still wondering how we got this product, check out this post: http://www.veritasprep.com/blog/2010/12 ... ly-number/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 15 May 2012 Posts: 41 Followers: 0 Kudos [?]: 5 [0], given: 94 Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink] 21 May 2013, 21:42 VeritasPrepKarishma wrote: GMATtracted wrote: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100? A. 240 B. 345 C. 270 D. 120 E. None of these To be divisible by 50, the factor must have $$2*5^2$$ and to be not divisible by 100, it must NOT have $$2^2*5^2$$. Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2) The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 270 1 - because 2 can have a power in only one way 6 - because 3 can have 6 different powers (0/1/2/3/4/5) 5 - because 5 can have 5 different powers (2/3/4/5/6) 9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8) If you are still wondering how we got this product, check out this post: http://www.veritasprep.com/blog/2010/12 ... ly-number/ Why can't 5 have 6 different powers. Because, you add 1 to the powers of 3 and 7, which I understand. Also, one 2 is understandable because more than one 2 will lead to divisibility by 100. Thanks! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5746 Location: Pune, India Followers: 1446 Kudos [?]: 7607 [0], given: 186 Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink] 21 May 2013, 22:50 Expert's post sharmila79 wrote: VeritasPrepKarishma wrote: GMATtracted wrote: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100? A. 240 B. 345 C. 270 D. 120 E. None of these To be divisible by 50, the factor must have $$2*5^2$$ and to be not divisible by 100, it must NOT have $$2^2*5^2$$. Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2) The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 270 1 - because 2 can have a power in only one way 6 - because 3 can have 6 different powers (0/1/2/3/4/5) 5 - because 5 can have 5 different powers (2/3/4/5/6) 9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8) If you are still wondering how we got this product, check out this post: http://www.veritasprep.com/blog/2010/12 ... ly-number/ Why can't 5 have 6 different powers. Because, you add 1 to the powers of 3 and 7, which I understand. Also, one 2 is understandable because more than one 2 will lead to divisibility by 100. Thanks! $$50 = 2*5^2$$ The power of 5 must be at least 2. It can be 3, 4, 5 or 6 as well. Here there are 5 different ways in which you can give a power to 5. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]  22 May 2013, 01:25
Expert's post
GMATtracted wrote:
If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240
B. 345
C. 270
D. 120
E. None of these

Similar question to practice: how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html

Hope it helps.
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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]  29 Jun 2014, 06:16
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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by   [#permalink] 29 Jun 2014, 06:16
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# If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by

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