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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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Updated on: 05 May 2013, 21:56
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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
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Originally posted by GMATtracted on 05 May 2013, 10:38.
Last edited by Bunuel on 05 May 2013, 21:56, edited 1 time in total.
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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by
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06 May 2013, 09:12
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 ... lynumber/
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Re: GMAT quant topic
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05 May 2013, 11:58
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




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Re: GMAT quant topic
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05 May 2013, 11:59
Another method can be factors of 50  factors of 100 18901620=270



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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by
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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 ... lynumber/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!



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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by
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21 May 2013, 22:50
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 ... lynumber/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.
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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by
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22 May 2013, 01:25



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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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31 Aug 2015, 08:11
here, we have to catch all factors for 50.
100=2^2*5^2 50=2*5^2
Now, to have number divisible by 50 and not by 100 we have to make sure that 2 is taken only once. As, if we pick 2 two's then it will form 100 with no. of 5 selected and since min 5 we have to select is 2, we cannot have 2 more than one.
Number= 2^7*3^5*5^6*7^8
So, there are 6 ways of picking 3, 9 ways of picking 7, 5 ways of picking 5 and only one way of picking 2 and that is 2^1.
Why only 5 ways for picking 5 when we have 7 ways(0,1,2,3,4,5,6)?
Reason: we need 50 so we have to pick 5^2 and greater than it. Also, we can pick only 2^1 because picking 2^2 will form 100 with 5 already selected. For example, 5^2*2^2=100. So, we cannot afford it.
Hence, total number of factors.
= 6(3's)*9(7's)*5(5's)*1(2's)=270
Hope it was helpful.
Thanks, Gaurav



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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by
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06 Apr 2016, 10:12
If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?
50=2*5^2 100=2^2*5^2 N=(2*2^6)*(5^2*5^4)*(3^5)*(7^8) =(2*5^2)*(2^6*5^4*3^5*7^8) We dont need any other power of 2 except (2*5^2) no of factors (power of 5+1)*(power of 3+1)*(power of 7+1)= 5*6*9=270
Hence C



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Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by
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25 Aug 2018, 17:06
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