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Re: If n is the greatest positive integer for which 2n is a fact [#permalink]

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25 Dec 2012, 09:18

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In the questions where you are supposed to find out the maximum power of a prime factor which is a factor of n!, keep one only on thing in mind: Let the prime number you are looking for be x, then: \(n/x + n/x^2 + n/x^3.........n/x^k\) where \(x^k\) <=n. If you apply this rule here, \(10/2 +10/2^2 +10/2^3\) or 5+2+1=8. Hence maximum power of 2 in \(10!\) will be 8.
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Re: If n is the greatest positive integer for which 2n is a fact [#permalink]

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09 Jan 2014, 12:22

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If n is the greatest positive integer for which 2^n is a factor of 10!, then n =?

A. 2 B. 4 C. 6 D. 8 E. 10

Solution: On a closer look this Qs is asking how many 2s are there in multiplication of 10! Now, as we know 10! can be written as -- 10.9.8.7.6.5.4.3.2.1

As this is a small multiplication ( To review , not to do), we can check how many 2s we have in this.., So, 10-1 (no.of 2s), 8-3(no.of 2s), 6-1(no.of 2s),4-2(no.of 2s),2-1(no.of 2s)

Re: If n is the greatest positive integer for which 2n is a fact [#permalink]

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14 May 2015, 14:05

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Many Test Takers get these types of questions wrong because they move too quickly through the work and don't do enough work on their pads. It's a relatively straight-forward prompt though...

We're essentially asked to find all the "2"s inside 10!

The 'key' to this question is to realize that some values have MORE THAN ONE 2 in them....

10! = (10)(9)(8)(7)(6)(5)(4)(3)(2)(1)

10 = 5x2 --> one 2 8 = 2x2x2 --> three 2s 6 = 3x2 --> one 2 4 = 2x2 --> two 2s 2 = 1x2 --> one 2

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