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If p is the product of the integers from 1 to 30, inclusive, wha [#permalink]

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11 Jul 2010, 02:36

okay, here is the thing !

I was solving 12 O.G book then this question pop up and I had no shortcut for it ! I mean it took me long to do it !

If p is the product of the integers from 1 to 30,inclusive, what is the greatest integer k for which 3^k is a factor of p ?

a 10 b 12 c 14 d 16 e 18

I mean if I keep multiplaying each value in order to hit the right one, it will take a lot of time where in GMAT you must spend 1 minute for each a question, is there any shortcut for this type of question ?

Re: Can I please know the simple method to solve this problem ? [#permalink]

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21 Jul 2010, 13:47

3

This post received KUDOS

sstudy wrote:

Can anybody please tell me how to solve this one ?

Que :

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18

Thanks, sstudy

I think it should be 3^k not 3k else the highest number i.e. 18 in the option should be the answer Now lets take 3^k

power 3 in 30! is 30/3 + 30/3^2 ..... till 3^k < 30

=> power of 3 in 30! = 30/3 + 30/3^2 + 30/27 = 10 + 3 + 1 = 14 ( ignore decimals think logically why)

Hence the answer is 14

The general formula is

\(\frac{n}{k}+ \frac{n}{k^2} +\frac{n}{k^3}\) till \(n>k^m\)
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Helpful Geometry formula sheet: http://gmatclub.com/forum/best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?

A)10 B)12 C)14 D)16 E)20

I need some great solution to solve this problem quickly.

The answer (30/3)+(30/9)+(30/27)=10+3+1=14 Answer - (c)

The logic You need to count the factors of 3 in the product. So just count all the numbers divisible by 3, which will be (30/3) {3,6,9,...,30} Now all the numbers divisible by 3^2 or 9 will have 2 factors of 3 so need to count them once more, this will be (30/9) {9,18,27} And then finally the numbers divisible by 3^3, need to be counted once more, which is (30/27) {27}

No higher powers of 3 will divide numbers <=30, so we are done
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