Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If p is the product of the integers from 1 to 30, inclusive, wha [#permalink]
11 Jul 2010, 01:36

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]
21 Jul 2010, 12:47

2

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

Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Re: Product of the Integers [#permalink]
12 Oct 2010, 23:17

monirjewel wrote:

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