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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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 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

Values which we are looking for are 3,6,9,12,.. all multiples till 30 now everything will give you atleast 1 power of 3 but there are values which will give more than 1 power of 3 and those values will be multiples of 9 9 -> will give 2 powers 18-> will give 2 powers 27 -> will give 3 powers NUmber of single powers of 3 = (30-3)/3 +1 - 1(for 9) - 1(for 18) -1(for 27) = 7 so total powers = 2(for 9) + 2(for 18) + 3(for 27) + 7 = 14

Thanks for your reply but this is the explaation as provided in the OG. I wanted to know if there are other ways of approaching the problem since this method might be cumbersome we encounter larger numbers.

nktdotgupta wrote:

If p is the product of 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

Values which we are looking for are 3,6,9,12,.. all multiples till 30 now everything will give you atleast 1 power of 3 but there are values which will give more than 1 power of 3 and those values will be multiples of 9 9 -> will give 2 powers 18-> will give 2 powers 27 -> will give 3 powers NUmber of single powers of 3 = (30-3)/3 +1 - 1(for 9) - 1(for 18) -1(for 27) = 7 so total powers = 2(for 9) + 2(for 18) + 3(for 27) + 7 = 14

This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?

Finding the number of powers of a prime number k, in the n!.

The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!.

Back to the original question: If p is the product of 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

Given \(p=30!\).

Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 14.

This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?

Finding the number of powers of a prime number k, in the n!.

The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!.

Back to the original question: If p is the product of 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

Given \(p=30!\).

Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 1.

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

Show Tags

22 Oct 2013, 12:20

Correct me if I am mistaken but understanding number properties seems to be in line with these type of problems.

Going back to basics - these type of problems seem to relate to the factor foundation rule - if a is a factor of b and b is a factor of c then a is a factor of c.

In these problems you are trying to find the greatest degree of a common prime- in this case- of 3 in 30!.

This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?

Finding the number of powers of a prime number k, in the n!.

The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!.

Back to the original question: If p is the product of 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

Given \(p=30!\).

Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 1.

Answer: C.

Hope it's clear.

Hi Bunuel,

This explanation makes complete sense but i'm curious as to why another method doesn't work.

I first found the number of integers (30-1+1) = 30 Average of 30+1/2 = 15.5

30 * 15.5 = 465

Factoring 465 only gave me one "3" and therefore I knew it was wrong but I had wasted valuable time. I realized that what I had done doesn't account for the product but rather the sumof the consecutive integers. Correct?

If the question was changed just a bit to say that "p" is the sum of the integers from 1 - 30 then my method would've worked and the answer would've been 1. Is that correct?

This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?

Finding the number of powers of a prime number k, in the n!.

The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!.

Back to the original question: If p is the product of 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

Given \(p=30!\).

Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 1.

Answer: C.

Hope it's clear.

Hi Bunuel,

This explanation makes complete sense but i'm curious as to why another method doesn't work.

I first found the number of integers (30-1+1) = 30 Average of 30+1/2 = 15.5

30 * 15.5 = 465

Factoring 465 only gave me one "3" and therefore I knew it was wrong but I had wasted valuable time. I realized that what I had done doesn't account for the product but rather the sumof the consecutive integers. Correct?

If the question was changed just a bit to say that "p" is the sum of the integers from 1 - 30 then my method would've worked and the answer would've been 1. Is that correct?

Yes, you are correct. If it were "p is the SUM of integers from 1 to 30, inclusive", then the answer would be 1.
_________________

This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?

Finding the number of powers of a prime number k, in the n!.

The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!.

Back to the original question: If p is the product of 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

Given \(p=30!\).

Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 1.

Answer: C.

Hope it's clear.

Br

Bunuel..."So the highest power of 3 in 30! is 1."...You mean 14 right?

This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?

Finding the number of powers of a prime number k, in the n!.

The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!.

Back to the original question: If p is the product of 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

Given \(p=30!\).

Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 1.

Answer: C.

Hope it's clear.

Br

Bunuel..."So the highest power of 3 in 30! is 1."...You mean 14 right?

Yes, it's a typo. Edited. Thank you.
_________________

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

Show Tags

14 Aug 2015, 21:30

ajju2688 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) 18

you have to do successive division by 3. The given product is 30! divide 30 by 3; the quotient is 10 divide 10 by 3; the quotient is 3 divide 3 by 3; the quotient is 1 add all the quotients, 10+3+1=14

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

Show Tags

30 Oct 2015, 09:58

Looking at the possible answers I understood what I should be looking for.

But was I the only one confused with the question, since "a factor is an integer that divides another integer without leaving a remainder"? What I mean is: having in mind the definition of a "factor", I could be looking for the biggest "factor", resulting from 3^k, and not the highest power of 3 in 30 (or the number of 3 in the prime factorization of p).

This is probably a dumb question, but I'm probably missing something that you guys are not and I want to be sure that I see it from now on.

gmatclubot

Re: If p is the product of integers from 1 to 30, inclusive
[#permalink]
30 Oct 2015, 09:58

Campus visits play a crucial role in the MBA application process. It’s one thing to be passionate about one school but another to actually visit the campus, talk...

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...