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:

2. Finding the number of powers of a prime number k, in the n!. What is the power of 3 in 35!...

In the same way as for 5? i.e., 35/3 + 35/9 + 35/27 = 11 + 3 + 1 = 15.

Am I right?

Absolutely, here is the way to calculate the number of powers of a prime number k, in n!. The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

What is the power of 2 in 25! \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\)

There is another formula finding powers of non prime in n!, but think it's not needed for GMAT.
_________________

Re: GMATPrep DS Product of first 30 integers [#permalink]

Show Tags

30 Nov 2009, 19:07

1

This post received KUDOS

I think the answer is C.

S1 by itself is not sufficient, coz if d=1 means 10 is a factor of 30!, true, if d =2, 100 is also a factor of 30!, d can be 1,2 or more... so insuff S2 by itself is not sufficient, coz d>6 means d can be 7,8,9 or anything - clearly insuff

combining the two however we can asnwer the question, because in 30! we have 7 powers of 10 as below:

1.2.3.4.5 has one power for 10 (2*5) 6.7.8.9.10 has one power for 10 (10) 11.12.13.14.15 has one power for 10 (15*14 or 15*12) 16.17.18.19.20 has one power for 10 (20) 21.22.23.24.25 has 2 powers for 10 (25*24) 26.27.28.29.30 has one power for 10 (30) total of 7 so \(10^7\) is the highest \(10^d\) being fact of 30! hence d=7
_________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?

(1) 10^d is a factor of f --> \(k*10^d=30!\).

First we should find out how many zeros \(30!\) has, it's called trailing zeros. It can be determined by the power of \(5\) in the number \(30!\) --> \(\frac{30}{5}+\frac{30}{25}=6+1=7\) --> \(30!\) has \(7\) zeros.

\(k*10^d=n*10^7\), (where \(n\) is the product of other multiples of 30!) --> it tells us only that max possible value of \(d\) is \(7\). Not sufficient.

(2) \(d>6\) Not Sufficient.

(1)+(2) \(d>6\), \(d_{max}=7\) --> \(d=7\).

Answer: C.

For trailing zeros see the link about factorials below.
_________________

(1) k*10^d=30! First we should find out how many zeros does 30! has, it's called trailing zeros. It can be determined by the power of 5 in the number 30! --> 1 in 5, 1 in 10, 1 in 15, 1 in 20, 2 in 25=(5*5) , 1 in 30 --> 1+1+1+1+2+1=7 --> 30! has 7 zeros. k*10^d=n*10^7, n>=k it tells us only that max possible value of d is 7. Not Sufficient.

(2) d>6 Not Sufficient.

(1)+(2) d>6, dmax=7 --> d=7

C.

Bunuel, I did not get why k*10^d=n*10^7, n>=k is the case? why is n>=k? If we said k*10^d = n*10^7, then why is k=n and d=7 wrong?
_________________

(1) k*10^d=30! First we should find out how many zeros does 30! has, it's called trailing zeros. It can be determined by the power of 5 in the number 30! --> 1 in 5, 1 in 10, 1 in 15, 1 in 20, 2 in 25=(5*5) , 1 in 30 --> 1+1+1+1+2+1=7 --> 30! has 7 zeros. k*10^d=n*10^7, n>=k it tells us only that max possible value of d is 7. Not Sufficient.

(2) d>6 Not Sufficient.

(1)+(2) d>6, dmax=7 --> d=7

C.

Bunuel, I did not get why k*10^d=n*10^7, n>=k is the case? why is n>=k? If we said k*10^d = n*10^7, then why is k=n and d=7 wrong?

30! is some huge number with 7 trailing zeros (ending with 7 zeros). Statement (1) says that \(10^d\) is factor of this number, but \(10^d\) can be 10 (d=1) or 100 (d=2) ... or 10,000,000 (d=7). Basically \(d\) can be any integer from 1 to 7, inclusive (if \(d>7\) then \(10^d\) won't be a factor of 30! as 30! has only 7 zeros in the end). So we can not determine single numerical value of \(d\) from this statement. Hence this statement is not sufficient.

Is there anyway to quickly determine if d>7 is not a factor of 30! ?

If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?

(1) 10^d is a factor of f --> \(k*10^d=30!\).

First we should find out how many zeros \(30!\) has, it's called trailing zeros. It can be determined by the power of \(5\) in the number \(30!\) --> \(\frac{30}{5}+\frac{30}{25}=6+1=7\) --> \(30!\) has \(7\) zeros.

\(k*10^d=n*10^7\), (where \(n\) is the product of other multiples of 30!) --> it tells us only that max possible value of \(d\) is \(7\). Not sufficient.

Side notes: 30! is some huge number with 7 trailing zeros (ending with 7 zeros). Statement (1) says that \(10^d\) is factor of this number, but \(10^d\) can be 10 (d=1) or 100 (d=2) ... or 10,000,000 (d=7). Basically \(d\) can be any integer from 1 to 7, inclusive (if \(d>7\) then \(10^d\) won't be a factor of 30! as 30! has only 7 zeros in the end). So we can not determine single numerical value of \(d\) from this statement. Hence this statement is not sufficient.

(2) \(d>6\) Not Sufficient.

(1)+(2) \(d>6\), \(d_{max}=7\) --> \(d=7\).

Answer: C.

For trailing zeros see the link about factorials in my signature.

Re: DS : product of first 30 positive integers [#permalink]

Show Tags

11 Aug 2010, 18:06

masland wrote:

Hmm think I might have got it.

30! has only 7 5s in it...

5, 10, 15, 20, 25 (2 fives), 30.

Yep, this is the easiest way to do it. 10^7 can't be a factor of 30! b/c 10^7 has more factors of 5 than 30!.

Whenever you see biiiig number like this and the word "factor" in the problem, this should be a big hint that prime factorization will get you to the final answer relatively quickly. Also, since 10^[any int] will only have prime factors of 2 and 5, and without much looking you can see that 30! will have a whole h*ll of a lot of factors of 2, you might be able to deduce quickly that the factor of 5 is going to be the limiting element here. So, figure out how many factors of 5 30! has in it, and you're made in the shade.

Re: DS : product of first 30 positive integers [#permalink]

Show Tags

11 Aug 2010, 18:07

madammepsychosis wrote:

masland wrote:

Hmm think I might have got it.

30! has only 7 5s in it...

5, 10, 15, 20, 25 (2 fives), 30.

Yep, this is the easiest way to do it. 10^7 can't be a factor of 30! b/c 10^7 has more factors of 5 than 30!.

Whenever you see biiiig number like this and the word "factor" in the problem, this should be a big hint that prime factorization will get you to the final answer relatively quickly. Also, since 10^[any int] will only have prime factors of 2 and 5, and without much looking you can see that 30! will have a whole h*ll of a lot of factors of 2, you might be able to deduce quickly that the factor of 5 is going to be the limiting element here. So, figure out how many factors of 5 30! has in it, and you're made in the shade.

Amend that: 10^7 does have enough 5's, but 10^8 wouldn't.

Re: DS : product of first 30 positive integers [#permalink]

Show Tags

12 Aug 2010, 17:46

It can be determined by the power of \(5\) in the number \(30!\) --> \(\frac{30}{5}+\frac{30}{25}=6+1=7\) --> \(30!\) has \(7\) zeros.

I don't understand the calculations that were performed here. How did you get to \(\frac{30}{5}+\frac{30}{25}=6+1=7\)? How did you know that the 5 was the factor needed? Thanks

Bunuel wrote:

masland wrote:

Is there anyway to quickly determine if d>7 is not a factor of 30! ?

If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?

(1) 10^d is a factor of f --> \(k*10^d=30!\).

First we should find out how many zeros \(30!\) has, it's called trailing zeros. It can be determined by the power of \(5\) in the number \(30!\) --> \(\frac{30}{5}+\frac{30}{25}=6+1=7\) --> \(30!\) has \(7\) zeros.

\(k*10^d=n*10^7\), (where \(n\) is the product of other multiples of 30!) --> it tells us only that max possible value of \(d\) is \(7\). Not sufficient.

Side notes: 30! is some huge number with 7 trailing zeros (ending with 7 zeros). Statement (1) says that \(10^d\) is factor of this number, but \(10^d\) can be 10 (d=1) or 100 (d=2) ... or 10,000,000 (d=7). Basically \(d\) can be any integer from 1 to 7, inclusive (if \(d>7\) then \(10^d\) won't be a factor of 30! as 30! has only 7 zeros in the end). So we can not determine single numerical value of \(d\) from this statement. Hence this statement is not sufficient.

(2) \(d>6\) Not Sufficient.

(1)+(2) \(d>6\), \(d_{max}=7\) --> \(d=7\).

Answer: C.

For trailing zeros see the link about factorials in my signature.

It can be determined by the power of \(5\) in the number \(30!\) --> \(\frac{30}{5}+\frac{30}{25}=6+1=7\) --> \(30!\) has \(7\) zeros.

I don't understand the calculations that were performed here. How did you get to \(\frac{30}{5}+\frac{30}{25}=6+1=7\)? How did you know that the 5 was the factor needed? Thanks

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...