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 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) From (2) d>6 and from (1) d_{max}=7 --> d=7.

Trailing zeros: Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}, where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!? \frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Re: If d is a positive integer and f is the product of the first [#permalink]
29 Jan 2012, 14:50

Hi Bunuel thanks - all makes sense apart from the concept of trailing zeros.

Am I right in saying this is how you said there will be 7 zero's.

30/5 + 30/25 = 6 + 1 (quotient) = 7. Where I am not clear is have you simply divided 30/25? I hope I am making myself clear, if I am not then please let me know. _________________

Re: If d is a positive integer and f is the product of the first [#permalink]
29 Jan 2012, 15:00

Expert's post

enigma123 wrote:

Hi Bunuel thanks - all makes sense apart from the concept of trailing zeros.

Am I right in saying this is how you said there will be 7 zero's.

30/5 + 30/25 = 6 + 1 (quotient) = 7. Where I am not clear is have you simply divided 30/25? I hope I am making myself clear, if I am not then please let me know.

Yes, you take only the integer part. For example how many trailing zeros does 126! have?

Re: If d is a positive integer and f is the product of the first [#permalink]
04 Jun 2013, 20:17

Hi. I think there is a solution without knowing the trailing zeros formula. Of course I. alone will not be enough (d= 10 or d=100 do the trick) and II. d>6 is vague. Now, to evaluate I and II together, like you know that 10 = 2*5, and 10^x = (2*5)^x = 2^x*5^x, if 10^d is a factor of f, like f = 1*2*3*4*5...*30, you need to see how many 2s and 5ves you can get. You have plenty of 2s, so lets focus on the 5ves. You actually get 7 fives between 1 and 30 (one in 5,10,15,20,30 and two in 25). So basically d could be any number between 1 and 7. Like II is "d>6", you know that d = 7.

(my 1st post, sorry for style... just trying to help, I suck at knowing formulas, although they save you time. (I learned a lot from this forum, Bunuel is my Guru!))

Re: If d is a positive integer and f is the product of the first [#permalink]
28 Jul 2014, 10:11

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...