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If d is a positive integer and f is the product of the first [#permalink]
28 Jan 2012, 17:18

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This post received KUDOS

Expert's post

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This post was BOOKMARKED

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 cannot 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

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Re: If d is a positive integer and f is the product of the first [#permalink]
19 Apr 2015, 09:11

Bunuel wrote:

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) \(d_{max}=7\) --> \(d=7\).

Answer: C.

Hope it helps.

there are not just 5 and 2 which multiplies to 10 but also 2*5.. then the number of 10's in the product can be greater than 7 isnt it?

Re: If d is a positive integer and f is the product of the first [#permalink]
20 Apr 2015, 03:44

Expert's post

SunthoshiTejaswi wrote:

Bunuel wrote:

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) \(d_{max}=7\) --> \(d=7\).

Answer: C.

Hope it helps.

there are not just 5 and 2 which multiplies to 10 but also 2*5.. then the number of 10's in the product can be greater than 7 isnt it?

Re: If d is a positive integer and f is the product of the first [#permalink]
27 Apr 2015, 20:31

in gmatprep there are many too difficult questions.

if we practice harder questions, take the questions form og or gmatprep not from prep companies. because the harder questions from prep companies are not basic. gmat is very basic.

We know that \(35\) has two factors \(5\) and \(7\) First impulse is to just take answer from previous question because of presence of \(5\) but we should calculate that number, that has less occurrences So \(5\) in \(30!\) meets \(7\) times but \(7\) in \(30!\) meets \(4\) times.

And we can infer that \(30!\) will be divisible by \(35\) only \(4\) times. So \(p = 4\)

We know that \(35\) has two factors \(5\) and \(7\) First impulse is to just take answer from previous question because of presence of \(5\) but we should calculate that number, that has less occurrences So \(5\) in \(30!\) meets \(7\) times but \(7\) in \(30!\) meets \(4\) times.

And we can infer that \(30!\) will be divisible by \(35\) only \(4\) times. So \(p = 4\)

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