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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]

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30 Nov 2009, 20:07

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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

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

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30 Nov 2009, 22:54

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Expert's post

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.

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

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11 Aug 2010, 16:26

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Expert's post

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.

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

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11 Aug 2010, 19:06

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]

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11 Aug 2010, 19:07

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]

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12 Aug 2010, 18:46

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.

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

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13 Aug 2010, 02:53

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Expert's post

estreet wrote:

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

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