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If d is a positive integer and f is the product of the first

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Re: If d is a positive integer and f is the product of the first [#permalink]

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New post 30 Dec 2015, 03:11
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 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) \(d_{max}=7\) --> \(d=7\).

Answer: C.

Hope it helps.


i understand the trailing zero part, but after that, i dont get it; especially why d has to range from 1 to 7

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Re: If d is a positive integer and f is the product of the first [#permalink]

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New post 04 Aug 2016, 17:52
Bunuel wrote:
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.

For more on this concept check Everything about Factorials on the GMAT: everything-about-factorials-on-the-gmat-85592.html


Hi Bunuel,

If i am following your explanation correctly then i guess there is a small typo in the post.

"where k must be chosen such that 5^(k+1)>n " should instead be "where k must be chosen such that 5^(k+1)<n "

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Re: If d is a positive integer and f is the product of the first [#permalink]

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New post 29 Aug 2017, 06:46
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 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) \(d_{max}=7\) --> \(d=7\).

Answer: C.

Hope it helps.


So the thing is:

10^d is not the only factor, it is one of the factors, that's why we cannot surely say that d=7. But if the question would have said that 10^d is the only factor then, "A" would be the right answer. Am i right?

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Re: If d is a positive integer and f is the product of the first [#permalink]

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New post 29 Aug 2017, 08:08
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saswatdodo 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 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) \(d_{max}=7\) --> \(d=7\).

Answer: C.

Hope it helps.


So the thing is:

10^d is not the only factor, it is one of the factors, that's why we cannot surely say that d=7. But if the question would have said that 10^d is the only factor then, "A" would be the right answer. Am i right?


1 is the only positive integer which has 1 factor. All other positive integers have more factors. It does not make sense to say that 10^d is the only factor of 30!.
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Re: If d is a positive integer and f is the product of the first [#permalink]

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New post 05 Sep 2017, 16:48
enigma123 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
(2) d>6


We are given that d is a positive integer and f = 30!. We need to determine the value of d.

Statement One Alone:

10^d is a factor of f

Since 10^1 and 10^2 could each divide into 30!, we do not have a unique value for d. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

d > 6

Since d could be 7, 8, or greater, statement two alone does not allow us to determine a unique value of d.

Statements One and Two Together:

Using both statements, since we know that d > 6, let’s determine the maximum value d can be given that 10^d divides into 30!. Essentially, we need to determine the maximum number of five-two pairs. (Recall that each five-two pair creates a factor of 10.) Since there are more twos than fives, let’s determine the number of fives.

The factors that are multiples of 5 in 30! are 5, 10, 15, 20, 25 = 5^2, and 30. So, we see there are 7 fives in 30!, and thus the maximum value of d is 7. Since d > 6, d must be 7.

Answer: C
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Re: If d is a positive integer and f is the product of the first   [#permalink] 05 Sep 2017, 16:48

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