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

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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

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

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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.

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

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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?

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!.
_________________

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
_________________

Jeffery Miller Head of GMAT Instruction

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