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

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

28 Jan 2012, 18:13

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52% (01:58) correct

47% (00:18) wrong

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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

[Reveal] Spoiler: OA

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Last edited by Bunuel on 30 Oct 2012, 01:42, edited 2 times in total.

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Re: +ve integer D [#permalink]

28 Jan 2012, 18:18

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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.
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Re: +ve integer D [#permalink]

28 Jan 2012, 18:19

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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
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Kudos [?]: 61 [0], given: 212

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

29 Jan 2012, 15:50

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.
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MGMAT 1 --> 530
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Kudos [?]: 9532 [0], given: 826

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

29 Jan 2012, 16:00

enigma123 wrote:

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

126/5+126/5^2+126/5^3=25+5+1=31.

Hope it's clear.
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Re: If d is a positive integer and f is the product of the first [#permalink] 29 Jan 2012, 16:00

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

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

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