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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.

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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