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I think the best way to solve the problem is this:

10 can only be made up of 2s and 5s (2x5). In the list of 1x2x3...x30, there are plenty of 2s, but only 7 5s (5, 10, 15, 20, 30 have one each, and 25 has 2 5s). So, f has 7 10s .. ie, f = k x 10^7

So, from statement one, d can be anything from 1 to 7

From statement two, d should be >6

Neither of the statements are sufficient in themselves, but together, they make d = 7

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

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09 Aug 2007, 21:26

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trahul4 wrote:

If d is a positive integer and f is the product if the first 30 positive integers, what is the value of d?

1. 10^d is a factor of f. 2. d>6

C.

Given: f=30!
(1) 10^d is a factor of f
Plug in d=1, 10 is a factor of f, Yes!
d=2, 100 is a factor of f, Yes! because 25*4 = 100
INSUFFICIENT.

(2) d>6. INSUFFICIENT

Together, plug in d=7, Is 10^7 is a factor of f?
5*2 = 10
10 = 10
15*12 = 180 = 18*10
20 = 2*10
25*4= 10*10
30 = 3*10
I don't think there is any more, SUFFICIENT.

1) 10^d is a factor of f so we have to find the powers of 10 in the 30! number of powers of 10 is equal to the number of 2 and 5 multiples of 5 less than or equal to 30 are 5,10, 15, 20, 25, 30. So number of powers of 5 in 30! = 7 As we have many multiple is 2, the maximum value of d is 7 (i.e. d can be 1 or 2 or 3 or 4 ...) 2) d>6. d can take any value. Clubbing 1 and 2 we get, d = 7 So answer is C

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.

(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 could you explain the trailing zeros bit in more detail please?

(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 could you explain the trailing zeros bit in more detail please?

If you are aiming for 700+ in GMAT you should know 2 important things about factorials:

1. 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.

2. Finding the number of powers of a prime number k, in the n!.

What is the power of 3 in 35!

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