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# If d is a positive factor and f is the product of the first

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

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15 May 2006, 07:03
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If d is a positive factor and f is the product of the first 30 positve integers, what is the value of d?

1. 10^d is a factor of f
2. d > 6
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15 May 2006, 07:11
I go with C.

Statement 1 gives d = 3, 6, 7

Statement 2 gives d = a lot of nos.

Combining, d = 7

Last edited by remgeo on 15 May 2006, 08:03, edited 1 time in total.
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15 May 2006, 07:18
My guess is C.

(1) f is a factor of 10, but 10^1, 10^2, 10^3.... are also factors of 10
INSUFF
(2) obviously insuff

together
in 30!, there are 3 factors that contain ten: 10:10*1, 20:10*2, 30:10*3, and 3 factors containing 5: 5, 15:5*3, 25:5*5. cobine each of those 5's with one of the many 2's in the factorization for another 4 10's, that makes 7 10's total. if d>6, d must be 7.
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15 May 2006, 08:10
If d is a positive factor and f is the product of the first 30 positve integers, what is the value of d?

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

From (1), d can be = 0, 1, 2, 3..
Not sufficient

From (2),
Not sufficient as we do not know anything about d

Taking both together:
f = 30! and we have to check if that # can be divided by 10^7 or higher.

(5x2) x 10 x 20 x 30 x (15x6) x (25x4=100) would make 10^7 a factor.

Hence C
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17 May 2006, 09:04

What if the value of d is 8,9,10....?

f = 30! = 265252859812191000000000000000000

Sorry. I just calculated it by excel...

so IMO, 10^8, 10^9, 10^10...also possible...
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17 May 2006, 09:40
jeunesis wrote:
:?
What if the value of d is 8,9,10....?

f = 30! = 265252859812191000000000000000000

Sorry. I just calculated it by excel...

so IMO, 10^8, 10^9, 10^10...also possible...

This is incorrect. There will be 7 zeros only. Thats how its calculated:

Numbers multiple of 5 = 5,15,25 = 3
Numbers multiple of 25 = 25 = 1
Numbers multiple of 10 = 10,20,30 = 3

So total zeros = 3+1+3 = 7

St1: d may be 1,2,3,4,5,6,7 : INSUFF
St2: d may be 6,7,....... and many more : INSUFF

Combined: d must be 7 : SUFF
So it must be C.
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17 May 2006, 20:32
I used the same approach as Dahiya, but stuck with 10^6.

When you counted 25 you counted it twice. Why?
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18 May 2006, 00:16
Natalya Khimich wrote:
I used the same approach as Dahiya, but stuck with 10^6.

When you counted 25 you counted it twice. Why?

Thats because 25 = 5x5.

We compute 5s and 2s in the product to determine the number of zeroes the product would have. Typically since the number of twos are sufficient, we can make our lives easier by counting the number of 5s alone. And 5s are counted by counting in all the factors of 5s (5, 10, 15 etc) but counting extra for powers of 5, like 25, 125 and 625 (which have 2, 3 and 4 5s respectively).

Hope that helps.
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18 May 2006, 00:16
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# If d is a positive factor and f is the product of the first

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