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

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Senior Manager
Joined: 27 Aug 2005
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If d is a positive integer and f is the product of the first [#permalink]  22 Sep 2005, 20:54
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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
Manager
Joined: 06 Aug 2005
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[#permalink]  23 Sep 2005, 01:14
I think we have had this one before.
But I can't find it in the search function.
Manager
Joined: 03 Aug 2005
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[#permalink]  23 Sep 2005, 03:36
I think it is C

1 is insufficient, 10, 100, 1000... are factors of f

2 is insufficient dcan be 7, 8, ...

If we take 1 and 2 together:

From 1 we have that 2^d * 5^d is a factor of f

In f there are 7 fives, so d cannot be bigger than 7

From 2 we have that d is bigger than 6, therefore, d = 7
Manager
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[#permalink]  23 Sep 2005, 04:10
This question uses similar logic.

http://www.gmatclub.com/phpbb/viewtopic ... ht=#124646
Current Student
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[#permalink]  23 Sep 2005, 07:40
how did you figure out that 30! has 7 5s in it?

30!=2^a*3^b*5^c...29

now, 30/5=6 5s

then you have the over 30/10=3 5s

30/15, gives you 2 5s

so IMHO there should be 6+3+2= 11 5s in 30!

jdtomatito wrote:
I think it is C

1 is insufficient, 10, 100, 1000... are factors of f

2 is insufficient dcan be 7, 8, ...

If we take 1 and 2 together:

From 1 we have that 2^d * 5^d is a factor of f

In f there are 7 fives, so d cannot be bigger than 7

From 2 we have that d is bigger than 6, therefore, d = 7
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

[#permalink]  23 Sep 2005, 07:43
Oops I take it back...

I realized there are 30/5=6 5s...

then we need to look at 25 which is another factor of 30!, so you get one more 5 there...TOtal is 7...
[#permalink] 23 Sep 2005, 07:43
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# If d is a positive integer and f is the product of the first

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