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# If K is a positive integer, and 5^k is the factor of the

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Senior Manager
Joined: 22 Nov 2005
Posts: 478
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Kudos [?]: 3 [0], given: 0

If K is a positive integer, and 5^k is the factor of the [#permalink]  28 May 2006, 22:38
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If K is a positive integer, and 5^k is the factor of the product of the numbers from 99 to 199, what is the value of k?

1). When divided by 2, the remainder is 1

2). When divided by 3, the remainder is 2
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 7

Kudos [?]: 37 [0], given: 0

Re: DS - Number Theory [#permalink]  30 May 2006, 11:43
gmat_crack wrote:
If K is a positive integer, and 5^k is the factor of the product of the numbers from 99 to 199, what is the value of k?

1). When divided by 2, the remainder is 1
2). When divided by 3, the remainder is 2

k could be 5 or 11 or 17 or 23 [5+(6n) because there are 23 5's in the product of integers between 99 and 199.

E make sense..
SVP
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

Kudos [?]: 46 [0], given: 0

E for me too.

1) k = 2q +1 not suff
2) K = 3y + 2 not suff

Both togther

K can be 5, 11 etc ........
Manager
Joined: 10 May 2006
Posts: 186
Location: USA
Followers: 1

Kudos [?]: 3 [0], given: 0

Re: DS - Number Theory [#permalink]  31 May 2006, 08:55
Professor wrote:
gmat_crack wrote:
If K is a positive integer, and 5^k is the factor of the product of the numbers from 99 to 199, what is the value of k?

1). When divided by 2, the remainder is 1
2). When divided by 3, the remainder is 2

k could be 5 or 11 or 17 or 23 [5+(6n) because there are 23 5's in the product of integers between 99 and 199.

E make sense..

Prof, how did you calculate 23 5's in the product of integers?
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 7

Kudos [?]: 37 [0], given: 0

Re: DS - Number Theory [#permalink]  31 May 2006, 09:42
tl372 wrote:
Professor wrote:
gmat_crack wrote:
If K is a positive integer, and 5^k is the factor of the product of the numbers from 99 to 199, what is the value of k?
1). When divided by 2, the remainder is 1
2). When divided by 3, the remainder is 2

k could be 5 or 11 or 17 or 23 [5+(6n) because there are 23 5's in the product of integers between 99 and 199.
E make sense..

Prof, how did you calculate 23 5's in the product of integers?

1) 200/5 - 100/5 = 20
2) 200/25 - 100/25 = 4
3) 200/125 - 100/125 = 1

sorry about hasty generalization. actually there are 25 (not 23) 5's.
Re: DS - Number Theory   [#permalink] 31 May 2006, 09:42
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