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# you have 1 min! 5^k is a factor of the product of odd

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CEO
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you have 1 min! 5^k is a factor of the product of odd [#permalink]  15 Sep 2003, 20:56
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you have 1 min!

5^k is a factor of the product of odd integers from 99 -199, what
is the greatest possible value of k?

Is it 20?

Thanks
praetorian
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Joined: 06 Sep 2003
Posts: 19
Location: island
Followers: 0

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

K=25

99*100*101..........*199

We find the figure are multiple of 5, so 100,105,...........,195, there are 20.
(195-100)/5 +1 =20

futher we should check 100=5*20, 105=5*21........,195=5*39, so we should pick up from 20,21,...............,39, which one can be the multiple of 5, there are 4.
(35-20)/5 +1=4
Futher 25=5*5, so should remember.
Combined above, 20+4+1=24

This is my answer, is it right?

Last edited by mystery on 16 Sep 2003, 19:54, edited 1 time in total.
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Re: PS : Factor [#permalink]  16 Sep 2003, 20:05
praetorian123 wrote:
you have 1 min!

5^k is a factor of the product of odd integers from 99 -199, what
is the greatest possible value of k?

Is it 20?

Thanks
praetorian

odd integers that are divisible by 5 are 105, 115, 125 ... 195. So, they are 10 in total. 125, 150 and 175 have 5^3, 5^2 and 5^2 as factors. So, k should be 10 + 2 + 1 + 1 = 14.

Intern
Joined: 06 Sep 2003
Posts: 19
Location: island
Followers: 0

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

I always misunderstand the question.

Mystery
CEO
Joined: 15 Aug 2003
Posts: 3467
Followers: 61

Kudos [?]: 717 [0], given: 781

Re: PS : Factor [#permalink]  18 Sep 2003, 10:35
edealfan wrote:
praetorian123 wrote:
you have 1 min!

5^k is a factor of the product of odd integers from 99 -199, what
is the greatest possible value of k?

Is it 20?

Thanks
praetorian

odd integers that are divisible by 5 are 105, 115, 125 ... 195. So, they are 10 in total. 125, 150 and 175 have 5^3, 5^2 and 5^2 as factors. So, k should be 10 + 2 + 1 + 1 = 14.

edeal..150 is even...

are we counting something extra here.
Re: PS : Factor   [#permalink] 18 Sep 2003, 10:35
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