OG 12th Ed. question #89

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OG 12th Ed. question #89 [#permalink]

14 Dec 2009, 21:21

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Could someone explain the solution for this question? I don't understand how the OG came up with t=201s or s= t.201?

If s is the product of integers from 100 to 200, inclusive and t is the product of the integers from 100 o 201, inclusive, what is 1/s = 1/t in term of t?

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Re: OG 12th Ed. question #89 [#permalink]

14 Dec 2009, 23:06

s=100*101*...........200=200!/99!(in terms of factorial)
t=201*200*199*..........100=201!/99!

now 1/s=1/t---->t/s=?
t=201!/99!
=201*200!/99!=201*s(as s=200!/99!)

HTH
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Re: OG 12th Ed. question #89 [#permalink]

27 Jan 2010, 11:50

Hi xcusemeplz2009:

This problem seems confusing but it's quite easy.

First, there are two sets of numbers: set "t" and set "s". The product of the set "s" is equal to The product of the set "t" EXCEPT this set has an additional number (201). This can be summarized like this:

Set "s": 101 * 102 * 103 *......200 = s

Set "t": 101 * 102 * 103 *......200 * 201 = t

Therefore:

t = s * 201 = 201s

and if t = 201s then s = t / 201 ========> this is the equation you have to plug into the stem equation.

Your next step is to solve the stem's equation:

If 1/s + 1/t then this equation is equal to 1/(t/201) + 1/t

So, applying the dobble C you get 201/t + 1/t = (201+ 1)/t = 202/t.

The answer is D.

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Re: OG 12th Ed. question #89 [#permalink]

12 Feb 2010, 18:57

Ans= 202/t

s=100*..*200
201s=(100*..*200)*201=t => 1/201s=1/t =>1/s=201/t

1/s+1/t=201/t+1/t=202/t

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Re: OG 12th Ed. question #89 [#permalink]

21 Nov 2011, 12:19

s = 100*...*200
t = 100*...*200*201
This means
s = t/201
Therefore,
1/s + 1/t = 201/t + 1/t = 202/t

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Re: OG 12th Ed. question #89 [#permalink]

21 Nov 2011, 12:54

s=10*101*102...*200
t=s*201
hence t/s=201..
for there onwards..this is a cakewalk
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Re: OG 12th Ed. question #89 [#permalink]

21 Nov 2011, 12:58

I hate this question
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Re: OG 12th Ed. question #89 [#permalink] 21 Nov 2011, 12:58

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OG 12th Ed. question #89

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OG 12th Ed. question #89

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