If S is the product of the integers from 1 to 100, inclusive

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If S is the product of the integers from 1 to 100, inclusive [#permalink]

27 Feb 2013, 13:39

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If S is the product of the integers from 1 to 100, inclusive, and T is the product of the integers from 1 to 101, inclusive, what is 1/S+1/T in terms of T ?

A)T/101
B)T/100+100
C)100+T/T
D)102/T
E)T+1/100

Answer:

Given: S=100! and T=101!.

1/S+1/T=1/100!+1/101!=101+1/101!
=102/T.

....

This part I fully understand: 1/100!+1/101!
But I am a little confused how this was determined: 101+1/101!
How are those "!" taken out on the numerator to get 101 and 1?

Can someone show me in detail please?

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Re: If S is the product of the integers from 1 to 100, inclusive [#permalink]

27 Feb 2013, 14:54

m24 q35

If S is the product of the integers from 1 to 100, inclusive, and T is the product of the integers from 1 to 101, inclusive, what is 1/S+1/T in terms of T ?

A. \frac{T}{101}
B. \frac{T}{100}+100
C. \frac{100+T}{T}
D. \frac{102}{T}
E. \frac{T+1}{100}

Given: S=100! and T=101!.

\frac{1}{S}+\frac{1}{T}=\frac{1}{100!}+\frac{1}{101!}=\frac{1}{100!}+\frac{1}{100!*101}=\frac{101+1}{100!*101}=\frac{101+1}{101!}=\frac{102}{T}.

Answer: D.
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Re: If S is the product of the integers from 1 to 100, inclusive [#permalink]

27 Feb 2013, 20:56

aaroncorr wrote:

You can also substitute T for S right away to get the expression in terms of T.

Given: S=100! and T=101! Therefore, S = T/101

1/S+1/T = 101/T + 1/T = 102/T
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Re: If S is the product of the integers from 1 to 100, inclusive [#permalink]

28 Feb 2013, 03:10

aaroncorr wrote:

Nothing much to add. Just another method which might work for those who don't have enough time to understand/solve this question. ALL the options are more than 1, except D. The given expression 1/S+1/T will also be much much less than 1. Thus answer is D.
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Re: If S is the product of the integers from 1 to 100, inclusive [#permalink] 28 Feb 2013, 03:10

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