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Let p = the product of all the odd integers between 500 and 598, and l
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11 Jan 2021, 09:51
11 Jan 2021, 09:51
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Let p = the product of all the odd integers between 500 and 598, and let q = the product of all the odd integers between 500 and 602. In terms of q, what is the value of \(\frac{1}{p}\)+ \(\frac{1}{q}\)?
A. \(\frac{1}{600q}\)
B. \(\frac{1}{359,999q}\)
C. \(\frac{1,200}{q}\)
D. \(\frac{360,000}{q}\)
E. \(359,999q\)
A. \(\frac{1}{600q}\)
B. \(\frac{1}{359,999q}\)
C. \(\frac{1,200}{q}\)
D. \(\frac{360,000}{q}\)
E. \(359,999q\)
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Joined: 10 Apr 2015
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Re: Let p = the product of all the odd integers between 500 and 598, and l
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11 Jan 2021, 10:24
11 Jan 2021, 10:24
3
Carcass wrote:
Let p = the product of all the odd integers between 500 and 598, and let q = the product of all the odd integers between 500 and 602. In terms of q, what is the value of \(\frac{1}{p}\)+ \(\frac{1}{q}\)?
A. \(\frac{1}{600q}\)
B. \(\frac{1}{359,999q}\)
C. \(\frac{1,200}{q}\)
D. \(\frac{360,000}{q}\)
E. \(359,999q\)
A. \(\frac{1}{600q}\)
B. \(\frac{1}{359,999q}\)
C. \(\frac{1,200}{q}\)
D. \(\frac{360,000}{q}\)
E. \(359,999q\)
p = (501)(503)(505)...(597)
q = (501)(503)(505)...(597)(599)(601)
So, q = (p)(599)(601)
So, 1/p + 1/q = 1/p + 1/(p)(599)(601) [replaced q with (p)(599)(601)]
= (599)(601)/(p)(599)(601) + 1/(p)(599)(601) [found common denominator]
= [(599)(601) + 1]/(p)(599)(601)
= 360,000/(p)(599)(601)
= 360,000/q [since q = (p)(599)(601)]
Answer: D
Re: Let p = the product of all the odd integers between 500 and 598, and l
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09 Aug 2023, 11:13
09 Aug 2023, 11:13
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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