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Let p = the product of all the odd integers between 500 and
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Updated on: 12 Jun 2013, 21:41
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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. \(359,999q\)
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Originally posted by fozzzy on 12 Jun 2013, 08:11.
Last edited by fozzzy on 12 Jun 2013, 21:41, edited 2 times in total.




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Re: Let p = the product of all the odd integers between 500 and
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12 Jun 2013, 09:05
fozzzy 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 1/p + 1/q?
A. 1/600q B. 1/359,999q C. 1,200/q D. 360,000/q E. 359,999q \(q=p*599*601=p(6001)(600+1)=p*(360,0001)=359,999p\) > \(p=\frac{q}{359,999}\). \(\frac{1}{p} + \frac{1}{q}=\frac{359,999}{q}+\frac{1}{q}=\frac{3600,000}{q}\). Answer: D.
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Re: Let p = the product of all the odd integers between 500 and
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12 Jun 2013, 08:18
\(p=501*503*...*597\) \(q=501*503*...*597*599*601\)
\(\frac{1}{p}+\frac{1}{q}=\frac{p+q}{pq}=\frac{p(1+599*601)}{pq}\)
\(\frac{1+599*601}{q}=\frac{360000}{q}\)




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Re: Let p = the product of all the odd integers between 500 and
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12 Jun 2013, 08:33
Let's formalise these expressions a bit : p = the product of all the odd integers between 500 and 598 Meaning that \(p = 501*503*...*597\) (1) Like wise for \(q\) being the product of all the odd integers between 500 and 602, we get, using (1) : \(q = 501*503*...597*599*601 = p*599*601\) (2) Since we are looking to express \(\frac{1}{p} + \frac{1}{q}\) in terms of \(q\), we get from (2) : \(p = \frac{q}{(599*601)}\) So then : \(\frac{1}{p} + \frac{1}{q}\) \(= \frac{1}{q/(599*601)} + \frac{1}{q} = \frac{1}{q} * (599*601 +1)\) Since\(599 = 600  1\) then \(599*601 + 1 = (600  1)*601 + 1 = 360600  601 + 1 = 360599 + 1 = 360000\) Which yields \(\frac{1}{p} + \frac{1}{q}\) = \(\frac{360000}{q}\) Which is answer choice D. Hope that helped



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Re: Let p = the product of all the odd integers between 500 and
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30 Aug 2016, 17:03
fozzzy 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. \(359,999q\) 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:
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Re: Let p = the product of all the odd integers between 500 and
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09 Jul 2017, 07:23
fozzzy 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. \(359,999q\) \(p = 501 * 503 * 505 * ............ * 597\) \(q = 501 * 503 * 505 * ......................* 599 * 601\) \(p = \frac{q}{599 * 601}\) \(\frac{1}{p} + \frac{1}{q}\) \(= \frac{1}{q/599 * 601} + \frac{1}{q}\) \(= \frac{599 * 601}{q} + \frac{1}{q}\) \(= \frac{599*601 + 1}{q}\) \(= \frac{(600  1) (600 + 1) + 1}{q}\) \(= \frac{((600)^2 + 600  600  1) + 1}{q}\) \(= \frac{(600)^2}{q}\) \(= \frac{360,000}{q}\) Hence, Answer is D
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Re: Let p = the product of all the odd integers between 500 and
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11 Jan 2014, 21:11
i think we can first simplify, expression 1/p +1/q to 1/q (q/p + 1). This way its easier to visualize that q/p will be only 599*601.



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Re: Let p = the product of all the odd integers between 500 and
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04 May 2016, 19:37
fozzzy 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. \(359,999q\) oh wow..good question..requires some outside the box thinking... p=q/599*601 1/p + 1/q = p+q/pq first thing: p+q q/599*601 + q = q+q(599*601)/599*601 pq = q^2/599*601 now [q+q(599*601)/599*601] * [599*601/q^2] we can simplify by 599*601 we get q+q(599*601)/q^2 we can factor out q in the numerator = q(1+599*601)/q^2 divide both sides by q 1+599*601/q 599*601 = (6001)(600+1) = 359,999 we add one and get 360,000 now...final step 360,000/q answer is D



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Re: Let p = the product of all the odd integers between 500 and
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23 Mar 2017, 05:00
I am struggling a bit with this question. I do not understand how 359,999pq=p∗599∗601=p(600−1)(600+1)=p∗(360,000−1)=359,999p > p=q359,999p=q359,999 .
This type of questions are just a big confusing in general. Any advice on where to revise this type of questions



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Re: Let p = the product of all the odd integers between 500 and
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23 Mar 2017, 05:37
vmelgargalan wrote: I am struggling a bit with this question. I do not understand how 359,999pq=p∗599∗601=p(600−1)(600+1)=p∗(360,000−1)=359,999p > p=q359,999p=q359,999 .
This type of questions are just a big confusing in general. Any advice on where to revise this type of questions We applied there \((ab)(a+b) = a^2  b^2\), thus \((6001)(600+1)=600^2  1^2=(360,0001)\). Theory on Algebra: http://gmatclub.com/forum/algebra101576.htmlAlgebra  Tips and hints: http://gmatclub.com/forum/algebratips ... 75003.htmlDS Algebra Questions to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=29PS Algebra Questions to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=50Hope it helps.
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Re: Let p = the product of all the odd integers between 500 and
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27 Mar 2017, 11:49
fozzzy 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. \(359,999q\) We are given that p = the product of the odd integers from 500 to 598, i.e., from 501 to 597 inclusive. We are also given that q = the product of the odd integers from 500 to 602, i.e., 501 to 601 inclusive. Thus: q = p(599)(601) Now we can evaluate 1/p + 1/q as: 1/p + 1/q = (599)(601)/[p(599)(601)] + 1/q = (599)(601)/q + 1/q = [(599)(601) + 1]/q Notice that (599)(601) = (600  1)(600 + 1) = 600^2  1. Thus, the numerator (599)(601) + 1 becomes 600^2  1 + 1, or simply 600^2. Therefore: 1/p + 1/q = [(599)(601) + 1]/q = 600^2/q = 360,000/q Answer: D
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Re: Let p = the product of all the odd integers between 500 and
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16 Apr 2017, 19:46
fozzzy 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. \(359,999q\) We can solve this question using algebra: p = (501)(503)...(595)(597). q = (501)(503)...(595)(597)(599)(601). The overlap between P and Q implies that q = (p)(599)(601) We could do this with another set of numbers (p is the odd integers between 2 and 8, q is the odd integers between 2 and 12) p= 3 x 5 x 7 q=3 x 5 x 7 x 9 x 11 105= p (11)(9) Anyways The answer choices are in terms of a variable so are result must be in the form of P+q/pq P =1. Q= (1)(599)(601) = (6001)(600+1) = 360000  1 = 359999. Therefore 1/p + 1/q = 1/1 + 1/359999 = 359999/359999 + 1/359999 = 360000/359999 = p + q/ q= Plug in q = 359999 into the answers to see which equals 360000/359999. 360000/q = 360000/359999. Thus D.



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Re: Let p = the product of all the odd integers between 500 and
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