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

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Let p = the product of all the odd integers between 500 and  [#permalink]

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New post 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. \(359,999q\)

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  [#permalink]

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New post 12 Jun 2013, 09:05
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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


\(q=p*599*601=p(600-1)(600+1)=p*(360,000-1)=359,999p\) --> \(p=\frac{q}{359,999}\).

\(\frac{1}{p} + \frac{1}{q}=\frac{359,999}{q}+\frac{1}{q}=\frac{3600,000}{q}\).

Answer: D.
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Re: Let p = the product of all the odd integers between 500 and  [#permalink]

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New post 12 Jun 2013, 08:18
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\(p=501*503*...*597\)
\(q=501*503*...*597*599*601\)

\(\frac{1}{p}+\frac{1}{q}=\frac{p+q}{pq}=\frac{p(1+599*601)}{pq}\)

\(\frac{1+599*601}{q}=\frac{360000}{q}\)
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Re: Let p = the product of all the odd integers between 500 and  [#permalink]

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New post 12 Jun 2013, 08:33
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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 :-D
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Re: Let p = the product of all the odd integers between 500 and  [#permalink]

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New post 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  [#permalink]

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New post 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 = (600-1)(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  [#permalink]

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New post 30 Aug 2016, 17:03
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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  [#permalink]

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New post 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  [#permalink]

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New post 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 \((a-b)(a+b) = a^2 - b^2\), thus \((600-1)(600+1)=600^2 - 1^2=(360,000-1)\).

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PS Algebra Questions to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=50

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Re: Let p = the product of all the odd integers between 500 and  [#permalink]

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New post 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  [#permalink]

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New post 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) = (600-1)(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  [#permalink]

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New post 09 Jul 2017, 07:23
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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  [#permalink]

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Re: Let p = the product of all the odd integers between 500 and   [#permalink] 27 Jul 2019, 04:27
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