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Director  Joined: 29 Nov 2012
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Let p = the product of all the odd integers between 500 and  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 69% (02:38) correct 31% (02:32) wrong based on 434 sessions

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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, 07:11.
Last edited by fozzzy on 12 Jun 2013, 20:41, edited 2 times in total.
Math Expert V
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Re: Let p = the product of all the odd integers between 500 and  [#permalink]

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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

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

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

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

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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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1
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 $$\frac{1}{p} + \frac{1}{q}$$ = $$\frac{360000}{q}$$

Hope that helped GMAT Club Legend  V
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Re: Let p = the product of all the odd integers between 500 and  [#permalink]

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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)]

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

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1
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. $$359,999q$$

$$p = 501 * 503 * 505 * ............ * 597$$

$$q = 501 * 503 * 505 * ......................* 599 * 601$$

$$p = \frac{q}{599 * 601}$$

$$\frac{1}{p} + \frac{1}{q}$$

$$= \frac{1}{q/599 * 601} + \frac{1}{q}$$

$$= \frac{599 * 601}{q} + \frac{1}{q}$$

$$= \frac{599*601 + 1}{q}$$

$$= \frac{(600 - 1) (600 + 1) + 1}{q}$$

$$= \frac{((600)^2 + 600 - 600 - 1) + 1}{q}$$

$$= \frac{(600)^2}{q}$$

$$= \frac{360,000}{q}$$

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

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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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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

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

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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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Posts: 62499
Re: Let p = the product of all the odd integers between 500 and  [#permalink]

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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 $$(600-1)(600+1)=600^2 - 1^2=(360,000-1)$$.

Theory on Algebra: http://gmatclub.com/forum/algebra-101576.html
Algebra - Tips and hints: http://gmatclub.com/forum/algebra-tips- ... 75003.html

DS Algebra Questions to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=29
PS Algebra Questions to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=50

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

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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. $$359,999q$$

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

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

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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. $$359,999q$$

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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