Veritas Test2 Q34

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Veritas Test2 Q34

by Abhijit K » Thu Feb 19, 2015 11:10 pm
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.1/359999q
C.1200/q
D.360000/q
E.359999*q

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by DavidG@VeritasPrep » Fri Feb 20, 2015 7:15 am
First, we know that p = 501*503...597 and q = 501*503...597*599*601

Notice that p is completely contained within q, so we can rewrite: q = p*599*601
Ultimately, we want our answer to be in terms of q, so we can rearrange the above as: p = q/(599*601)

Now, let's substitute that into 1/p+1/q

We'll get 1/[q/(599*601)] + 1/q.

Dividing by a fraction is the same as multiplying by the reciprocal, so this becomes

599*601/q + 1/q

(599*601 +1)/q

At this point, you could just say we have some huge number over q and see the answer must be D, or we can do a little fun work with difference of squares to calculate precisely. 599*601 = (600-1)(600+1), so we're looking for

[(600-1)(600+1) + 1]/q
[600^2 - 1^2 + 1]/q
[360,000 -1 + 1]/q
360,000/q

The answer is D.
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by DavidG@VeritasPrep » Fri Feb 20, 2015 7:22 am
And if you catch the difference of squares earlier, you could also think about it like this:

q = p*599*601

q = (600-1)(600+1)p

q = (360,000 -1)p

p = q/(360,000-1)

1/p + 1/q = (360,000-1)/q + 1/q = 360,000/q
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