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| OG - DS - #207 |
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GAMT 2007
Guest
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p>2, so p can be 3,5,7,11 and so on...
n =4p so n can be 12,20,28,44 and so on.. for all values of p, n will always be an even number also n = 2*2*p hence for all the possible values of p - n will have exactly 3 even divisors 2, 4 and n itself. So answer is (B) Hope it helps GMAT 2007 |
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GMAT 2007
Guest
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Sorry missed one. 2p will also be an even divisor of 4p, so that makes the total different even divisors to 4. Answer is C.
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Ron Purewal
MGMAT STAFF
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The preceding post is correct; the most systematic way to approach this problem, however, is with 'prime boxes.'
The prime box for 4p contains 2, 2, and p. The prime box for the divisor, since the divisor is even, must contain at least one of the 2's. There are four different ways to do this: 2 2, p 2, 2 2, 2, p These are the four solutions mentioned in the above posts. |
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| OG - DS - #207 |
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