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by eyelikecheese » Mon Dec 20, 2010 3:03 pm
The prime factors of 48 are:2,2,2,2,3

So some combination of these digits has to equal 48.

Also, by this prime factorization, it says 48 has:2^4 3^1 total factors. which is (4+1)*(1+1)=10 total factors, with 5 factor pairs.Those 5 factors are 2*24 4*12 6*8 1*48 16*3. So we must use the 2 stems to see which pair of factors this is

1)states that K&P are consecutive even integers. This eliminates 16*3 because 3 is odd. Also, they must be consecutive integers; therefore, all but (6*8) remains.BUT this 6 & 8 could also be negative, thus (-6*-8=48) INSUFFICIENT

2) K&P are both positive integers. This doesn't narrow the possible values down, but only tells us that K & P are both positive. INSUFFICIENT

C.) By combining both statements 1 and 2, it tells us that the answer must be +6*+8=48
Thus 6+8=14
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by [email protected] » Mon Dec 20, 2010 4:52 pm
I got at the same answer. However, according to Grockit there are 2 possible choices 8,6 & -8,-6. Hence answer choice E. I was always taught that while considering factors, always consider positive numbers.
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by [email protected] » Tue Dec 21, 2010 7:01 am
Can anyone explain this?
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by Haaress » Thu Dec 23, 2010 9:48 pm
IMO the correct answer should be C.

Given kp=48, what is the value of k+p?

Paraphrase : Is there a unique value of K + P?

1. k & p are consecutive even integers.

If KP=48, then if K = 8 , P = 6 , So, K+P= 14 or If K = -8 , P = -6 , So, K+P=-14 ..Insuff

2, k & p are both positive integers.

If K = 8 , P= 6, then K + P = 14, however, K = 2 , P = 24, making the sum of K and P = 26...thus insuff.

Combining 1 and 2. Both K and P are consecutive even integers( from 1) that are both positive ( from 2) results in K + P = 8 +6 OR 6+8 =14. Therefore, both stmts are Sufficient. [/list]
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