GMAT Prep 2- Quant 5
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Pl help me with solution.
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The answer is B and this is how.
We have to see if n is odd.
St1: n can divisible by 3 and still be even. eg. 3,6,9,12.....
no enough
St2: 2n is divisible by twice as many number as by n. This is only possible if n is odd.
eg. if n=4; 2n=8 and they would both be divisible by only 2
if n=9 2n=18, now 2n is divisible by both 2 and 3 vs. only 3 with n.
We have to see if n is odd.
St1: n can divisible by 3 and still be even. eg. 3,6,9,12.....
no enough
St2: 2n is divisible by twice as many number as by n. This is only possible if n is odd.
eg. if n=4; 2n=8 and they would both be divisible by only 2
if n=9 2n=18, now 2n is divisible by both 2 and 3 vs. only 3 with n.
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Picking a few numbers and seeing a patttern is a good way to solve.
Conceptually this is what happens:
If n was even 2 would be one of its factor and so is 1 (1 is a factor of every integer)
If u mutliply by 2 each of the factors the factors will not double since 2*1 =2 is already a factor of n so 2n misses out on the double the factors part
So n cant be even
If n was odd then all its factors are odd.
When u multiply by 2 each odd factor of n u get a new even factor for 2n thereby doubling its factors (even*odd=EVEN)
Hope this helps too!
Regards,
CR
Conceptually this is what happens:
If n was even 2 would be one of its factor and so is 1 (1 is a factor of every integer)
If u mutliply by 2 each of the factors the factors will not double since 2*1 =2 is already a factor of n so 2n misses out on the double the factors part
So n cant be even
If n was odd then all its factors are odd.
When u multiply by 2 each odd factor of n u get a new even factor for 2n thereby doubling its factors (even*odd=EVEN)
Hope this helps too!
Regards,
CR
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I just came across this question, and got the right OA.
However reading the explanation above (Cramya's in particular) I am a little confused.
If n = 3, then its factors are 1 and 3 (thus it has 2 factors). So 2n = 6, and its factors are 1,3 and 2. But this does not mean that 2n is "divisible by twice as many positive factors", as you have 3 factors now instead of 2.
So for me, the only answer to this question is when n = 1, as the only factor it has is 1, and therefore 2n = 2 has 2 factors.
Can someone please tell me how I have misinterpreted the question?
Thanks
However reading the explanation above (Cramya's in particular) I am a little confused.
If n = 3, then its factors are 1 and 3 (thus it has 2 factors). So 2n = 6, and its factors are 1,3 and 2. But this does not mean that 2n is "divisible by twice as many positive factors", as you have 3 factors now instead of 2.
So for me, the only answer to this question is when n = 1, as the only factor it has is 1, and therefore 2n = 2 has 2 factors.
Can someone please tell me how I have misinterpreted the question?
Thanks
GMAX
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6 has four factors (1, 2, 3, and 6), not three.Baldini wrote: If n = 3, then its factors are 1 and 3 (thus it has 2 factors). So 2n = 6, and its factors are 1,3 and 2. But this does not mean that 2n is "divisible by twice as many positive factors", as you have 3 factors now instead of 2.
So for me, the only answer to this question is when n = 1, as the only factor it has is 1, and therefore 2n = 2 has 2 factors.
Can someone please tell me how I have misinterpreted the question?
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