IS the integer n odd?
1) n is divisible by 3.
2) 2n is divisible by twice as many positive integers as n
First stmt is clear to me, Can someone explain the 2nd one.
Even-Odd
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to understand B clearly, assure two different value sets for n and then arrive on 2n
let me try;
Assume n as Odd - let's say 5, divisible by 2 different integers (factors); 1 & 5
so 2n is 10 - having 4 factors; 1,2,5,10
Now Assure even - let's say 6 - 4 factors; 1,2,3,6
so 2n is 12 - 6 factors; 1,2,3,4,6,12
Tried with couple of other odd and even nos and found the same result.
IMO B
What is OA?
let me try;
Assume n as Odd - let's say 5, divisible by 2 different integers (factors); 1 & 5
so 2n is 10 - having 4 factors; 1,2,5,10
Now Assure even - let's say 6 - 4 factors; 1,2,3,6
so 2n is 12 - 6 factors; 1,2,3,4,6,12
Tried with couple of other odd and even nos and found the same result.
IMO B
What is OA?
I am I
i think this is B because to have twice the number of factors if multiplied by 2, n has to have factors which are distinct and prime (not including 2)
meaning that it will be odd since all prime numbers other than 2 are odd.
meaning that it will be odd since all prime numbers other than 2 are odd.
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OA is B but can you explain.robust wrote:i think this is B because to have twice the number of factors if multiplied by 2, n has to have factors which are distinct and prime (not including 2)
meaning that it will be odd since all prime numbers other than 2 are odd.
I am not clear how you proved that n has to have factors which are distinct and prime
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Ron Purewal posted a good explanation of the reason here:
www.beatthegmat.com/few-quetions-which- ... 10004.html
www.beatthegmat.com/few-quetions-which- ... 10004.html
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I tried using a table for the second seems to help gather my ideas quite clearly..
n 2n nf 2nf
3 6 2 4
4 8 3 4
5 10 2 4
9 18 3 6
here nf = factors of n
2nf - factors of 2n
you see a pattern emerges. only for odd numbers it seems true.[/list]
n 2n nf 2nf
3 6 2 4
4 8 3 4
5 10 2 4
9 18 3 6
here nf = factors of n
2nf - factors of 2n
you see a pattern emerges. only for odd numbers it seems true.[/list]
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I am not clear how you have calculated nf and 2nf and what pattern emergesvish150783 wrote:I tried using a table for the second seems to help gather my ideas quite clearly..
n 2n nf 2nf
3 6 2 4
4 8 3 4
5 10 2 4
9 18 3 6
here nf = factors of n
2nf - factors of 2n
you see a pattern emerges. only for odd numbers it seems true.[/list]