abhasjha wrote:Is the integer n odd?

1. n is divisible by 3

2. 2n is divisible by twice as many positive integers as n

The integer is not odd.-You can only make this out considering both the statements.
Independently insufficient.
Statement 1:Independently ,6 and p both are divisible by 9. Hence,insufficient.
Statement 2: This can only happen when n is a prime number.So,if you multiply it by 2 ,you have 2 as the additional 2nd divisor.

IMO B.

harsh.champ wrote:

abhasjha wrote:Is the integer n odd?

1. n is divisible by 3

2. 2n is divisible by twice as many positive integers as n

The integer is not odd.-You can only make this out considering both the statements.
Independently insufficient.
Statement 1:Independently ,6 and p both are divisible by 9. Hence,insufficient.
Statement 2: This can only happen when n is a prime number.So,if you multiply it by 2 ,you have 2 as the additional 2nd divisor.

IMO B.

It can happen with any odd number not only prime numbers

For example 9 has 3 factors 1,3 and 9

18 on the other hand has 1,2,3,6,9 and 18 - 6 factors [ I will post a detailed proof why this is happening on request from purists (so called)

B is the answer

ajith wrote:

harsh.champ wrote:

abhasjha wrote:Is the integer n odd?

1. n is divisible by 3

2. 2n is divisible by twice as many positive integers as n

The integer is not odd.-You can only make this out considering both the statements.
Independently insufficient.
Statement 1:Independently ,6 and p both are divisible by 9. Hence,insufficient.
Statement 2: This can only happen when n is a prime number.So,if you multiply it by 2 ,you have 2 as the additional 2nd divisor.

IMO B.

It can happen with any odd number not only prime numbers

For example 9 has 3 factors 1,3 and 9

18 on the other hand has 1,2,3,6,9 and 18 - 6 factors [ I will post a detailed proof why this is happening on request from purists (so called)

B is the answer

Can you plz explain how you came to the answer? Why didnt you try to look at even number? Im stuck, because i thought we would need some more numbers to test.

sadullaevd wrote:
Can you plz explain how you came to the answer? Why didnt you try to look at even number? Im stuck, because i thought we would need some more numbers to test.

There is a way to find out no of factors of a number say n

If it can be expressed as n = x^a*y^b*z*c.....
Where x,y,z...each are prime numbers and a,b,c,d ... etc are integers

No of factors of that number n = (a+1)(b+1) .....

I will demonstrate it by an example 12 = 2^2*3^1
No of factors of 12 = (2+1)(1+1) = 6
And let us try to list them - 1,2,3,4,6 and 12


That is the background you need to attempt this question

Now option 2 says - 2n is divisible by twice as many positive integers as n
Now imagine when n is odd, if you multiply by 2 the new number has a new prime factor 2 and it will have 2 (1+1) times more factors and if n is even, this factor is not 2 (b becomes b+1) and the factor is b+1/b

I will again demonstrate by example

12 =2^2*3^1 ; no of factors = 2*3 =6
24 = 12*2 = 2^3*3^1 , no of factors = 4*2 =8
(I hope you notice number of factors did not double)

Now let us take an odd number 15

15 = 3*5 ; no of factors = 2*2 =4
30 = 2*3*5 ; no of factors = 2*2*2 = 8

Hope you understood, I am glad that you asked the question

thank you, now i totally understand it.

OA is B