DS Question 13

Is the integer n odd?

(1) n is divisible by 3.

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

Answer - B

Statement (1) - insufficient as n can be both even and odd.
Statement (2) - sufficient - Here n will have a unique prime factorisation as a product of

p1 ^ q1 * p2 ^ q2 * ....... pk ^qk

where p1 is less than p2 , p2 is less than p3 and so on.

Now p1 is either equal to 2 or it is an odd integer greater than 2.
Thus if n is odd then 2n has a unique prime factorisation of

2 * p1^q1 * p2^q2 * .... * pk ^ qk

but if n is even then 2n can be factorised as

p1^(q1+1) * p2^q2 * ... where p1=2, q1 >= 1

Hence , now the total number of factors is based on choice of prime powers for each prime in term.

Thus if n is odd, then 2n must have twice as many factors as n.

Alternatively in simpler language

If n is even --- 2n will not have twice as many divisors as n
e.g . For n=2; 2n or 4 has 1,2,4 as divisors ,
For n = 4 divisors are 1,2,4; 2n=8 has 1,2,4,8

But if n is odd
then if n=3 divisors are 1,3 and 2n=6 divisors are 1,2,3,6
n=5 divisors are 1,5 and 2n=10 divisors are 1,2,5,10

Thus if n is odd the 2nd condition is satisfied. Hence ans is B