amanvermagmat
A positive integer X has 'n' total factors, another positive integer Y has 'm' total factors. Is X+Y odd?
(1) m = 2*n.
(2) 2*X has 2*n total factors.
The sum (X + Y) will be odd if one of X or Y is odd and the other is even. The sum will be even if both are odd or both are even.
Given: X has n factors. Y has m factors.
Statement 1: m = 2*n
Y has twice as many factors as X. It is possible with one being odd and the even or otherwise.
Example: X = 2. Y = 6
X has 2 factors and Y has 4 factors. X+Y is even.
Counter example: X = 3. Y = 6
X has 2 factors. Y has 4 factors. X + Y is odd.
Statement 1 alone is NOT sufficient.
Statement 2: 2*X has 2*n
If 2X has 2n factors and X has n factors, it is possible only when 2 is NOT a prime factor of X. So, we can infer that X is odd.
But we do not know anything about Y.
If Y is also odd, (X + Y) will be even. If Y is even, (X + Y) will be odd.
Statement 2 alone is NOT sufficient.
Combining the statements: m = 2*n and 2*X has 2*n
From statement 2, X is odd.
Example: X = 5 and Y = 6.
X has 2 factors. 2X = 10 has 4 factors. So, satisfies statement 2.
X has 2 factors and Y has 4 factors. So, satisfies statement 1.
Sum (X + Y) is odd.
Example: X = 5 and Y = 15.
X has 2 factors. 2X = 10 has 4 factors. So, satisfies statement 2.
X has 2 factors and Y has 4 factors. So, satisfies statement 1.
Sum (X + Y) is even.
Statements together NOT sufficient. Choice E.
Theory behind number of factors: If a number N can be prime factorised as a^p * b^q, number of factors of N will be (p + 1)(q + 1).
If N is even, then one of a or b will be 2. Let us say a = 2. p > 0.
2N will take power p to (p + 1).
Therefore, number of factors of 2N = (p + 2)(q + 1)
Number of factors of 2N/Number of factors of N = (p + 2)/(p + 1)
If this value has to be 2, it is possible only when p = 0. If that is the case, then 2 could not have been a factor of N.
Inference: If multiplying a prime c to a number N doubles the number of factors of N, then c is not a factor of N.