aadikamagic wrote:
If x represents the number of positive factors of integer y, is x odd?
(1) y = n! where n is a positive integer greater than 1
(2) y = m^2 − 1 where m is a positive integer greater than 1
This is a very good question!
Logic used Number of factors of any number = product of the number of factors of its prime factors
Eg: 6 = 3*2;
Number of factors of 3 is 2 {1,3}
Number of factors of 2 is 2 {1,2}
Therefore, number of factors of 6 = 2*2 =4 {1,2,3,6}
Now evaluating the statements:
Statement 1: y = n! and n is an integer greater than 1
Number of factors of 2 is 2 (even)
Therefore, number of factors of n!, with n>1, = 2*(some integer) = even
Therefore, no matter what, the total number of factors of n! with n>1 is even
Statement 1 is sufficient
Statement 2: y = m^2 − 1 where m is a positive integer greater than 1
y=(m-1)*(m+1)
Please note that m is greater than 1
Therefore, m-1>0
Case 1: m=2
Therefore, y=(2-1)(2+1) = 1*3
Number of factors of 3 {1,3} is 2. Therefore, even
Case 2: m>2
Therefore, m-1>1
In this case, undoubtedly, y is a product of prime numbers (if y is a product of composite numbers, then composite numbers are a product of prime numbers)
Therefore, y=prime*(any integer)
Number of factors of y = (number of factors of a prime number)*(some integer)
= 2*(some integer)
= even
Therefore, no matter what, number of factors of y are ever i.e. x is even
Statement 2 is sufficient
Therefore, (D)
Mool Mantra: Only 1 and perfect even powers of prime numbers have odd factors. Rest all have even factors