BTGmoderatorDC wrote: ↑Sun Dec 29, 2019 4:42 pm
If x is a positive integer, how many positive integers less than x are divisors of x ?
(1) x^2 is divisible by exactly 4 positive integers less than x^2.
(2) 2x is divisible by exactly 3 positive integers less than 2x.
OA
A
Source: Official Guide
Solution:
If we can determine the number of divisors of x, then if we subtract 1 from this number, the result will be the number of divisors of x that are less than x.
Statement One Only:
x^2 is divisible by exactly 4 positive integers less than x^2.
This means x^2 has 5 divisors (including x^2 itself). In order to have 5 divisors, x^2 = p^4 for some prime number p. Therefore, x = p^2 and hence x has 2 + 1 = 3 divisors (including x itself). So there are 2 divisors of x that are less than x.
Statement one alone is sufficient.
Statement Two Only:
2x is divisible by exactly 3 positive integers less than 2x.
This means 2x has 4 divisors (including 2x itself). In order to have 4 divisors, 2x = p^3 for some prime number p or 2x = q * r for some distinct prime numbers q and r. In the former case, x = p^3/2. However, since x is a positive integer and p is a prime, p must be 2. Therefore, x = 2^3/2 = 4, and hence x has 3 divisors (namely, 1, 2 and 4). So there are 2 divisors of x that are less than x. In the latter case, x = (q * r)/2. Again, since x is a positive integer, either q or r is 2. If q = 2, then x = r and if r = 2, then x = q. Either way, x is a prime and has 2 divisors. So there is only 1 divisor of x that is less than x.
Statement two alone is not sufficient.
Answer: A
