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If x is a positive integer, how many positive integers less

Expert replies
by BTGmoderatorDC » 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
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Source: — Data Sufficiency |

by Jay@ManhattanReview » Tue Dec 31, 2019 12:07 am
BTGmoderatorDC wrote: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
Let's take each statement one by one.

(1) x^2 is divisible by exactly 4 positive integers less than x^2.

=> x^2 is divisible by exactly 5 positive integers, including x^2.

Thus, x^2 = a^4, where a is a prime number. The reason is that the no. of factors of a^4 = 4 + 1 = 5. For example, Say a^4 = 2^4 = 16. The no. of factors of 16 are 1, 2, 4, 8 and 16.

Thus, x = a^2. The no. of factors of a^2 are 1, a, and a^2: there are 3 factors. Thus, there are 2 positive integers less than x are divisors of x. Sufficient.

(2) 2x is divisible by exactly 3 positive integers less than 2x.

The correct answer: A

Hope this helps!

-Jay
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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

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