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.
\(x^2\) will be divisible by exactly 5 integers including \(x^2\), only when x is a perfect square of a prime number..
for example \(x=a^2..x^2=a^4\)... Number of factors = \((4+1)=5\)
So, x or \(a^2\) will have (2+1) or 3 factors, and our answer is 2 as we have to exclude x itself.
Suff

(2) 2x is divisible by exactly 3 positive integers less than 2x.
This can have two cases for 4 factors including 2x itself..
a) x is any other prime number other than 2...\(2^1x^1\) will have factors as (1+1)(1+1)=2*2=4.. x will have (1+1) or 2 factors and just 1 factor less than x itself.
b) x is also in terms of 2 or \(x=2^2\), so \(2x=2^3\)...factors =(3+1) or 4, and x or \(2^2\) will have (2+1) or 3 factors, and our answer is 2 as we have to exclude x itself.
Insuff


A

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Bunuel Could you please provide a simplified answer for this question??

Here is my solution. priya2810

Statement 1: x^2 = 16 has only 4 divisors, then x = 4 has only two divisors 1 and 2 excluding 4. Now, x^2 = 25, 36, or 49 all have less or more divisors than 4 so they do not work. Thus, the number of positive integers less than x that are divisors of x is 2.

Statement 2: take 2x=6 and 2x=8. The first x will equal 3 and second will equal 4. 3 has only 1 divisor less 3 and that is 1. But 4 has 1 and 2 as divisors less than 4. This makes this statement insufficient.

Answer is A.

Does that help?

Bunuel
chetan2u
VeritasKarishma
can u help me with this?

say x=p^a*q^b where p and q are prime numbers

now no of factors excluding x = (a+1)(b+1)-1= a+b+ab

1)x^2=p^2a*q^2b
now no of factors of x^2 excluding x^2
= (2a+1)*(2b+1)-1=4
implies a+b+2ab=2

so how to solve further
required is a+b+ab but i have a+b+2ab ?

Plz help

Why do you assume that x=p^a*q^b ?
It is possible that x = p^a or x = p^a*q^b*r^c...
etc

You just need the number of factors of x and you can subtract 1 out of it to get the number of factors less than x.

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

So x^2 has 5 factors. 5 cannot be a product of two numbers greater than 1 since it is a prime number.
So x^2 = p^4
to give x = p^2
So x will have total 3 factors and 2 factors less than x.
Sufficient

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

So 2x has 4 total factors.
4 = 4*1 = 2*2
2x = p*q (in this case, x is not 2 but some other prime number. So it will have exactly 1 factor smaller than itself)
or
2x = p^3 (in this case x = 2^2 so it has 2 factors smaller than itself)
We don't know whether x has 1 or 2 factors less than itself.
Not sufficient

Answer (A)
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VeritasKarishma: Thanks for the explaination.

Very high quality question from OG Advanced!

Statement 1: x² has 5 positive factors
I.e. x = a² where a is prime
x has 3 factors and 2 if them are less than x
*Sufficient*

Statement 2 says that x may be 2² or an odd prime number.
So x may have 3 factors or 2 factors hence
*Not sufficient*

Answer: Option A
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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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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.

For x^2 to have exactly 5 factors, it needs to be a the square of a prime number.

For example \(x = 2^2; x^2 = 4^2 = 16\)

There are 2 factors of x that are less than x (1 and 2). SUFFICIENT.

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

This statement tells us that 2x has 4 divisors. x may be \(2^2\) or an odd prime number -- x may have 2 or 3 factors. INSUFFICIENT.

Answer is A.
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