IMO B

OA A

sivaelectric wrote:1. If sqrt x is a positive integer, is sqrt x a prime number?

• A. x is divisible by exactly 3 positive numbers.
  B. All positive factors of x are odd.

Statement 1: x is divisible by exactly 3 positive numbers.
To determine the number of factors of an integer:

1. Prime-factorize the integer
2. Add 1 to each exponent
3. Multiply

For example:
24 = 2^3 * 3^1.
Adding 1 to each exponent and multiplying, we determine that 24 has (3+1)*(1+1)= 8 factors.

Thus, in order for x to have exactly 3 positive factors, x = (prime number)^2. Adding 1 to the exponent, we can see that x will have exactly 2+1=3 factors.
Since x is the square of a prime number, √x is prime.
Sufficient.

Statement 2: All positive factors of x are odd.
The factors of x=9 are 1,3,9. √9 = 3, which is prime.
The factors of x=81 are 1,3,9,27,81. √81=9, which is not prime.
Insufficient.

The correct answer is A.

Given: x^1/2 is a positive integer
To Find: is x^1/2 a prime number


Evaluating Statement 1:

X is divisible by exactly 3 positive number

The numbers that are divisible be exactly three numbers are always a square of prime numbers

2^2 = 4 -> 1,2,4
3^3 = 9 -> 1,3,9
5^5 = 25 -> 1,5,25
7^7 = 49 -> 1,7,49
and so on

from this we can determine that sqrt(x) will always be a prime number if x has exactly three positive factors

Therefore sufficient

Evaluating Statement 2:

All +ve factors of x are odd

Now, by testing numbers we can prove that this is insufficient

49 -> sqrt(49)=7 a +ve integer -> Factors 1,7,49-> odd no of factors -> sqrt(x) is a prime
81 -> sqrt(81)=9 a +ve integer -> Factors 1,3,9,27,81-> odd no of factors -> sqrt(x) is not prime

Therefore insufficient

Answer should be A

IMO A