If p is an integer, then p is divisible by how many positive integers?

we are looking for the number of factors of p.

Statement 1: \(p = 2^x\) , where x is prime. check it.

\(p = 2^2 or 2^3 or 2^5\)

when \(2^2\), number of factors will be 2 + 1 = 3

when \(2^3\) , number of factors will be 3 + 1 = 4.

So, Not sufficient.

Statement 2:\(p = x^2\), x is a prime.

\(2^2 or 3^2 or 5^2\).

when \(2^2\) , factors = 2 + 1 = 3

when \(3^2\) factors = 2 + 1 = 3.

Sufficient.

The best answer is B.

St1:- p = 2^x, where x is a prime number
Possible values of p: \(2^2,2^3,2^5,2^7\)
No of factors of p: 3,4,6,8 etc.
Insufficient

St2:- p = x^2, where x is a prime number
Here x can't be factorized further and power of x is a constant.
Hence, the no of factors is constant i.e, 3.
Sufficient.

Ans. (B)

For those who know the "Unique Factors Trick" statement 2 should be really quick without any number picking. In the Unique Factors Trick you break a number down to its primes and express them as exponents (so like 12 = 2^2 * 3^1), then add 1 to each exponent and multiply the exponents together (so with 12 = 2^2 * 3^1 you'd add 1 to each exponent and have (2+1)(1+1) = 3 * 2 = 6 factors (and those factors are 1, 2, 3, 4, 6, and 12).

Well here even if you don't know x, you do know that it's prime, so you've already taken the first step of the Unique Factors Trick: you have its prime base, x, taken to an exponent. Just add one to that exponent and you're set. x^2 --> 2 + 1 factors = 3 factors.

For this one, too, it's not a bad idea to know the "rule" of statement 1 in reverse: if an integer has exactly three unique factors, it's the square of a prime number.