GMATBeast wrote:

If the integer n has exactly three positive divisors, including 1 and n, how many positive divisors does n^2 have?

(A) 4

(B) 5

(C) 6

(D) 8

(E) 9

OG 11 #241.

Would someone mind explaining? I'm not satisfied with the explanation in the OG.

Important property: the

number of distinct factors of a perfect square is ALWAYS ODD.

The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square. (A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square).

Hence, since given that n has 3 (odd) divisors then

n is a perfect square, specifically square of a prime. The divisors of \(n\) are: \(1\), \(\sqrt{n}=prime\) and \(n\) itself. So, \(n\) can be 4, 9, 25, ... For example divisors of 4 are: 1, 2=prime, and 4 itself.

Now, \(n^2=(\sqrt{n})^4=prime^4\), so it has 4+1=5 factors (check below for that formula).

Answer: B.

Else you can just plug some possible values for \(n\): say \(n=4\) then \(n^2=16=2^4\) --> # of factors of 2^4 is 4+1=5.

Answer: B.

Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\).

NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

So, the # of factors of x=a^2*b^3, where a and b are different prime numbers is (2+1)(3+1)=12.

Hope it's clear.

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