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*5^2Total 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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60% (01:37) correct
