If the integer n has exactly three positive divisors, including 1 and n, how many positive divisors does n^2 have?
(E) 9OG 11
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
itself. So, n
can be 4, 9, 25, ... For example divisors of 4 are: 1, 2=prime, and 4 itself.
, so it has 4+1=5 factors (check below for that formula).
Else you can just plug some possible values for n
: say n=4
--> # of factors of 2^4 is 4+1=5.
Answer: B.Finding the Number of Factors of an Integer
First make prime factorization of an integer n=a^p*b^q*c^r
, where a
, and c
are prime factors of n
, and r
are their powers.
The number of factors of n
will be expressed by the formula (p+1)(q+1)(r+1)
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
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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