Question

n is a positive integer with exactly two different divisors greater than 1, how many positive factors does n^2 have? A. 4 B. 5 C. 6 D. 8 E. 9

Answer 1

Answer:

C. 5

Step-by-step explanation:

Remember that for any integer $n$, the integers $1 \text{and} n$ are both divisors (or factors) of $n$. First, we will prove that n is a square, and then we will compute the factors of the n².

In this case, the integer n has exactly two different divisors greater than 1. It's impossible that $n=1$, since 1 doesn't have positive factors greater than 1. Then $n>1$, therefore $n$ itself is one of the required divisors. Denote by $a$ the other divisor greater than 1, and note that to satisfy the condition on the divisors, $a<n$.

Because a divides n, there exists some integer $k$ such that $n=ak$. We must have that $k>1$, if not, then $k\leq 1$, which implies that  $ak=n\leq a$, which contradicts the part above.

Now, $k>1$ and, by definition of divisibility, k divides n. Then k must be equal either to n or a, since we can't have three different divisors of this kind. If $k=n$ then $n=an$ and by cancellation, $1=a$ which is a contradiction. Therefore $k=a$ and $n=a^2$.

We have that $n=(a^2)^2=a^4$. We can write n as $n=a^3a=a^2a^2=a^4 \cdot 1$. From the first equation, a divides n and a³ divides n. From the second equation, a² divides, and from the last one, 1 divides n and a⁴=n divides n. Thus n has exactly 5 positive factors.