If positive integer number n has exactly 8 divisors, what is the least number of divisors can n2 have?
To solve this problem, we need to find the least number of divisors that n^2 can have, given that n has exactly 8 divisors.
First, let's understand the relationship between the number of divisors of a number and the prime factorization of that number.
The prime factorization of a number represents the number expressed as a product of its prime factors. For example, the prime factorization of 12 is 2^2 * 3^1, which means 12 can be expressed as a product of two 2s and one 3.
The number of divisors of a number can be determined using its prime factorization. Suppose a number can be expressed as a product of prime factors raised to certain powers. If the prime factorization of the number is a^x * b^y * c^z, then the number of divisors is given by (x+1) * (y+1) * (z+1).
Now, let's apply this knowledge to solve the problem. We know that n has exactly 8 divisors. So, let's assume the prime factorization of n as a^x * b^y.
Since n has 8 divisors, we have (x+1) * (y+1) = 8.
To find the least number of divisors n^2 can have, we need to minimize the value of (x+1) * (y+1).
Since n^2 = (a^x * b^y)^2 = a^(2x) * b^(2y), the prime factorization of n^2 will have double the exponents of a and b.
If x+1 = 2(k+1) for some positive integer k, then 2x = 2k+2, and (x+1) will have k+1 prime factors.
Similarly, if y+1 = 2(l+1) for some positive integer l, then 2y = 2l+2, and (y+1) will have l+1 prime factors.
So, to minimize (x+1) * (y+1) and find the least number of divisors for n^2, we should choose the smallest possible values for k and l.
Since k+1 and l+1 should be prime factors, we choose k=0 and l=0.
Thus, x+1 = 2(0+1) = 2, y+1 = 2(0+1) = 2.
So, the minimum number of divisors for n^2 is (x+1) * (y+1) = 2 * 2 = 4.
Therefore, the least number of divisors n^2 can have is 4.