Solution: Divisors or factors of a number, say M, are the integers that can divide into M without a remainder. That is, if we have a number M, such that M = ab and a and b are positive integers, then a and b are factors of M.
When M is expressed as a product of its prime factors only, then we say that we have prime factorized M. If we prime factorize a positive integer, M, as M = \({p_1}^{t_1} \) *\({p_2}^{t_2} \) * ……*\({p_n}^{t_n} \), where \(p_i\) stands for different prime numbers, and \(t_i\) are positive integers and stands for the exponents of the different prime factors or divisors, then the number of factors of M = (\(t_1\) + 1)·(\(t_2\) + 1)......(\(t_n\) + 1).
The important part here is the word “different.”
In the question, n(A) denotes the number of positive divisors of a natural number A. We are required to find the total number of A’s that satisfy n(A) = 3 between 1 and 50, inclusive.
To have 3 positive divisors, A must have a single prime factor with the highest power of 2. This is possible when prime numbers are squared.
=> \(2^2\) = 4 → 3 divisors → 1, 2, and 4.
=> \(3^2\) = 9 → 3 divisors → 1, 3, and 9.
=> \(5^2\) = 25 → 3 divisors → 1, 5, and 25.
=> \(7^2\) = 49 → 3 divisors → 1, 7, and 49.
Hence, there are 4 A’s that satisfy n(A) = 3 between 1 and 50, inclusive.
Therefore, A is the correct answerAnswer A