In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a perfect square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is

According to question, in the given 3-digit number N, the digits are non-zero and distinct.

So, the possible digits in 3-digit number = $$2,3,5,6,7,8$$

It is also given that only one of the digits is a prime number.

So, the minimum possible value of N = 268

(2 is the smallest prime digit, and the non-prime digits has to be 6 and 8)

Now, $$268=4\times\ 67=2^2\times\ 67$$

So, the number of factors = $$\left(2+1\right)\left(1+1\right)=3\times\ 2=6$$