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

The correct answer is 6.

To find the minimum possible value of the 3-digit number N, let's break down the given conditions step-by-step to figure out what digits can be used.

Step 1: Identify the pool of available digits.
The digits of N must be non-zero, meaning they can only be drawn from {1, 2, 3, 4, 5, 6, 7, 8, 9}.
Additionally, none of the digits can be a perfect square. The perfect square digits from 1 to 9 are 1, 4, and 9. Excluding these leaves us with the possible digits: {2, 3, 5, 6, 7, 8}.

Step 2: Apply the prime number condition.
From our remaining pool of {2, 3, 5, 6, 7, 8}, we can categorize the numbers into primes and non-primes (composites):
Prime numbers: {2, 3, 5, 7}
Non-prime numbers: {6, 8}

The problem states that exactly one of the three digits in N is a prime number. Because the digits must be distinct, the other two digits must come from the non-prime group. Since there are exactly two non-prime digits available ({6, 8}), N must contain both the digit 6 and the digit 8.

Therefore, the three distinct digits of N will be {P, 6, 8}, where P is one of the prime numbers from {2, 3, 5, 7}.

Step 3: Determine the minimum possible value of N.
To make the 3-digit number N as small as possible, we should place the smallest available digits in the highest place values (hundreds and tens).
To minimize N, we must choose the smallest possible prime digit for P, which is 2. This makes our set of three digits {2, 6, 8}.
To form the smallest number using the digits 2, 6, and 8, we arrange them in ascending order: 2 for the hundreds place, 6 for the tens place, and 8 for the units place. Thus, the minimum possible value of N is 268.

Step 4: Find the number of factors of N.
Now we need to find the number of factors of 268. First, let's find its prime factorization by dividing it by the smallest prime factors.

$268÷2=134$

$134÷2=67$

Since 67 is a prime number itself, the prime factorization of 268 is:

$268=2^2\times {67}^1$

To find the total number of factors of a number given its prime factorization, we take the exponents of each prime factor, add 1 to each exponent, and multiply those sums together:

$\text{Number of factors}=(2+1)\times (1+1)$

$\text{Number of factors}=3\times 2=6$

Therefore, the number of factors of the minimum possible value of N is 6.