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 digits are non-zero, so the available digits are $1$ through $9$.

Removing perfect squares ($1, 4, 9$), the allowed digits are: $2, 3, 5, 6, 7, 8$

Among these:

• Prime digits: $2, 3, 5, 7$
• Non-prime digits: $6, 8$

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Since exactly one digit must be prime, the other two digits must be non-prime.

The only non-prime allowed digits are $6$ and $8$, so these two must be used.

The third digit (the prime one) can be: $2, 3, 5,$ or $7$

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To minimize $N$, the hundreds digit should be as small as possible.

The smallest available prime is $2$, so the digits are $\{2, 6, 8\}$.

Arranging in ascending order to get the smallest 3-digit number:

$N_{\min} = 268$

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Prime factorisation of $268$:

$268 = 2 \times 134$

$= 2 \times 2 \times 67$

$= 2^2 \times 67$

Here, $67$ is prime since it is not divisible by $2, 3, 5,$ or $7$.

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If $N = p^a \times q^b$, then number of factors $= (a+1)(b+1)$

Number of factors of $268$

$= (2+1)(1+1)$

$= 3 \times 2$

$= 6$