Question

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:

Hint:

When minimizing a multi-digit number under digit constraints: \begin{itemize} \item First fix which digits must appear (here, two composites and one prime), \item Then choose the smallest possible digits, \item Finally arrange them in ascending order for the minimum value. \end{itemize}

Answer 1

Correct Answer: 6

Step 1: Build \(N\) digit by digit, smallest first. Hundreds digit \(1\) is forbidden (it's a perfect square), so try \(2\): it is non-zero, not a perfect square, and prime, so use it as the one allowed prime digit.
Step 2: With \(2\) already used as the prime, the remaining two digits must both be non-prime and non-square. From \(\{3,4,5,6,7,8,9\}\), removing squares \(\{4,9\}\) and primes \(\{3,5,7\}\) leaves only \(\{6,8\}\).
Step 3: Arrange the smallest remaining digits next: tens \(=6\), units \(=8\), giving \(N=268\). Factorising, \(268=2^2\times67\), which has \((2+1)(1+1)=6\) factors.
\[ \boxed{6} \]