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: List allowed digits. Digits are non-zero and not perfect squares. Perfect squares from 1 to 9: \[ 1 = 1^2,\quad 4 = 2^2,\quad 9 = 3^2. \] So allowed digits: \[ \{2,3,5,6,7,8\}. \]
Step 2: Classify digits as prime / non-prime. Primes: \[ \{2,3,5,7\}. \] Non-primes (composite): \[ \{6,8\}. \] We need a 3-digit number with distinct digits and \emph{exactly one} prime digit. Thus, 1 prime digit + 2 non-prime digits. But the only available non-primes are 6 and 8, so they must both be used. The remaining digit must be a prime, and to minimize $N$, choose the smallest prime: \(2\). So the digits of $N$ are: \[ \{2,6,8\}. \]
Step 3: Form the minimum possible number. Arrange digits in ascending order to get the smallest 3-digit number: \[ N = 268. \]
Step 4: Find number of factors of $N$. Prime factorize $268$: \[ 268 \div 2 = 134,\quad 134 \div 2 = 67,\quad 67 \text{ is prime}. \] So: \[ 268 = 2^2 \times 67^1. \] If \[ N = p_1^{a_1} p_2^{a_2} \cdots, \] then the number of positive factors is: \[ (a_1 + 1)(a_2 + 1)\cdots \] Here: \[ a_1 = 2,\; a_2 = 1 \Rightarrow \text{Number of factors} = (2+1)(1+1) = 3 \times 2 = 6. \] Therefore, the number of factors of the minimum possible $N$ is \[ \boxed{6}. \]