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

To solve the problem, we need to determine the minimum 3-digit number $N$ where each digit is non-zero, distinct, not a perfect square, and only one digit is a prime number. Then, we count the factors of this number.

Step 1: Identify possible digits.
- The digits must be non-zero: {1, 2, 3, 4, 5, 6, 7, 8, 9}.
- Digits that are perfect squares: {1, 4, 9}.
- Remaining digits: {2, 3, 5, 6, 7, 8}.

Step 2: Determine prime digits from the possibilities.
- Prime digits: {2, 3, 5, 7}.
- Only one digit should be prime.

Step 3: Construct the minimum number with the conditions.
- Since only one digit should be prime, choose the smallest prime digit: 2.
- Choose the two smallest non-prime digits: 6 and 8.
- Arrange in ascending order to form the smallest number: 268.

Step 4: Verify and compute the number of factors for 268.
- Confirm digits: 2 (prime), 6 and 8 (non-square, distinct).
- Factorize 268: \(268 = 2^2 × 67^1\).
- Total factors: \((2+1) × (1+1) = 3 × 2 = 6\).

Conclusion: The number of factors of the minimum possible value of \(N = 268\) is 6.
Verification: The result, 6, is within the given range (6, 6).