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:
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Correct Answer: 6
Solution and Explanation
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).
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