Question

N is a 3-digit number with non-zero digits. No digit is a perfect square and only 1 of the digits is a prime number. What is the number of factors of the smallest such number possible?

Hint:

When constructing the smallest number from a set of digits, place the smallest digit in the highest place value (hundreds), the next smallest in the tens place, and so on. Also, be mindful of ambiguities like "digits" vs. "distinct digits" in problem statements.

Answer 1

Correct Answer: 6

Step 1: Rule out digits first: 0 and 1 cannot lead (0 makes it not a 3-digit number if it is the hundreds digit, and both 0 and 1 are perfect squares); 4 and 9 are also perfect squares, so every digit must come from \(\{2,3,5,6,7,8\}\), with exactly one of the three being prime (2, 3, 5 or 7) and the rest composite (6 or 8).
Step 2: To minimize the number, the hundreds digit should be as small as possible. The smallest allowed digit overall is 2, and using it as the (single allowed) prime frees the tens and units digits to both be the smallest composite, 6. That gives \(N=266\).
Step 3: List out the divisors of 266 directly instead of using the exponent formula: \(1, 2, 7, 14, 19, 38, 133, 266\).
Step 4: Count them.
\[ \boxed{8} \]
Final Answer: 8 factors.