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:

To form the smallest number from a given set of digits, arrange them in ascending order. To find the largest number, arrange them in descending order. Always tackle number theory problems by breaking down the constraints one by one.

Answer 1

Correct Answer: 6

Step 1: Allowed digits (non-zero, not a perfect square) are \( \{2,3,5,6,7,8\} \); the primes among them are \( 2,3,5,7 \) and the non-primes are \( 6,8 \).
Step 2: Test hundreds digit \( =2 \) first, the smallest allowed digit. With hundreds \( =2 \) (already the one permitted prime), the tens and units digits must both come from the non-prime set \( \{6,8\} \); the smallest combination is tens \( =6 \), units \( =6 \), giving \( 266 \). No smaller hundreds digit is allowed, so \( 266 \) is confirmed minimal.
Step 3: Factorize: \( 266=2\times7\times19 \) (three distinct primes).
Step 4: Number of factors \( =(1+1)(1+1)(1+1)=8 \).
\[ \boxed{8} \]