Determining Numbers with 15 Factors
Given that a number \( N \) has 15 factors.
The factor count of 15 can be expressed as \( 15 = 1 \times 15 = 3 \times 5 \).
Therefore, the possible exponent combinations for the prime factorization of \( N \) are:
- \( (p+1)(q+1) = 3 \times 5 \Rightarrow p = 2, q = 4 \)
- This implies \( N \) can be represented as \( N = a^2 \cdot b^4 \) or \( N = a^4 \cdot b^2 \), where \( a \) and \( b \) are distinct prime numbers.
Scenario 1: \( N = 2^4 \times 3^2 \)
\( 2^4 = 16 \), \( 3^2 = 9 \)
\( \Rightarrow N = 16 \times 9 = 144 \)
Scenario 2: \( N = 2^2 \times 3^4 \)
\( 2^2 = 4 \), \( 3^4 = 81 \)
\( \Rightarrow N = 4 \times 81 = 324 \)
Sum of the Two Smallest Numbers Found:
\( 144 + 324 = 468 \)
Correct Answer: (C): \( \boxed{468} \)