Wait, what? Why can't there exist such an \(a\)? Am I missing something? Let a = 81, then:
\(\displaystyle
81 \equiv 3 \mod 6 \\
81 \equiv 7 \mod 4 \\
81 \equiv 6 \mod 15
\)
I agree that the Chinese Remainder Theorem won't help here, but I found this solution, among infinitely many others, simply by considering the system of equations:
\(\displaystyle
a = 6K + 3 \\
a = 7L + 4 \\
a = 15M + 6 \\
\{K, L, M\} \in \mathbb{Z}
\)