a = 3 mod 6

a = 4 mod 7

a = 6 mod 15

but I find a = 3 mod 6 is 105n

_{1}= 3n

_{1}= mod 6

and their gcd is 3 not 1 so in this question is it no solutuion?

- Thread starter peterso
- Start date

a = 3 mod 6

a = 4 mod 7

a = 6 mod 15

but I find a = 3 mod 6 is 105n

and their gcd is 3 not 1 so in this question is it no solutuion?

- Joined
- Jan 27, 2012

- Messages
- 5,178

\(\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}

\)