About Chinese Remainder Theorem

[peterso]

the question is find all possible integers a which satisfy the following system of congruences

a = 3 mod 6
a = 4 mod 7
a = 6 mod 15

but I find a = 3 mod 6 is 105n₁ = 3n₁ = mod 6
and their gcd is 3 not 1 so in this question is it no solutuion?

 

[HallsofIvy]

Strictly speaking the "Chinese remainder theorem" does not apply here because 6 and 15 are not mutually prime. There does not exist such an a.

 

[ksdhart2]

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}
\)

 

[ksdhart2]

Whoops! I've just been informed I made a big error and reversed the numbers. I meant to type "4 mod 7" to be consistent with the information given in the problem. That's my bad, sorry about that.