Number Theory - Remainders A number is of the form $13k_1+12$ and of the form $11k_2+7$
That is $N = 13k_1 + 12 = 11k_2 + 7$
Now why must N also equal $(13 \times 11)k_3 + 51$ ?
Thanks
 A: Alternately, $N=13k_1+12 = 11k_2+7 \implies N-51 = 13(k_1-3)=11(k_2-4)$  
Thus $N-51$ must be a multiple of both $13$ and $11$.
A: Let me give a trick. We have $N + 92 = 13k_1 + 104 = 13(k_1 + 8)$ and $N+92 = 11k_2 + 99 = 11(k_2 + 9)$.
Then, $N+92 = 11\times 13 k_3$ or $N = 143k_3 -92$ or $N=143k_4 + 51$.
See also link
A: $N$ as defined is $12$ mod $13$ by first definition and $7$ mod $11$ by second definition, since both $13$ and $11$ are relatively prime (in fact primes) thus $N$ should have same mod ($13 \times 11$) as well, rest follows by manipulating the $12$, $7$ remainders with $13$, $11$
A: Finding a Solution
To find an $N$ so that
$$
\begin{align}
N&\equiv12&\pmod{13}\\
N&\equiv7&\pmod{11}
\end{align}\tag{1}
$$
we can start by solving
$$
13x+11y=1\tag{2}
$$
using the Extended Euclidean Algorithm. The implementation described in this answer gives
$$
\begin{array}{r}
&&1&5&2\\\hline
1&0&1&-5&11\\
0&1&-1&6&-13\\
13&11&2&1&0\\
\end{array}\tag{3}
$$
which implies
$$
13(-5)+11(6)=1\tag{4}
$$
Using $(4)$, we can show
$$
\begin{align}
66&\equiv1&\pmod{13}\\
66&\equiv0&\pmod{11}
\end{align}\tag{5}
$$
and
$$
\begin{align}
-65&\equiv0&\pmod{13}\\
-65&\equiv1&\pmod{11}
\end{align}\tag{6}
$$
$12$ times $(5)$ plus $7$ times $(6)$ gives
$$
\begin{align}
337&\equiv12&\pmod{13}\\
337&\equiv7&\pmod{11}
\end{align}\tag{7}
$$
Subtracting $2\cdot11\cdot13=286$ from the left sides of $(7)$ gives
$$
\begin{align}
51&\equiv12&\pmod{13}\\
51&\equiv7&\pmod{11}
\end{align}\tag{8}
$$

Finding a General Solution
If $N_1$ and $N_2$ are any two solutions to $(1)$, then
$$
\begin{align}
N_1-N_2&\equiv0&\pmod{13}\\
N_1-N_2&\equiv0&\pmod{11}
\end{align}\tag{9}
$$
Therefore,
$$
N_1-N_2\equiv0\pmod{11\cdot13}\tag{10}
$$
Putting together $(8)$ and $(10)$, we get that all the solutions of $(1)$ are given by
$$
N\equiv51\pmod{143}\tag{11}
$$
