Find the smallest positive integer that satisfies the system of congruences $N \equiv 2 \pmod{11}, N \equiv 3 \pmod{17}. $ Find the smallest positive integer that satisfies the system of congruences
\begin{align*}
N &\equiv 2 \pmod{11}, \\
N &\equiv 3 \pmod{17}.
\end{align*}
The only way I know to solve this problem is by listing it all out, and so far, it's not working.  Is there a faster way?  Thanks for posting a solution!
 A: By the second constraint $N$ is a number of the form $17k+3$. We may now impose the first constraint:
$$ 17k+3\equiv 6k+3 \equiv 3(2k+1) \equiv 2\pmod{11} $$
leading to $2k+1\equiv 8\pmod{11}$, equivalent to $k\equiv 9\pmod{11}$. It follows that the smallest positive number fulfilling both constraints is given by
$$ \color{red}{N}=17\cdot 9+3 = \color{red}{156}.$$
A: How it can be done using 'Euclid's algorithm': $N\equiv 3$ mod 17 can be written as N= 17i+ 3.  Since 17= 11+ 6 is equivalent to 6 mod 11, we can write $N\equiv 2$ (mod 11) as $N\equiv 17i+ 3\equiv 6i+ 3= 2$ (mod 11) which is the sae as $6i= 2- 3= -1= 10$ (mod 11) or, dividing by 2, $3i\equiv 5$ (mod 11).  
That is the same as 3i= 5+ 11j or 3i- 11j= 5.  Euclids algorithm: 3 divides into 11 three times with remainder 2: 11- 3(3)= 2.  2 divides into 3 once with remainder 1: 3- 2= 1.  Replace that "2" with 11- 3(3): 3- (11- 3(3))= 4(3)- 1(11)= 1.  Multiplying both sides by 5, 20(3)- 5(11)= 5.  
So one solution to 3i- 11j= 5 is i= 20, j= 5.   Since we had before N= 17ii+ 3, N= 17(20)+ 3= 340+ 3= 343.  
To check: 11 divides into 343 31 times with remainder 2: 343= 2 mod 11.  17 divides into 343 20 times with remainder 3: 343= 3 mod 17.
A: Let's write
$$N=11x+2=17y+3$$
or
$$11x-17y=1$$
and solve it for $x$ and $y$.
Taking mod $11$,
$$5y\equiv 45\pmod{11}$$
For $y=9$ we have $$11x-153=1$$
and $x=14$.
You can also use Bezout's identity.
A: You have to find a Bézout's  relation between $11$ and $17$, either using the Extended Euclidean algorithm or finding an obvious relation. Here, it's easy to find
$$2\cdot 17-3\cdot 11=1.$$
Once you have this relation, it's easy to check  that a solution to the system of congruences $$\;\begin{cases}N\equiv a\mod11,\\N\equiv b\mod 17\end{cases}\quad \text{is}\quad N\equiv\color{red}{2a\cdot 17-3b\cdot 17\mod 11\cdot 17}.$$
