<s>[INST] 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!
 [/INST] 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.
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