Discrete Math Chinese Remainder theorem simple problem with division involved

Question

I am having trouble understanding how to finish this problem.

"Which integers leave a remainder of 1 when divided by 2 and also leave a remainder of 1 when divided by 3?".

I started out with M =(2)(3) =6 Then I went and tried to state that A1=1, A2=1 Then M1=2 M2=3 Now I am having trouble of coming up with Y1= ? and Y2= ? I am left with A1*M1*Y1 + A2+M2+Y2 = ... How do I find Y1 and Y2 ?

If anyone can show me please show a process and answers ,because I have encountered other problems involving Chinese remainder theorem However they usually involve A1,A2,A3... M1,M2,M3 etc

Answer 1

Asserting that $x\equiv1\pmod2$ and that $x\equiv1\pmod3$ is the same thing as asserting that $2\mid x-1$ and that $3\mid x-1$. Obviously, this is the same thing as asserting that $6\mid x-1$. So, the solutions of your problem are the numbers of the form $6k+1$.

Answer 2

If $m$ leaves a remainder of $1$ when divided by $2$, then $m-1$ must be a multiple of $2$.

If $m$ leaves a remainder of $1$ when divided by $3$, then $m-1$ must be a multiple of $3$.

If $m-1$ is a multiple of $2$ and of $3$, then $m-1$ must be a multiple of $6$.

Hence $m-1=6t$ for any integer $t$.

Hence $m = 6t + 1$ for any integer $t$.

Check:

$6t+1 = 2(3t)+1$

$6t+1 = 3(2t)+1$