[tex]\begin{cases}x\equiv2\pmod3\\x\equiv1\pmod5\\x\equiv3\pmod{29}\end{cases}[/tex]
Let's start by supposing [tex]x=2+3+3=11[/tex]. Modulo 3, we end up with 2 as needed.
But modulo 5, we want to get 1, so we'd need to multiply the first and last terms by 5 and the second term by the inverse of 3 modulo 5. We have [tex]3\times2\equiv6\equiv1\pmod5[/tex], so we multiply by 2:
[tex]x=2\times5+3\times2+3\times5=31[/tex]
But now modulo 3, the first term gives a remainder of 1, so simply multiply by 2:
[tex]x=2\times5\times2+3\times2+3\times5=41[/tex]
Next, modulo 29, we can force the first two terms to vanish by multiplying them by 29, but the last term still yields 15. We want to get 3 on its own, so we could just multiply the third term by the inverse of 5 modulo 29. We have [tex]5\times6\equiv30\equiv1\pmod{29}[/tex].
[tex]x=2\times5\times2\times29+3\times2\times29+3\times5\times6=844[/tex]
Now, [tex]844\equiv1\pmod3[/tex], so we need to multiply the first term by 2 one more time; [tex]844\equiv4\pmod5[/tex], so we need to multiply the second term by the inverse of 4 modulo 5, which would be 4 since [tex]4^2\equiv16\equiv1\pmod5[/tex]; and [tex]844\equiv3\pmod{29}[/tex], so the last term is okay.
[tex]x=2\times5\times2\times29\times2+3\times2\times29\times4+3\times5\times6=1946[/tex]
We know 1946 is a possible solution because we engineered it that way, but it's not the smallest positive solution. We have
[tex]1946\equiv206\pmod{3\times5\times29}\equiv206\pmod{435}[/tex]
The general solution to the system is then [tex]x=206+435n[/tex], where [tex]n\in\mathbb Z[/tex].
