Chinese Remainder Theorem Finding the Modulo

Question

Find numbers $t,u,v$ so that $33t+2 = 20u+13 = 29v-1 $

This is a Chinese Remainder Theory problem, but the problem I am having is finding what are the appropriate modulo. I figure it is easiest to start with solving $33t+2=29v-1$ so I could rewrite $33t +3\equiv 29v(\mod ????) $ Or I could just see where multiplying values of $T$ gets me something that will divide $29$ but I feel like this is the wrong way to approach the problem. If someone could give me help finding what modulo I am looking for, it would be much appreciated.

Answer 1

Hint: $33t+2 = x$ means that $x \equiv 2 \pmod{33}$.

Answer 2

Let $33t+2=20u+13=29v−1=x$.

Then $x \cong 2 (mod 33)$, $x \cong 13 (mod 20)$, $x \cong 1 (mod 29)$.

Let's handle the two first.

33t+2=20u+13, so 33t-11=20u, and 11(3t-1)=20u. Hence $u$ is divisible by $11$. However, remember that we are dealing with $(mod 33)$.

If we write $u \cong 0 (mod 33)$, then it is wrong because $11(3t-1)=33t-11 \cong -11 (mod 33)$, and $20u \cong 0 (mod 33)$.

If we write $u \cong 11 (mod 33)$, then it is correct because $11(3t-1)=33t-11 \cong 22 (mod 33)$, and $20y \cong 22 (mod 33)$.

If we write $u \cong 22 (mod 33)$, then it is wrong beause $20u \cong 11 (mod 33)$.

Thus $u \cong 22 (mod 33)$ and $x \cong 22\cdot20+13=493 (mod 660)$ (substitute back)

Then continue it for the third equation.

Answer 3

As was mentioned the equations are x = 2(mod 33), x = 13(mod 20), and x = -1(mod 29). Solving the first two equations simultaneously, you get x = 233(mod 660). Solving this result with the third equation, you get x = 14093(mod 19140) which is the final answer.