Find all numbers for which the system of congruences

has a solution.

\begin{align*}

x &\equiv r \pmod{6}, \\

x &\equiv 9 \pmod{20}, \\

x &\equiv 4 \pmod{45}

\end{align*}

pls explain it in a simple way

Dysoncloud May 31, 2024

#1**+1 **

I don't know anything about modular math, but here's an answer from a "Guest."

This one is relatively simple. You have to take each modulus x some constant + remainder = each other as follows:

(6 * a) + r =(20 * b) + 9 =(45 * c ) + 4=x. You simply try, by iteration, to find the values of a, b, and c.

If you look at (45*c) + 4 and try c= 1, then it becomes: 45*1+4=49. and if you look at 20b+9, you will see that b= 2, and it becomes 20*2 + 9 =49. Check and see if it balances 6*a +r. If you take 49 and divide by 6, you will get 8, because 6*8 =48. And so 49 mod 6 =1. Therefore, r can only =1.

Since the LCM of 6, 20, 45 =180, therefore x will be:

x =180C + 49, and C =0, 1, 2, 3, 4, 5.......etc. So:

x =180*0 + 49 =49

x =180*1 + 49 =229

x =180*2 +49 =409

x =180*3+49 =589..........and so on. All these values of x satisfy the congruences.

CPhill May 31, 2024