Find all numbers for which the system of congruences

\(\begin{align*} x &\equiv r \pmod{6}, \\ x &\equiv 9 \pmod{20}, \\ x &\equiv 4 \pmod{45} \end{align*}\)

has a solution.

Guest Jun 16, 2018

#1**+1 **

Because the numbers involved are very small, if you look at the last 2 congruences, and just by simple inspection, you will see that x = 49 works. Substitute it in the first and you should get r = 1.

So, the simple answer seems to be: x = 49 and r = 1.

Since the LCM of [6, 20, 45] = 180, then:

x =180n + 49, where n =0, 1, 2, 3.........etc.

So, x can take infinite values such as: 49, 229, 409, 589........etc.

Guest Jun 16, 2018

edited by
Guest
Jun 16, 2018