+0

# system of congruences

0
105
3

Find all numbers r for which the system of congruences

x = r (mod 6)

x = 10 (mod 20)

x = 5 (mod 45)

has a solution.

Dec 28, 2020

#1
+5
+1

Is there not infinite solutions for r?

I think you are missing some restrictions.

Dec 28, 2020
#2
+117
+2

Do you mean $$=$$ or $$\equiv$$? If you mean $$\equiv$$ then I can help you.

From the second equation, since $$x \equiv 9 \pmod{20}$$, we know $$x$$ is odd. Then $$r$$ can be one of $$1$$$$3$$, and $$5$$. From the third equation, $$x \equiv 4 \pmod{45}$$. We see that $$x \equiv 1 \pmod{3}$$ (since $$x = 45y+4$$, for some integer $$y$$). Then we see that $$r$$ can only be $$\boxed1$$ (since if $$x \equiv 3 \mod{6}$$, it will have remainder of $$0$$ when divided by $$3$$ and if $$x \equiv 5 \mod{6}$$, it will have a remainder of $$2$$ when divided by $$3$$.

Dec 29, 2020
#3
+112827
+2

I have the same answer as Crypto, my method is just a bit different.

x=10mod 20  ....   x=10, 30, 50, ............ , 230 ......

x=5mod45     ....   x= 5, 50, 95, 140, 185, 230 ......

since these are both true and 230-50 = 180

x= 50+180k    where k is an integer great or equal to 0

$$\frac{x}{6}=8+30k+\frac{2}{6}\\ so\\ x\mod6 = 2$$

Dec 29, 2020