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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
avatar+5 
+1

Is there not infinite solutions for r?

I think you are missing some restrictions.

 Dec 28, 2020
 #2
avatar+86 
+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
avatar+112029 
+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

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