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# Help ASAP

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Find all numbers $r$ 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.

May 26, 2018

### Best Answer

#2
+102765
+2

This is the question.

Find all numbers r 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.

I get r=1

Lets see....  (All the pronumerals I will use are integers)

x/20 = k remainder 9 so    x=20k+9

x/45 = t remainder 4 so     x=45t+4

so

20k+9 = 45t+4

45t-20k = 9-4

45t - 20k = 5

9t - 4k = 1

one obvious solution is t=1 and k=2

9( 1 ) - 4 (2) =1

9(1+4n)-4(2+9n)=1      (because I have added 36n and then taken 36n away again)

so a general solution is   t=1+4n   and  k=2+9n

x=20(2+9n)+9 = 40+180n+9 = 49+180n

or

x=45(1+4n)+4 = 45+180n+4 = 49+180n      Good they are the same.

Now we have x is equivalent to r mod 6

So

(49+180n)/6 = Q remainder r

49+180n = 6Q+r

48+180n+1=6Q+r

6(8+30n)+1=6Q+r

r = 1

May 26, 2018

### 2+0 Answers

#1
0

Sorry, can't read your question.

May 26, 2018
#2
+102765
+2
Best Answer

This is the question.

Find all numbers r 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.

I get r=1

Lets see....  (All the pronumerals I will use are integers)

x/20 = k remainder 9 so    x=20k+9

x/45 = t remainder 4 so     x=45t+4

so

20k+9 = 45t+4

45t-20k = 9-4

45t - 20k = 5

9t - 4k = 1

one obvious solution is t=1 and k=2

9( 1 ) - 4 (2) =1

9(1+4n)-4(2+9n)=1      (because I have added 36n and then taken 36n away again)

so a general solution is   t=1+4n   and  k=2+9n

x=20(2+9n)+9 = 40+180n+9 = 49+180n

or

x=45(1+4n)+4 = 45+180n+4 = 49+180n      Good they are the same.

Now we have x is equivalent to r mod 6

So

(49+180n)/6 = Q remainder r

49+180n = 6Q+r

48+180n+1=6Q+r

6(8+30n)+1=6Q+r

r = 1

Melody May 26, 2018