Question

Find the complete solution for x≡7 (mod 10) and x≡5 (mod 12). Please show steps. @-------College Al…

Answer 1

This is what I have so far:

Answer 2

x≡5 (mod 12) -> x= 5 + 12t

Answer 3

and then I substituted that equation into x≡ 7 (mod 10) to give 5 + 12t = 7(mod 10)

Answer 4

which gave 12t ≡ 2 (mod 10)

Answer 5

i'm not sure if that's right, but now i'm stuck. :(

Answer 6

damn i did it the other way. i said
\[x\equiv 7(10)\implies x = 7+10k\]
\[7+10k\equiv5(12)\implies 10k\equiv -2(12)\equiv 10(12)\]
\[ \implies k \equiv 1(12)\implies k=1+10j\] back substitute to get
\[x=7+10k=7+10(1+12j)=7+10+120j=17+120j\]

Answer 7

i can retry it starting with
\[x\equiv 5(12)\] if you like. should work out the same

Answer 8

that would actually be great, thx :)

Answer 9

yes it will
do what you did exactly. when you get to the step that says
\[12t\equiv 2(10)\] then add ten to the left to get
\[12t\equiv 12(10)\] so
\[t\equiv 1(10)\]

Answer 10

i meant add ten to the right

Answer 11

now you have
\[t=1+10j\] so
\[5+12t=5+12(1+10j)=17+120j\] just like i got

Answer 12

you can check for a few values of j to see that this is right. for example if
\[j=1\] you get 137 which works since it has a remainder of 7 when divided by 10 and a remainder of 5 when divided by 12

Answer 13

Thank you so much! I've been trying to understand this for weeks! :)

Answer 14

hope that helped. i think the key step might have been starting with
\[12t\equiv 2(10)\] and then getting
\[112t\equiv 1(10)\]

Answer 15

yw

Answer 16

x=17+60n
If you use 120, where I have 60, you will miss solutions such as x=77.

Answer 17

Thanks, Stacey! I just checked it over.

Answer 18

The error occurs with 12t=12 (mod 10)
When you divide by 12 the modulus is divided by any common factor.
t=1 (mod 5)