Question

When N is divided by 10, the remainder is a. When N is divided by 13, the remainder is b. What is N modulo 130, in terms of a and b?

(I need my answer in the form ra+sb, where r and s are replaced by nonnegative integers less than 130.)

Answer 1

familiar with congruences ?

Answer 2

Somewhat

Answer 3

somewhat is enough :)

Answer 4

We're given
\[N\equiv a\pmod{10}\\N\equiv b\pmod{13}\]

second congruence is same as \(N=13k+b\tag{1}\)

substitute that in first congruence and solve \(k\)

Answer 5

So that gives me \[10x+a=13k+b\]?

Answer 6

How do I solve it from there?

Answer 7

yes, your goal is to solve \(k\)

\(10x+a=13k+b\)

\(10x + a = 10k+3k+b\)

\(10(x-k)+a-b = 3k\)

\(10y+a-b = 3k\)

fine so far ?

Answer 8

Yes, but how am I supposed to solve it from there?

Answer 9

multiply both sides by "7"

Answer 10

Why?

Answer 11

\(10y+a-b=3k\)

multiply both sides by 7 and get
\(10z+7(a-b) = 20k+k\)

\(10w+7(a-b)=k\)

Answer 12

substitute that in equation \((1)\) and you're done!

Answer 13

Is \[N=13k+b\] equation 1?

Answer 14

Yep

Answer 15

So \[13(10w+7(a-b))\]?

Answer 16

well \[N=13(102+7(a-b))+b\]?

Answer 17

10w I menat

Answer 18

gah so many typos today \[N=13(10w+7(a-b))+b\]

Answer 19

\[\begin{align}N &= 13k+b\\~\\
&=13[10w+7(a-b)] +b\\~\\
&=130w+\color{blue}{91(a-b)+b}
\end{align}\]

Answer 20

and to put it in the form of \[ra+sb\]? \[130w+91a-90b\]

Answer 21

How do I put that into the ra+sb form?

Answer 22

91a - 90b

is the remainder when u divide N by 130

Answer 23

r = 91
s = -90

Answer 24

Oh... thank you!

Answer 25

ofcourse if u want positive numbers
-90 is same as 40 in mod 130

Answer 26

r = 91
s = 40

looks more better

Answer 27

yeah :)