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?

(Your answer should be in the form ra+sb, where r and s are replaced by nonnegative integers less than 130.)
Problem Hints:
There should be a number n such that na\equiv a\pmod{10} and na\equiv0\pmod{13}. What is n and how does it help?

Answer 1

actually the problem hint is: There should be a number n such that na==a(mod 10) and na==0(mod 13) what is n and how does it help? the"==" sign means congruent...

Answer 2

again CRT

Answer 3

\(\large \color{black}{\begin{aligned}
\normalsize \text{If }\hspace{.33em}\\~\\
N& \equiv a\pmod {10}\hspace{.33em}\\~\\
N& \equiv b\pmod {13} \hspace{.33em}\\~\\
\gcd(10,13)& =1\hspace{.33em}\\~\\
\normalsize \text{then }\hspace{.33em}\\~\\
N& \equiv 10bx+13ay\pmod {130} \hspace{.33em}\\~\\
\normalsize \text{where }\hspace{.33em}\\~\\
10x+13y&=1\hspace{.33em}\\~\\
\implies (10x+13y)\pmod 2&=1\pmod 2\hspace{.33em}\\~\\
\implies y\pmod 2&=1\hspace{.33em}\\~\\
y=-3,x=4\hspace{.33em}\\~\\
\implies N&\equiv 40b-39a\pmod {130} \hspace{.33em}\\~\\
\end{aligned}}\)

Answer 4

Well I still am not able to fig. the ans

Answer 5

40b-39a, is that the ans

Answer 6

it is given that \(\text{ ra+sb should be non-negative}\)

Answer 7

so u have to just convert the negative remainder to positive

Answer 8

\(\large \color{black}{\begin{aligned}

N&\equiv 40b-39a\pmod {130} \hspace{.33em}\\~\\
\implies N&\equiv 40b+(130-39)a\pmod {130} \hspace{.33em}\\~\\
\implies N&\equiv 40b+91a\pmod {130} \hspace{.33em}\\~\\
\end{aligned}}\)

Answer 9

oh.

Answer 10

thanks!