+239 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.)
+984 Hey Rollingblade!
When N is divided by 10, the remainder is a,
When N is divided by 13, the remainder is b.
From these two statements, we can setup the modular arithmetic systems:
\(\begin{align*} x &\equiv 1 \pmod{10}, \\ x &\equiv 0 \pmod{13}. \end{align*}\)
The solution is: \(X\equiv91\pmod{130}\)
Then we solve:
\(\begin{align*} y &\equiv 0 \pmod{10}, \\ y &\equiv 1 \pmod{13}. \end{align*}\)
The solution is: \(y\equiv40\pmod{130}\)
Therefore, the answer is \(N\equiv91a+40b\pmod{130}\)
I hope this helped,
Gavin
+984 Hey Rollingblade!
When N is divided by 10, the remainder is a,
When N is divided by 13, the remainder is b.
From these two statements, we can setup the modular arithmetic systems:
\(\begin{align*} x &\equiv 1 \pmod{10}, \\ x &\equiv 0 \pmod{13}. \end{align*}\)
The solution is: \(X\equiv91\pmod{130}\)
Then we solve:
\(\begin{align*} y &\equiv 0 \pmod{10}, \\ y &\equiv 1 \pmod{13}. \end{align*}\)
The solution is: \(y\equiv40\pmod{130}\)
Therefore, the answer is \(N\equiv91a+40b\pmod{130}\)
I hope this helped,
Gavin
Rollingblade, PLEASE make sure that your questions are readable, this is unbearable. You can't expect this forum to help you if you can't write your question instead of copying and pasting it.
+239 this is very readable, I don’t understand what you mean
another user already answer, which means it is readable
Of course this specific question is readable, but you copy and pasted it. some of your other questions haven't been answered, because no one has deciphered them. Stop copying your questions because sometimes it turns into a mess that no one can read. You post unreadable questions and don't even respond when users ask you to clarify.
