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# Help ASAP

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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.)

May 26, 2018

#1
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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

May 26, 2018

#1
+988
+2

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

GYanggg May 26, 2018
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Thank you so much!

#2
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May 26, 2018
edited by Guest  May 26, 2018
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this is very readable, I don’t understand what you mean