+278 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.)
+26367 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.)
\(\begin{array}{|rclcl|} \hline N &\equiv& a \pmod{10} \quad &\text{or}& \quad N= a+10n,\ n \in Z \\ N &\equiv& b \pmod{13} \quad &\text{or}& \quad N= a+13m,\ m \in Z \\ \\ \hline \\ N = a+10n &=& b+13m \\ 10n &=& 13m+b-a \\\\ \mathbf{n} & \mathbf{=}& \mathbf{\dfrac{13m+b-a}{10}} \\\\ n &=& \dfrac{10m+3m+b-a}{10} \\\\ n &=& \dfrac{10m}{10} +\dfrac{3m+b-a}{10} \\\\ n &=& m +\underbrace{\dfrac{3m+b-a}{10}}_{=c} \\\\ c &=& \dfrac{3m+b-a}{10} \\\\ 10c &=& 3m+b-a \\\\ 3m &=& 10c+a-b \\\\ m &=& \dfrac{10c+a-b}{3} \\\\ \mathbf{m} & \mathbf{=}& \mathbf{\dfrac{10c+a-b}{3}} \\\\ m &=& \dfrac{9c+c+a-b}{3} \\\\ m &=& \dfrac{9c}{3} + \dfrac{c+a-b}{3} \\\\ m &=& 3c + \underbrace{\dfrac{c+a-b}{3}}_{=d} \\\\ d &=& \dfrac{c+a-b}{3} \\\\ 3d &=& c+a-b \\\\ \mathbf{c} & \mathbf{=}& \mathbf{3d+b-a} \\ \hline \end{array} \)
\(\mathbf{m=\ ?}\)
\(\begin{array}{|rcll|} \hline \mathbf{m} & \mathbf{=}& \mathbf{\dfrac{10c+a-b}{3}} \quad & | \quad \mathbf{c} & \mathbf{=}& \mathbf{3d+b-a} \\\\ m &=& \dfrac{10(3d+b-a)+a-b}{3} \\\\ m &=& \dfrac{30d+10b-10a+a-b}{3} \\\\ m &=& \dfrac{30d+9b-9a}{3} \\\\ \mathbf{m} & \mathbf{=}& \mathbf{10d+3b-3a} \\ \hline \end{array}\)
\(\mathbf{N=\ ?}\)
\(\begin{array}{|rcll|} \hline N &=& b+13m \quad & | \quad \mathbf{m} & \mathbf{=}& \mathbf{10d+3b-3a} \\ N &=& b+13(10d+3b-3a) \\ N &=& 130d -39a + 40b \\ \hline \end{array}\)
\(\mathbf{N \pmod{130} =\ ?}\)
\(\begin{array}{|rcll|} \hline N &=& 130d -39a + 40b \quad & | \quad \pmod{130} \\ N &\equiv& 0 -39a + 40b \pmod{130} \\ N &\equiv& -39a + 130a + 40b \pmod{130} \\ \mathbf{N} & \mathbf{\equiv} & \mathbf{91a + 40b \pmod{130}} \\ \hline \end{array}\)
\(N \pmod{130} = 91a + 40b\)
+26367 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.)
\(\begin{array}{|rclcl|} \hline N &\equiv& a \pmod{10} \quad &\text{or}& \quad N= a+10n,\ n \in Z \\ N &\equiv& b \pmod{13} \quad &\text{or}& \quad N= a+13m,\ m \in Z \\ \\ \hline \\ N = a+10n &=& b+13m \\ 10n &=& 13m+b-a \\\\ \mathbf{n} & \mathbf{=}& \mathbf{\dfrac{13m+b-a}{10}} \\\\ n &=& \dfrac{10m+3m+b-a}{10} \\\\ n &=& \dfrac{10m}{10} +\dfrac{3m+b-a}{10} \\\\ n &=& m +\underbrace{\dfrac{3m+b-a}{10}}_{=c} \\\\ c &=& \dfrac{3m+b-a}{10} \\\\ 10c &=& 3m+b-a \\\\ 3m &=& 10c+a-b \\\\ m &=& \dfrac{10c+a-b}{3} \\\\ \mathbf{m} & \mathbf{=}& \mathbf{\dfrac{10c+a-b}{3}} \\\\ m &=& \dfrac{9c+c+a-b}{3} \\\\ m &=& \dfrac{9c}{3} + \dfrac{c+a-b}{3} \\\\ m &=& 3c + \underbrace{\dfrac{c+a-b}{3}}_{=d} \\\\ d &=& \dfrac{c+a-b}{3} \\\\ 3d &=& c+a-b \\\\ \mathbf{c} & \mathbf{=}& \mathbf{3d+b-a} \\ \hline \end{array} \)
\(\mathbf{m=\ ?}\)
\(\begin{array}{|rcll|} \hline \mathbf{m} & \mathbf{=}& \mathbf{\dfrac{10c+a-b}{3}} \quad & | \quad \mathbf{c} & \mathbf{=}& \mathbf{3d+b-a} \\\\ m &=& \dfrac{10(3d+b-a)+a-b}{3} \\\\ m &=& \dfrac{30d+10b-10a+a-b}{3} \\\\ m &=& \dfrac{30d+9b-9a}{3} \\\\ \mathbf{m} & \mathbf{=}& \mathbf{10d+3b-3a} \\ \hline \end{array}\)
\(\mathbf{N=\ ?}\)
\(\begin{array}{|rcll|} \hline N &=& b+13m \quad & | \quad \mathbf{m} & \mathbf{=}& \mathbf{10d+3b-3a} \\ N &=& b+13(10d+3b-3a) \\ N &=& 130d -39a + 40b \\ \hline \end{array}\)
\(\mathbf{N \pmod{130} =\ ?}\)
\(\begin{array}{|rcll|} \hline N &=& 130d -39a + 40b \quad & | \quad \pmod{130} \\ N &\equiv& 0 -39a + 40b \pmod{130} \\ N &\equiv& -39a + 130a + 40b \pmod{130} \\ \mathbf{N} & \mathbf{\equiv} & \mathbf{91a + 40b \pmod{130}} \\ \hline \end{array}\)
\(N \pmod{130} = 91a + 40b\)
