When N is divided by 10, the remainder is a. When N is divided by 13, the remainder is b. If a = 5 and b = \(10\), what is N modulo 130?
+9519 Here we have the conditions \(\begin{cases}N\equiv 5 \pmod {10}\\N\equiv 10\pmod{13}\end{cases}\).
One way without using Chinese Remainder Theorem is to list all numbers with \(N \leq 130\) and \(N \equiv 10 \pmod{13}\).
You can start from 13 + 10 = 23, and add 13 each time you progress in the list.
The list is {23, 36, 49, 62, 75, ...}.
Now, note that 75 also satisfies \(N \equiv 5 \pmod{10}\). Then this is the answer we want.
Note that \(75 \operatorname{mod} 130 = \boxed{75}\). That's the answer.
Remark: This method is not rigorous at all. For a solution with more rigor, you would need Chinese Remainder Theorem.
