How many positive integers $n$ from 1 to 5000 satisfy both congruences, $n\equiv 5\pmod{12}$ and $n\equiv 11\pmod{13}$?

Use the Chinese remainder theorem. Suppose we set . Then clearly taken modulo 12, the second term vanishes, and incidentally ; taken modulo 13, the first term vanishes, but the second term leaves a remainder of 2. To counter this, we can multiply the second term by the inverse of 12 modulo 13, which is 12 since .

So, we found that , but the least positive solution is , and in general we can have for any integer .

Now, since , or , we know that there are 32 possible integers that satisfy the congruences.

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