Find all pairs $(m, n)$ of natural numbers such that $200 m+6 n=2006$.
First we divide both sides of the equation by 2 and get: $100 m+3 n=1003$. Since $m$ and $n$ are natural numbers we immediately get that $m \leq 10$. Since $3 n$ is divisible by 3 and 1003 gives remainder 1 upon division by 3 , we conclude that $100 m$ must also give the remainder 1 upon division by 3 . Since $100 m=99 m+m$ and $99 m$ is divisible by 3 we see that $m$ must give remainder 1 when divided by 3 . Thus $m$ has to be one of the numbers $1,4,7,10$. Corresponding $n$ s are 301, 201,101 and 1 , respectively.