Two dice appear to be normal dice with their faces numbered from $1$ to $6$, but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$. The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc.  Then there are $21^2=441$ possible rolls. There are $2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56$ rolls that will result in a seven.  The odds are therefore $\frac{56}{441}=\frac{8}{63}$.  The answer is $8+63=\boxed{071}$
See also [2006 AMC 12B Problems/Problem 17](https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_17)