Actually, your answer was wrong. Here is the solution that AoPS gave.
There are a total of $6^3=216$ possible sets of dice rolls. If one of the re-rolled dice matches the pair we set aside and the other two form a pair, we will have a full house. But we will also have a full house if all three re-rolled dice come up the same.
Consider the first case. There are $3$ ways to pick which of the three dice will match a pair, and then $5$ ways to pick a value for the other two dice so that they form a pair (but don't match the first three dice), for a total of $3\cdot 5=15$ possible outcomes, plus the outcome that all five dice match.
In the second case, we need all three dice to match each other. There are $5$ ways to pick which value the three dice will have so that they don't match the first pair, plus the outcome that all five dice match.
So there are a total of $15+5=20$ ways to get a full house without all five dice matching, added to the possibility that all five dice match, which makes $21$ ways to get a full house. So, the probability is successful outcomes/total outcomes= 21/216= 7/72
Let $p$ and $q$ be the solutions the the equation $10x^2 - mx + 420 = 0$. We use the fact that the sum and product of the roots of a quadratic equation $ax^2+bx+c = 0$ are given by $-b/a$ and $c/a$, respectively, so $p+q = m/10$ and $pq = 420/10 = 42$. Since $m = 10(p+q)$, we minimize $m$ by minimizing the sum $p+q$. Since $p$ and $q$ are integers and multiply to 42, the possible values of $(p,q)$ are $(1,42),(2,21),(3,14),(6,7),(7,6),(14,3),(21,2),(42,1)$. (Note that if $p$ and $q$ are both negative, then $p+q$ is negative, so $m$ would be negative, which is excluded by the problem.) The sum $p+q$ is minimized when $(p,q) = (6,7)$ or $(7,6)$. In either case, $m = 10(p+q) = 10(6+7) =130
A number is a perfect square and a perfect cube if and only if it is a perfect sixth power. Note that 10^2=100 and 4^3<100<5^3, while 2^6<100<3^6=9^3. Hence, there are 10 squares and 4 cubes between 1 and 100, inclusive. However, there are also 2 sixth powers, so when we add 10+4 to count the number of squares and cubes, we count these sixth powers twice. However, we don't want to count these sixth powers at all, so we must subtract them twice. This gives us a total of 10+4-2*2=10 different numbers that are perfect squares or perfect cubes, but not both. Thus, our probability is 10/100=1/10.
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