1. If the cubic equation $x^{3}-10 x^{2}+P x-30=0$ has three positive integer roots, determine the value of $P$.
Solution 1: If the equation has roots $x_{1}, x_{2}, x_{3}$, then $x_{1}+x_{2}+x_{3}=10, x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=P$ and $x_{1} x_{2} x_{3}=30$. Suppose $x_{1}=1$ is a root. Then $x_{2}+x_{3}=9$ and $x_{2} x_{3}=30$. For any pair of integers which multiply to 30 , their sum is at least 11 , so this is not possible, and 1 is not a root. The only other possibility is that the roots are $2,3,5$, which gives $P=2 \cdot 3+3 \cdot 5+5 \cdot 2=31$.

Solution 2: The roots of the equation are integers and have a product of 30 . The possible sets of roots are $(1,1,30),(1,2,15),(1,3,10),(1,5,6),(2,3,5)$. Of these, only $(2,3,5)$ sums to 10 , so it must be our set of roots. To find $\mathrm{P}$, we substitute $x=2$ into the equation to get $8-40+2 P-30=0$, which yields $P=31$.