Difference between revisions of "2001 AMC 8 Problems/Problem 2"

(solution to 2001 AMC 8 Problem 2)
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Revision as of 18:10, 19 May 2011

Let the numbers be $x$ and $y$. Then we have $x+y=11$ and $xy=24$. Solving for $x$ in the first equation yields $x=11-y$, and substituting this into the second equation gives $(11-y)(y)=24$. Simplifying this gives $-y^2+11y=24$, or $y^2-11y+24=0$. This factors as $(y-3)(y-8)=0$, so $y=3$ or $y=8$, and the corresponding $x$ values are $x=8$ and $x=3$. These are essentially the same answer: one number is $3$ and one number is $8$, so the largest number is $8, \boxed{\text{D}}$.