# 2001 AMC 8 Problems/Problem 2

## Problem

I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number? $\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

## Solution 1

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 $8, \boxed{\text{D}}$.

## Solution 2

Use the answers to attempt to "reverse engineer" an appropriate pair of numbers. Looking at option $A$, guess that one of the numbers is $3$. If the sum of two numbers is $11$ and one is $3$, then other must be $11 - 3 = 8$. The product of those numbers is $3\cdot 8 = 24$, which is the second condition of the problem, so our number are $3$ and $8$.

However, $3$ is the smaller of the two numbers, so the answer is $8$ or $\boxed{D}$.

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