2001 AMC 8 Problems/Problem 2
Contents
[hide]Problem
I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
Solution 1
Let the numbers be and . Then we have and . Solving for in the first equation yields , and substituting this into the second equation gives . Simplifying this gives , or . This factors as , so or , and the corresponding values are and . These are essentially the same answer: one number is and one number is , so the largest number is .
Solution 2
Use the answers to attempt to "reverse engineer" an appropriate pair of numbers. Looking at option , guess that one of the numbers is . If the sum of two numbers is and one is , then other must be . The product of those numbers is , which is the second condition of the problem, so our number are and .
However, is the smaller of the two numbers, so the answer is or .
Solution 3
We go through the divisor pairs of to see which two numbers sum to . The numbers clearly cannot be negative. If one was negative, then the other must also be negative in order to multiply to a positive product, but it would be impossible for the numbers to add up to a positive sum. So, we look at the positive divisor pairs of , namely and , and , and , and and . The only pair of numbers that sums to is and . The larger number is , so the answer is .
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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All AJHSME/AMC 8 Problems and Solutions |
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