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?
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 numbers 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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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