2000 AMC 12 Problems/Problem 1
- The following problem is from both the 2000 AMC 12 #1 and 2000 AMC 10 #1, so both problems redirect to this page.
Contents
Problem
In the year , the United States will host the International Mathematical Olympiad. Let and be distinct positive integers such that the product . What is the largest possible value of the sum ?
Solution 1 (Verifying the Statement)
First, we need to recognize that a number is going to be lowest only if, of the factors, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like . It becomes much more clear that this is true, and in this situation, the value of would be . Now, we use this process on to get as our factors. Hence, we have
Solution By: armang32324
Solution 2
The sum is the highest if two factors are the lowest.
So, and .
Solution 3 (Answer Choices)
We see since is divisible by , we can eliminate all of the first answer choices because they are way too small and get as our final answer.
See Also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by First Question |
Followed by Problem 2 |
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 AMC 12 Problems and Solutions |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.