Difference between revisions of "2000 AMC 12 Problems/Problem 1"
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First, we need to recognize that a number is going to be lowest only if, of the <math>3</math> [[factor]]s, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like <math>30</math>. It becomes much more clear that this is true, and in this situation, the value of <math>I + M + O</math> would be <math>18</math>. Now, we use this process on <math>2001</math> to get <math>667 * 3 * 1</math> as our <math>3</math> factors. | First, we need to recognize that a number is going to be lowest only if, of the <math>3</math> [[factor]]s, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like <math>30</math>. It becomes much more clear that this is true, and in this situation, the value of <math>I + M + O</math> would be <math>18</math>. Now, we use this process on <math>2001</math> to get <math>667 * 3 * 1</math> as our <math>3</math> factors. | ||
Hence, we have <math>667 + 3 + 1 = \boxed{\text{(E) 671}}</math> | Hence, we have <math>667 + 3 + 1 = \boxed{\text{(E) 671}}</math> | ||
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+ | Solution By: armang32324 | ||
== Solution == | == Solution == |
Revision as of 00:25, 7 April 2023
- The following problem is from both the 2000 AMC 12 #1 and 2000 AMC 10 #1, so both problems redirect to this page.
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 a 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
The sum is the highest if two factors are the lowest.
So, and .
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.