Difference between revisions of "2000 AMC 12 Problems/Problem 1"
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== Solution 1 (Verifying the Statement)== | == Solution 1 (Verifying the Statement)== | ||
| − | First, we need to recognize that a number is going to be | + | First, we need to recognize that a number is going to be largest 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>. |
| − | + | ~armang32324 | |
== Solution 2== | == Solution 2== | ||
| Line 17: | Line 17: | ||
The sum is the highest if two [[factor]]s are the lowest. | The sum is the highest if two [[factor]]s are the lowest. | ||
| − | So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \boxed{\text{(E)}}</math>. | + | So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \boxed{\text{(E) 671}}</math>. |
== Solution 3 (Answer Choices) == | == Solution 3 (Answer Choices) == | ||
| − | We see since <math>2 + 0 + 0 + 1</math> is divisible by <math>3</math>, we can eliminate all of the first <math>4</math> answer choices because they are way too small and get <math>\boxed{\text{E}}</math> as our final answer. | + | We see since <math>2 + 0 + 0 + 1</math> is divisible by <math>3</math>, we can eliminate all of the first <math>4</math> answer choices because they are way too small and get <math>\boxed{\text{(E) 671}}</math> as our final answer. |
| − | == Video Solution | + | == Solution 4 (Faster Way) == |
| + | |||
| + | Notice <math>2001 = 3 * 29 * 23</math>, so we can just maximize this with <math>667 * 3 * 1</math>, which has a sum of <math>671</math>. Our answer is <math>\boxed{\text{(E) 671}}</math>. | ||
| + | |||
| + | -itsj | ||
| + | |||
| + | ==Video Solution by Daily Dose of Math== | ||
https://www.youtube.com/watch?v=aSzsStkkYeA | https://www.youtube.com/watch?v=aSzsStkkYeA | ||
| − | ~ | + | ~Thesmartgreekmathdude |
==See Also== | ==See Also== | ||
Latest revision as of 23:53, 17 November 2024
- 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 largest 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 
.
~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.
Solution 4 (Faster Way)
Notice 
, so we can just maximize this with , which has a sum of 
. Our answer is .
-itsj
Video Solution by Daily Dose of Math
https://www.youtube.com/watch?v=aSzsStkkYeA
~Thesmartgreekmathdude
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. 
