Difference between revisions of "2000 AMC 12 Problems/Problem 1"
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<math>\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671</math> | <math>\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671</math> | ||
− | == Solution 1 (Verifying | + | == 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 == | + | == Solution 2== |
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) == | ||
+ | |||
+ | 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 by Daily Dose of Math== | ||
+ | |||
+ | https://www.youtube.com/watch?v=aSzsStkkYeA | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== |
Latest revision as of 08:46, 23 July 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.
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.