Difference between revisions of "2000 AMC 12 Problems/Problem 1"
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| − | + | {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #1]] and [[2000 AMC 10 Problems|2000 AMC 10 #1]]}} | |
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| − | and 1+3+667=671 | + | ==Problem== |
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| + | In the year <math>2001</math>, the United States will host the [[International Mathematical Olympiad]]. Let <math>I,M,</math> and <math>O</math> be distinct [[positive integer]]s such that the product <math>I \cdot M \cdot O = 2001 </math>. What is the largest possible value of the sum <math>I + M + O</math>? | ||
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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> | ||
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| + | == Solution == | ||
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| + | The sum is the highest if two [[factor]]s are the lowest. | ||
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| + | So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \boxed{\text{(E)}}</math>. | ||
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| + | ==See Also== | ||
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| + | {{AMC12 box|year=2000|before=First<br />Question|num-a=2}} | ||
| + | {{AMC10 box|year=2000|before=First<br />Question|num-a=2}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 21:18, 23 November 2020
- 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
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. 
