Difference between revisions of "2000 AMC 12 Problems/Problem 1"

Revision as of 14:24, 17 October 2007

Problem

In the year $2001$ , the United States will host the International Mathematical Olympiad. Let $\displaystyle I,M,$ and $\displaystyle O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $\displaystyle I + M + O$ ?

$\mathrm{(A) \ 23 } \qquad \mathrm{(B) \ 55 } \qquad \mathrm{(C) \ 99 } \qquad \mathrm{(D) \ 111 } \qquad \mathrm{(E) \ 671 }$

Solution

The sum is the highest if two factors are the lowest! So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \mathrm{(E)}$ .

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

@@ Line 9: / Line 9: @@
 ==See Also==
-* [[2000 AMC 12]]
+{{AMC12 box|year=2000|before=First Question|num-a=2}}
-* [[2000 AMC 12 Problems/Problem 2 | Next problem]]
 [[Category:Introductory Algebra Problems]]

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Difference between revisions of "2000 AMC 12 Problems/Problem 1"

Revision as of 14:24, 17 October 2007

Problem

Solution

See Also