2000 AMC 12 Problems/Problem 1: Difference between revisions

Revision as of 10:10, 15 October 2006

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 $1+3+667=671 \Longrightarrow \mathrm{(E)}$ .

@@ Line 6: / Line 6: @@
 == Solution ==
 The sum is the highest if two factors are the lowest!
-So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671</math>, (E).
+So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \mathrm{(E)}</math>.
+==See Also==
+* [[2000 AMC 12/Problem 2 | Next problem]]
+* [[2000 AMC 12]]

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2000 AMC 12 Problems/Problem 1: Difference between revisions

Revision as of 10:10, 15 October 2006

Solution

See Also