2000 AMC 12 Problems/Problem 1: Difference between revisions

Revision as of 12:10, 13 November 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 $1+3+667=671 \Longrightarrow \mathrm{(E)}$ .

Page

Toolbox

Search

2000 AMC 12 Problems/Problem 1: Difference between revisions

Revision as of 12:10, 13 November 2006

Solution

See Also

Revision as of 18:19, 5 November 2006 view source Xantos C. Guin (talk \| contribs) editor 580 edits added category ← Older edit	Revision as of 12:10, 13 November 2006 view source JBL (talk \| contribs) Administrators 2,310 edits m 2000 AMC 12/Problem 1 moved to 2000 AMC 12 Problems/Problem 1 Newer edit →
(No difference)