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| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 1]] |
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| − | In the year 2001, the United States will host the International Mathematical Olympiad. Let <math>I</math>, <math>M</math>, and <math>O</math> be distinct positive integers 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>\mathrm{(A)}\ 23 \qquad\mathrm{(B)}\ 55 \qquad\mathrm{(C)}\ 99 \qquad\mathrm{(D)}\ 111 \qquad\mathrm{(E)}\ 671</math>
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| − | ==Solution==
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| − | <math>2001=1\cdot 3\cdot 667=3\cdot 23\cdot 29</math>
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| − | <math>1+3+667=671</math>
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| − | <math>3+23+29=55</math>
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| − | <math>1+29+69=99</math>
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| − | <math>1+23+87=111</math>
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| − | Clearly, <math>671</math>, or <math>\boxed{E}</math> is the largest.
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| − | ==See Also==
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| − | {{AMC10 box|year=2000|before=First question|num-a=2}}
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