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| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 9]] |
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| − | ==Solution==
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| − | 71, 76, 80, 82, 91.
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| − | The sum of the first 2 must be even, so we must choose 2 evens or the 2 odds.
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| − | Let us look at the numbers (mod 3).
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| − | 2,1,2,1,1.
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| − | If we choose the two odds, the next number must be a multiple of 3, of which there is none.
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| − | Similarly, if we choose 76,80 or 80,82, the next number must be a multiple of 3, of which there is none.
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| − | So we choose 76,82 first.
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| − | The next number must be 1 (mod 3), of which only 91 remains.
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| − | The sum of these is 249. This is equal to 1 (mod 4).
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| − | Thus, we need to choose one number that is 3 (mod 4). 71 is the only one that works.
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| − | Thus, 80 is the last score entered.
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| − | C.
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| − | ==See Also==
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| − | {{AMC10 box|year=2000|num-b=13|num-a=15}}
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