Difference between revisions of "2000 AMC 10 Problems/Problem 14"

Latest revision as of 23:50, 26 November 2011

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