Difference between revisions of "2000 AMC 10 Problems/Problem 11"

Latest revision as of 23:42, 26 November 2011

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-==Problem==
+#REDIRECT [[2000 AMC 12 Problems/Problem 6]]
-==Solution==
-Two prime numbers between <math>4</math> and <math>18</math> are both odd.
-odd*odd=odd.
-odd-odd-odd=odd.
-Thus, we can discard the even choices.
-<math>ab-a-b=(a-1)(b-1)-1</math>.
-<math>(a-1)(b-1)-1=21,119,231</math>.
-Both of these factors are even, so the number +1 must be a multiple of <math>4</math>.
-<math>120</math> is the only possiblity.
-<math>11,13</math> satisfy this, <math>11*13-11-13=143-24=119</math>.
-C
-==See Also==
-{{AMC10 box|year=2000|num-b=10|num-a=12}}