Difference between revisions of "2014 AIME II Problems/Problem 15"

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==Solution==
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Note that in base 2 <math>a_1</math> = 2, <math>a_10</math> = 3, <math>a_11</math> = 2 * 3 = 6, ..., <math>a_10010101</math> = 2 * 5 * 11 * 19 = 2090. Thus, t = 10010101 (base 2) = <math>\boxed{149}</math>.

Revision as of 09:52, 28 March 2014

Solution

Note that in base 2 $a_1$ = 2, $a_10$ = 3, $a_11$ = 2 * 3 = 6, ..., $a_10010101$ = 2 * 5 * 11 * 19 = 2090. Thus, t = 10010101 (base 2) = $\boxed{149}$.