Difference between revisions of "2014 AIME II Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | Note that in base 2 <math>a_1</math> = 2, <math> | + | Note that (in base 2 for the indices, base 10 for the values) <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 09:53, 28 March 2014
Solution
Note that (in base 2 for the indices, base 10 for the values) = 2, = 3, = 2 * 3 = 6, ..., = 2 * 5 * 11 * 19 = 2090. Thus, t = 10010101 (base 2) = .