Mock AIME 5 Pre 2005 Problems/Problem 15

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We need to find $[log_2{53}]+[log_2{54}]+[log_2{55}]+...+[log_2{64}], and since$2^{\frac{11}{2}}=\sqrt{2048}<46$, and we have$[log_2{53}]=[log_2{54}]=[log_2{55}]=...=[log_2{64}]=6$, and the answer is$6\times12=\boxed{72}$. ~AbbyWong