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Difference between revisions of "Mock AIME 5 Pre 2005 Problems/Problem 15"

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

Latest revision as of 00:11, 15 February 2024

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

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