Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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}]=[...")
 
 
Line 1: Line 1:
−
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}$.
+
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 23:11, 14 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_5_Pre_2005_Problems/Problem_15&oldid=215520"