1994 AIME Problems/Problem 4

Problem

Find the positive integer $n\,$ for which $\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994$ (For real $x\,$ , $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$ )

Solution

Notice that $\lfloor\log_2{a}\rfloor$ is equal to $\log_2{2^{x}}$ , where $2^x$ is the largest power of $2$ less than or equal to $a$ . Using this fact, the equation can be rewritten as $\log_2{1}+\log_2{2}+\log_2{2}+\log_2{4}+\cdots+\log_2{n}=1994$ Simplifying the logarithms, we get $0+1+1+2+2+2+2+\ldots+2^m\cdot m=1994$ Template:Incomplete

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1994 AIME Problems/Problem 4

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