Question

Find the positive integer $n$ for which $\begin{aligned}&\lfloor\log_2 1\rfloor+\lfloor\log_2 2\rfloor+\lfloor\log_2 3\rfloor\\&\quad+\cdots+\lfloor\log_2 n\rfloor=1994.\end{aligned}$ (For real $x,$ $\lfloor x\rfloor$ is the greatest integer not exceeding $x.$ )

Answer

Answer: 312

Concepts:floor and ceiling functions logarithm power of 2

Difficulty rating: 1980

Small Hint:

Group integers having the same value of $\lfloor\log_2 k\rfloor$

Big Hint:

Compute the sum through $255$ , after which every new term initially contributes $8$

Solution:

For $2^j\leq k<2^{j+1},$ the summand is $j.$ Thus the sum through $255=2^8-1$ is $\sum_{j=0}^7j2^j=(8-2)2^8+2=1538.$ The remaining $1994-1538=456$ is $57\cdot8,$ so we include $57$ more integers beginning with $256.$ Hence $n=255+57=312.$

1994 AIME Problem 4

Answer: 312

Difficulty rating: 1980

Small Hint:

Big Hint:

Solution:

Problem 4 in Other Years