1994 AIME Problems/Problem 4


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 $\le x.\,$)


Note that if $2^x \le a<2^{x+1}$ for some $x\in\mathbb{Z}$, then $\lfloor\log_2{a}\rfloor=\log_2{2^{x}}=x$.

Thus, there are $2^{x+1}-2^{x}=2^{x}$ integers $a$ such that $\lfloor\log_2{a}\rfloor=x$. So the sum of $\lfloor\log_2{a}\rfloor$ for all such $a$ is $x\cdot2^x$.

Let $k$ be the integer such that $2^k \le n<2^{k+1}$. So for each integer $j<k$, there are $2^j$ integers $a\le n$ such that $\lfloor\log_2{a}\rfloor=j$, and there are $n-2^k+1$ such integers such that $\lfloor\log_2{a}\rfloor=k$.

Therefore, $\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor= \sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1994$.

Through computation: $\sum_{j=0}^{7}(j\cdot2^j)=1538<1994$ and $\sum_{j=0}^{8}(j\cdot2^j)=3586>1994$. Thus, $k=8$.

So, $\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}$.

Alternatively, one could notice this is an arithmetico-geometric series and avoid a lot of computation.

See also

1994 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS