2007 AIME I Problems/Problem 7

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Problem

Let $N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil  - \lfloor \log_{\sqrt{2}} k \rfloor )$

Find the remainder when $N$ is divided by 1000. ($\lfloor{k}\rfloor$ is the greatest integer less than or equal to $k$, and $\lceil{k}\rceil$ is the least integer greater than or equal to $k$.)

Solution

The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer); otherwise, it is equal to 1. Thus, we need to find when or not $\log_{\sqrt{2}} k$ is an integer.

The change of base formula shows that $\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}$. For the $\log 2$ term to cancel out, $k$ is a power of $2$. Thus, $N$ is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from $2^0 = 1$ to $2^9 = 512$.

The formula for the sum of an arithmetic sequence and the sum of a geometric sequence yields that our answer is $\left[\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9)\right] \mod{1000}$.

Simplifying, we get $\left[1000\left(\frac{1000+1}{2}\right) -1023\right] \mod{1000} \equiv [500-23] \mod{1000} \equiv 477 \mod{1000}.$ The answer is $\boxed{477}$

See also

2007 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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