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 $\displaystyle \log 2$ 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 $\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9) = 500500 - \frac{2^{10}-1}{2-1} = 499477$ . The solution is $477$ .

2007 AIME I (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2007 AIME I Problems/Problem 7

From AoPSWiki

Problem

Solution

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