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Difference between revisions of "2007 AIME I Problems/Problem 7"

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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 <math>\log_{\sqrt{2}} k</math> is an integer.
 
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 <math>\log_{\sqrt{2}} k</math> is an integer.
  
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The change of base formula shows that <math>\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}</math>. For the <math>\displaystyle \log 2</math> term must cancel out, it is only an integer if <math>k</math> is a [[exponent|power]] of <math>2</math>. Thus, <math>N</math> is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from <math>2^0 = 1</math> to <math>2^9 = 512</math>.  
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The change of base formula shows that <math>\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}</math>. For the <math>\log 2</math> term to cancel out, <math>k</math> is a [[exponent|power]] of <math>2</math>. Thus, <math>N</math> is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from <math>2^0 = 1</math> to <math>2^9 = 512</math>.  
  
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The formula for the sum of an [[arithmetic sequence]] and the sum of a [[geometric sequence]] yields that <math>\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9) = 500500 - \frac{2^{10}-1}{2-1} = 499477</math>. The solution is <math>477</math>.
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The formula for the sum of an [[arithmetic sequence]] and the sum of a [[geometric sequence]] yields that our answer is <math>\left[\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9)\right] \mod{1000}</math>.
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Simplifying, we get
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<math>\left[1000\left(\frac{1000+1}{2}\right) -1023\right] \mod{1000} \equiv [500-23] \mod{1000} \equiv 477 \mod{1000}.</math> The answer is <math>\boxed{477}</math>
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 21:20, 6 August 2019

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 (Problems • Answer Key • Resources)
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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