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

(terribly worded solution, someone plz fix)
 
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== Solution ==
 
== Solution ==
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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]]) or 1 otherwise. <math>\log_{\sqrt{2} k</math> 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, and then subtract all powers of 2.  
+
The ceiling of a number minus the floor of a number is either equal to zero (if the number is an [[integer]]) or 1 otherwise. <math>\log_{\sqrt{2}} k</math> 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, and then subtract all powers of 2.  
  
−
The formula for the sum of an [[arithmetic series]] 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>.
+
The formula for the sum of an [[arithmetic series]] 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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2007|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2007|n=I|num-b=6|num-a=8}}

Revision as of 19:25, 14 March 2007

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) or 1 otherwise. $\log_{\sqrt{2}} k$ is only an integer if $k$ is a power of $2$. Thus, $N$ is equal to the sum of all the numbers from 1 to 1000, and then subtract all powers of 2.

The formula for the sum of an arithmetic series 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 $477$.

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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