Difference between revisions of "2007 iTest Problems/Problem 29"

(Solution to Problem 29)
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==Solution==
 
==Solution==
  
The list of numbers is an [[arithmetic sequence]] with <math>2007</math> terms, first term <math>1</math>, and last term <math>2007</math>.  Using the arithmetic series sum formula, <math>S = \frac{2007(1+2007}{2} = 2015028</math>.  The remainder when <math>S</math> is divided by <math>1000</math> is <math>\boxed{28}</math>.
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The list of numbers is an [[arithmetic sequence]] with <math>2007</math> terms, first term <math>1</math>, and last term <math>2007</math>.  Using the arithmetic series sum formula, <math>S = \frac{2007(1+2007)}{2} = 2015028</math>.  The remainder when <math>S</math> is divided by <math>1000</math> is <math>\boxed{28}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 18:16, 10 June 2018

Problem

Let $S$ be equal to the sum $1+2+3+\cdots+2007$. Find the remainder when $S$ is divided by $1000$.

Solution

The list of numbers is an arithmetic sequence with $2007$ terms, first term $1$, and last term $2007$. Using the arithmetic series sum formula, $S = \frac{2007(1+2007)}{2} = 2015028$. The remainder when $S$ is divided by $1000$ is $\boxed{28}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 28
Followed by:
Problem 30
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