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Difference between revisions of "2013 AMC 8 Problems/Problem 3"

(Created page with "==Problem== ==Solution== ==See Also== {{AMC8 box|year=2013|before=First Problem|num-a=2}} {{MAA Notice}}")
 
 
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==Problem==
 
==Problem==
 +
What is the value of <math>4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)</math>?
 +
 +
<math>\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000</math>
  
 
==Solution==
 
==Solution==
 +
We group the addends inside the parentheses two at a time:
 +
<cmath>
 +
\begin{align*}
 +
-1 + 2 - 3 + 4 - 5 + 6 - 7 + \ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\
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&= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\
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&= 500.
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\end{align*}
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</cmath>
 +
Then the desired answer is <math>4 \times 500 = \boxed{\textbf{(E)}\ 2000}</math>.
 +
 +
==Video Solution==
 +
https://youtu.be/4Gy1CrYTDHA ~savannahsolver
 +
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/TkZvMa30Juo?t=3074
 +
 +
~ pi_is_3.14
 +
  
 
==See Also==
 
==See Also==
−
{{AMC8 box|year=2013|before=First Problem|num-a=2}}
+
{{AMC8 box|year=2013|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:01, 26 December 2022

Problem

What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$?

$\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$

Solution

We group the addends inside the parentheses two at a time: \begin{align*} -1 + 2 - 3 + 4 - 5 + 6 - 7 + \ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\ &= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\ &= 500. \end{align*} Then the desired answer is $4 \times 500 = \boxed{\textbf{(E)}\ 2000}$.

Video Solution

https://youtu.be/4Gy1CrYTDHA ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=3074

~ pi_is_3.14


See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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