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

(Solution)
(more detailed solution)
 
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==Solution==
 
==Solution==
  We see that if we group two numbers at a time, we see that it would make 1 on each. There are 500 pairs so the answer inside the parentheses is 1 times 500 equals 500. Now we just multiply that by 4 to get 2000. So the answer is E.
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We group the addends inside the parentheses two at a time:
 +
<cmath>
 +
\begin{align*}
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-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>
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Then the desired answer is <math>4 \times 500 = \boxed{\textbf{(E)}\ 2000}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=2|num-a=4}}
 
{{AMC8 box|year=2013|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:47, 29 March 2021

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 $4 \times 500 = \boxed{\textbf{(E)}\ 2000}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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

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