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

(Problem)
(Solution)
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==Solution==
 
==Solution==
Notice that we can pair up every two numbers to make a sum of 1:
 
<cmath> \begin{eqnarray*}(-1 + 2 - 3 + 4 - \cdots + 1000) &=& ((-1 + 2) + (-3 + 4) + \cdots + (-999 + 1000)) \\ &=& (1 + 1 + \cdots + 1) \\  &=& 500\end{eqnarray*}</cmath>
 
  
Therefore, the answer is <math>4 \cdot 500= \boxed{\textbf{(E)}\ 2000}</math>.
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  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 is -1 times 500 equals to \cdot 500= \boxed{\textbf{(E)}\ 2000}$.
  
 
==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}}

Revision as of 16:50, 2 November 2020

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 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 is -1 times 500 equals to \cdot 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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