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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.
+
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) \\
 +
&= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\
 +
&= 500.
 +
\end{align*}
 +
</cmath>
 +
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}}

Revision as of 13:47, 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 (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

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