Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
 
(15 intermediate revisions by 8 users not shown)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
Notice that we can pair up every two numbers to make a sum of 1:
+
We group the addends inside the parentheses two at a time:  
−
<cmath> \begin{eqnarray*}(-1 + 2 - 3 + 4 - \cdots + 1000) &=& ((-1 + 2) + (-3 + 4) + \cdots + (-999 + 1000)) \\ &=& (1 + 1 + \cdots + 1) \\ &=& 500\end{eqnarray*}</cmath>
+
<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>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/4Gy1CrYTDHA ~savannahsolver
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/TkZvMa30Juo?t=3074
 +
 
 +
~ pi_is_3.14
  
−
Therefore, the answer is <math>4 \cdot 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 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_3&oldid=184853"