Difference between revisions of "2013 AMC 8 Problems/Problem 3"
Notmyopic667 (talk | contribs) (→Solution) |
Scrabbler94 (talk | contribs) (more detailed solution) |
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==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) \\ | ||
+ | &= \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 14:47, 29 March 2021
Problem
What is the value of ?
Solution
We group the addends inside the parentheses two at a time: Then the desired answer is .
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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