Difference between revisions of "2018 AMC 8 Problems/Problem 5"
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Rearranging the terms, we get <math>(1-2)+(3-4)+(5-6)+...(2017-2018)+2019</math>, and our answer is <math>-1009+2019=\boxed{\textbf{(E) }1010}</math>. | Rearranging the terms, we get <math>(1-2)+(3-4)+(5-6)+...(2017-2018)+2019</math>, and our answer is <math>-1009+2019=\boxed{\textbf{(E) }1010}</math>. | ||
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| + | If you were stuck on this problem, refer to AOPS arithmetic lessons. | ||
| + | |||
| + | ~Nivaar | ||
==Solution 2== | ==Solution 2== | ||
| − | We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So, do the second last ones and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get <math>\boxed{\textbf{(E) }1010}</math> | + | We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So, do the second last ones and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get <math>\boxed{\textbf{(E) }1010}</math>. |
~avamarora | ~avamarora | ||
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It is similar to the Solution 1: | It is similar to the Solution 1: | ||
| − | Rearranging the terms, we get <math>1+(3-2)+(5-4)+(6-5)...(2017-2016)+(2019-2018)</math>, and our answer is <math>1+1009=\boxed{\textbf{(E) }1010}</math> | + | Rearranging the terms, we get <math>1+(3-2)+(5-4)+(6-5)...(2017-2016)+(2019-2018)</math>, and our answer is <math>1+1009=\boxed{\textbf{(E) }1010}</math>. |
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~LarryFlora | ~LarryFlora | ||
==Solution 4== | ==Solution 4== | ||
| − | Note that the sum of consecutive odd numbers can be expressed as a square, namely <math>1+3+5+7+...+2017+2019 = 1010^2</math>. We can modify the negative numbers in the same way by adding 1 to each negative term, factoring a negative sign, and accounting for the extra 1's by subtracting 1009. We then have <math>1010^2-1009^2-1009</math>. Using difference of squares, we obtain <math>(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{1010}</math> | + | Note that the sum of consecutive odd numbers can be expressed as a square, namely <math>1+3+5+7+...+2017+2019 = 1010^2</math>. We can modify the negative numbers in the same way by adding 1 to each negative term, factoring a negative sign, and accounting for the extra 1's by subtracting 1009. We then have <math>1010^2-1009^2-1009</math>. Using difference of squares, we obtain <math>(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{1010}</math>. |
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~SigmaPiE | ~SigmaPiE | ||
| + | |||
| + | == Video Solution (CRITICAL THINKING!!!)== | ||
| + | https://youtu.be/uMo2Jlbm7WY | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
| − | ==See Also== | + | ==See Also== y |
{{AMC8 box|year=2018|num-b=4|num-a=6}} | {{AMC8 box|year=2018|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 11:12, 23 January 2024
Contents
Problem
What is the value of 
?
Solution 1
Rearranging the terms, we get 
, and our answer is 
.
If you were stuck on this problem, refer to AOPS arithmetic lessons.
~Nivaar
Solution 2
We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So, do the second last ones and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get 
.
~avamarora
Solution 3
It is similar to the Solution 1:
Rearranging the terms, we get 
, and our answer is 
.
~LarryFlora
Solution 4
Note that the sum of consecutive odd numbers can be expressed as a square, namely 
. We can modify the negative numbers in the same way by adding 1 to each negative term, factoring a negative sign, and accounting for the extra 1's by subtracting 1009. We then have 
. Using difference of squares, we obtain 
.
~SigmaPiE
Video Solution (CRITICAL THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
==See Also== y
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
