2018 AMC 8 Problems/Problem 5

Problem

What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$ ?

$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$

Solution 1

Rearranging the terms, we get $(1-2)+(3-4)+(5-6)+...(2017-2018)+2019$ , and our answer is $-1009+2019=\boxed{\textbf{(E) }1010}$ .

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 $\boxed{\textbf{(E) }1010}$ .

~avamarora

Solution 3

It is similar to the Solution 1: Rearranging the terms, we get $1+(3-2)+(5-4)+(6-5)...(2017-2016)+(2019-2018)$ , and our answer is $1+1009=\boxed{\textbf{(E) }1010}$ .

~LarryFlora

Solution 4

Note that the sum of consecutive odd numbers can be expressed as a square, namely $1+3+5+7+...+2017+2019 = 1010^2$ . 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 $1010^2-1009^2-1009$ . Using difference of squares, we obtain $(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{1010}$ .

~SigmaPiE

Solution 5

1. Identify the Number of Terms

Odd numbers (1 to 2019):

 The \(k\)-th odd number is given by:
 : \(2k - 1\).  
 Setting \(2k - 1 = 2019\) gives:
 : \(2k = 2020 \Rightarrow k = 1010\).  
 Therefore, there are **1010** odd numbers.

Even numbers (2 to 2018):

 The \(k\)-th even number is:
 : \(2k\).  
 Setting \(2k = 2018\) gives:
 : \(k = 1009\).  
 Hence, there are **1009** even numbers.

2. Sum of the Odd Numbers

It is a well-known fact that the sum of the first $n$ odd numbers is:

$n^2$.

Thus, the sum of odd numbers is:

$1010^2 = 1010 \times 1010$.

3. Sum of the Even Numbers

The sum of the first $n$ even numbers can be calculated as:

$n(n+1)$.

For $n = 1009$, the sum is:

$1009 \times (1009 + 1) = 1009 \times 1010$.

4. Calculate the Difference

The original expression is the difference between the two sums:

\[ 1010 \times 1010 - 1009 \times 1010. \] Factor out $1010$:

\[ 1010(1010 - 1009) = 1010 \times 1 = 1010. \]

Final Answer

\[ \boxed{1010} \]

Video Solution (CRITICAL THINKING!!!)

https://youtu.be/uMo2Jlbm7WY

~Education, the Study of Everything

Video Solution

https://youtu.be/ykNMFdRMd0o

~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.

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