Difference between revisions of "2018 AMC 8 Problems/Problem 5"

Latest revision as of 13:17, 6 June 2025

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

The sum of odd numbers can be represented as the number of consecutive odds starting from one, squared. Since 2019 is (2019+1)/2, there are 1010 odd numbers. Therefore, the sum of the odd numbers is $1010^2 \quad \text{or} \quad 1010 \times 1010.$

The sum of even numbers up to 2018 is calculated by noting that there are 2018/2 = 1009 even numbers. Using the formula for the sum of the first n even numbers, n(n+1), we get $1009 \times (1009+1) \quad \text{or} \quad 1009 \times 1010.$

The difference between these two sums is $1010 \times 1010 - 1009 \times 1010 = 1010.$

Thus, the correct answer is $-1009+2019=\boxed{\textbf{(E) }1010}$ .

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 33: / Line 33: @@
 ==Solution 5==
-=== 1. Identify the Number of Terms ===
+The sum of odd numbers can be represented as the number of consecutive odds starting from one, squared. Since 2019 is (2019+1)/2, there are 1010 odd numbers. Therefore, the sum of the odd numbers is
-* '''Odd numbers (1 to 2019):'''
+<cmath>1010^2 \quad \text{or} \quad 1010 \times 1010.</cmath>
-  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 sum of even numbers up to 2018 is calculated by noting that there are 2018/2 = 1009 even numbers. Using the formula for the sum of the first n even numbers, n(n+1), we get
-  The \(k\)-th even number is:
+<cmath>1009 \times (1009+1) \quad \text{or} \quad 1009 \times 1010.</cmath>
-  : \(2k\).
-  Setting \(2k = 2018\) gives:
-  : \(k = 1009\).
-  Hence, there are **1009** even numbers.
-=== 2. Sum of the Odd Numbers ===
+The difference between these two sums is
-It is a well-known fact that the sum of the first \(n\) odd numbers is:
+<cmath>1010 \times 1010 - 1009 \times 1010 = 1010.</cmath>
-: \(n^2\).
-Thus, the sum of odd numbers is:
-: \(1010^2 = 1010 \times 1010\).
-=== 3. Sum of the Even Numbers ===
+Thus, the correct answer is <math>-1009+2019=\boxed{\textbf{(E) }1010}</math>.
-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:
-:
-\[
-\times 1010 - 1009 \times 1010.
-\]
-Factor out \(1010\):
-:
-\[
-(1010 - 1009) = 1010 \times 1 = 1010.
-\]
-== Final Answer ==
-\[
-\boxed{1010}
-\]
 == Video Solution (CRITICAL THINKING!!!)==
@@ Line 91: / Line 59: @@
 {{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

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