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Difference between revisions of "2014 UNCO Math Contest II Problems/Problem 2"

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== Solution ==
 
== Solution ==
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This is asking for the sum of 1-2+3-4...+289. We know that the sum of the first N even numbers is N^2+N. Because there are 144 even numbers less than 289, we take 144^2+144 and get 20880. Because we're subtracting these, we have -20880. We are also adding the first 145 odd numbers, and the sum of the first N odd numbers is N^2. This is 145^2, so we get 21025. Adding these gives us 145, which is the answer.
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It is evident that we are being asked to find the sum of the sequence <math>1-2+3-4+...-288+289</math>.
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A straight-forward approach is to find the sum of all the odd numbers from 1 to 289, inclusive, and subtract the sum of all the even numbers from 2 to 288, inclusive. Doing so would give you <math>145^2 - (144^2 + 144) = (145 - 144)(145 + 144) - 144 = (1)(145 + 144) - 144 = \boxed{145}</math>.
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However, a faster method would be to see that you can group the terms in the sequence <math>1-2+3-4+...-288+289</math> into <math>1 + (-2 + 3) + (-4 + 5) +...+ (-288 + 289)</math>. Using this, we can reduce many pairs of positive and negative numbers to <math>1</math>s. Since there are as many <math>1</math>s as there are odd numbers between 1 and 289, inclusive, and there are 145 odd numbers, the answer is <math>(1)(145) = \boxed{145}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 11:28, 24 December 2019

Problem

Define the Cheshire Cat function $\fbox{:)}$ by

\begin{align*} \fbox{:)}(x) &= -x \quad \text {if } x \text{ is even and} \\ \fbox{:)}(x) &= x \quad \text{ if }x \text{ is odd} \end{align*}

Find the sum $\fbox{:)}(1) + \fbox{:)}(2) + \fbox{:)}(3) + \fbox{:)}(4) + . . .+ \fbox{:)}(289)$


Solution

It is evident that we are being asked to find the sum of the sequence $1-2+3-4+...-288+289$.

A straight-forward approach is to find the sum of all the odd numbers from 1 to 289, inclusive, and subtract the sum of all the even numbers from 2 to 288, inclusive. Doing so would give you $145^2 - (144^2 + 144) = (145 - 144)(145 + 144) - 144 = (1)(145 + 144) - 144 = \boxed{145}$.

However, a faster method would be to see that you can group the terms in the sequence $1-2+3-4+...-288+289$ into $1 + (-2 + 3) + (-4 + 5) +...+ (-288 + 289)$. Using this, we can reduce many pairs of positive and negative numbers to $1$s. Since there are as many $1$s as there are odd numbers between 1 and 289, inclusive, and there are 145 odd numbers, the answer is $(1)(145) = \boxed{145}$.

See also

2014 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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