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Difference between revisions of "2023 WSMO Team Round Problems/Problem 3"

(Created page with "==Problem== In the figure below, there are <math>1023</math> total circles. The area between circles alternate between shaded and non-shaded. If the area of the shaded region...")
 
 
Line 17: Line 17:
  
 
==Solution==
 
==Solution==
 +
Note that
 +
<cmath>\begin{align*}
 +
k &= 2023^2-2022^2+2021^2-2019^2\ldots+3^2-2^1+1^2\\
 +
&= (2023-2022)(2023+2022)+\ldots+(3-2)(3+2)+1^2\\
 +
&= (2023+2022)+(2021+2019)+\ldots+(3+2)+1\\
 +
&= \frac{2023\cdot2024}{2}\implies\\
 +
k\pmod{1000}&\equiv\boxed{276}.
 +
\end{align*}</cmath>
 +
 +
~pinkpig

Latest revision as of 13:55, 13 September 2025

Problem

In the figure below, there are $1023$ total circles. The area between circles alternate between shaded and non-shaded. If the area of the shaded region is $k\pi,$ find the remainder when $k$ is divided by $1000.$ [asy] size(5cm); fill(circle((0,0),10),black); fill(circle((0,0),9),white); fill(circle((0,0),8),black); fill(circle((0,0),7),white); fill(circle((0,0),3),black); fill(circle((0,0),2),white); fill(circle((0,0),1),black); dot((0,6)); dot((0,5)); dot((0,4)); [/asy]

Solution

Note that \begin{align*} k &= 2023^2-2022^2+2021^2-2019^2\ldots+3^2-2^1+1^2\\ &= (2023-2022)(2023+2022)+\ldots+(3-2)(3+2)+1^2\\ &= (2023+2022)+(2021+2019)+\ldots+(3+2)+1\\ &= \frac{2023\cdot2024}{2}\implies\\ k\pmod{1000}&\equiv\boxed{276}. \end{align*}

~pinkpig

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