Difference between revisions of "2021 AMC 10A Problems/Problem 1"
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Knowing that <math>\sqrt{2} \approx 1.41</math> and <math>\sqrt{3} \approx 1.73,</math> we get <cmath>(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2) \approx 0.59\cdot 3.41 -1.26\cdot 4.73 + 2 \cdot 6 | Knowing that <math>\sqrt{2} \approx 1.41</math> and <math>\sqrt{3} \approx 1.73,</math> we get <cmath>(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2) \approx 0.59\cdot 3.41 -1.26\cdot 4.73 + 2 \cdot 6 | ||
=8.0521,</cmath> which is closest to <math>\boxed{\textbf{(D) } 8}.</math> | =8.0521,</cmath> which is closest to <math>\boxed{\textbf{(D) } 8}.</math> | ||
| + | |||
| + | ==Solution 5 == | ||
| + | |||
| + | Let <math>g(z)=z^2-z</math>. The function <math>\pi\csc(\pi z)</math> has simple poles at the integers <math>n</math> with residue | ||
| + | |||
| + | <cmath> | ||
| + | \operatorname{Res}\big(\pi\csc(\pi z),z=n\big)=(-1)^n, | ||
| + | </cmath> | ||
| + | |||
| + | because <math>\sin(\pi z)\sim (-1)^n\pi(z-n)</math> near <math>z=n</math>. | ||
| + | |||
| + | Hence the residue of the product | ||
| + | |||
| + | <cmath> | ||
| + | F(z)=\pi\csc(\pi z)\,g(z) | ||
| + | </cmath> | ||
| + | |||
| + | at <math>z=n</math> is | ||
| + | |||
| + | <cmath> | ||
| + | \operatorname{Res}(F,z=n)=(-1)^n g(n)=(-1)^n(n^2-n). | ||
| + | </cmath> | ||
| + | |||
| + | Therefore the finite alternating sum we want equals the sum of those residues for <math>n=2,3,4</math>: | ||
| + | |||
| + | \begin{align*} | ||
| + | &(2^2-2)-(3^2-3)+(4^2-4) \\ | ||
| + | &=(-1)^2(2^2-2)+(-1)^3(3^2-3)+(-1)^4(4^2-4). \\ | ||
| + | & = (4-2)- (9-3) + (16-4) = 2-6+12 = 8. | ||
| + | \end{align*} | ||
| + | |||
| + | So the value is <math>\boxed{\textbf{(D) } 8}.</math> | ||
| + | |||
==Video Solution 1 (Lightning Fast)== | ==Video Solution 1 (Lightning Fast)== | ||
Revision as of 02:21, 21 September 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Overkill: Just for Fun)
- 5 Solution 4 (When you have too much time)
- 6 Solution 5
- 7 Video Solution 1 (Lightning Fast)
- 8 Video Solution 3 (Very Very Quick Computation)
- 9 Video Solution 4 (Quick Computation)
- 10 Video Solution 5 by OmegaLearn (Arithmetic Computation)
- 11 Video Solution 6
- 12 Video Solution 7
- 13 Video Solution 8 (Problems 1-3)
- 14 Video Solution 9
- 15 See Also
Problem
What is the value of ![\[(2^2-2)-(3^2-3)+(4^2-4)\]](http://latex.artofproblemsolving.com/f/b/f/fbf9a3889b146c23eb653b9dc16255c03a229dfe.png)
Solution 1
This corresponds to 
-happykeeper
Solution 2
We have
~MRENTHUSIASM
Solution 3 (Overkill: Just for Fun)
We have
-PureSwag
Solution 4 (When you have too much time)
Using the difference of squares, we have ![\[(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2).\]](http://latex.artofproblemsolving.com/d/c/e/dce654c8a197d06312c02b9cf3b07066856347c3.png)
Knowing that 
and 
we get ![\[(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2) \approx 0.59\cdot 3.41 -1.26\cdot 4.73 + 2 \cdot 6 =8.0521,\]](http://latex.artofproblemsolving.com/9/2/c/92c78ba7e8054466f98fc098f06c66d0c2ab6367.png)
which is closest to
Solution 5
Let 
. The function 
has simple poles at the integers 
with residue
![\[\operatorname{Res}\big(\pi\csc(\pi z),z=n\big)=(-1)^n,\]](http://latex.artofproblemsolving.com/b/5/d/b5d6538a3019fcc7485230d054323a6ebc5bd6f6.png)
because 
near 
.
Hence the residue of the product
![\[F(z)=\pi\csc(\pi z)\,g(z)\]](http://latex.artofproblemsolving.com/5/6/3/563a2c03958e6ae973050fd576134e130d72a5e1.png)
at is
![\[\operatorname{Res}(F,z=n)=(-1)^n g(n)=(-1)^n(n^2-n).\]](http://latex.artofproblemsolving.com/2/a/a/2aacc8bdf31fa6cad85cf26afc270337b4d4aa42.png)
Therefore the finite alternating sum we want equals the sum of those residues for 
:
\begin{align*} &(2^2-2)-(3^2-3)+(4^2-4) \\ &=(-1)^2(2^2-2)+(-1)^3(3^2-3)+(-1)^4(4^2-4). \\ & = (4-2)- (9-3) + (16-4) = 2-6+12 = 8. \end{align*}
So the value is
Video Solution 1 (Lightning Fast)
~ Education, the Study of everything
Video Solution 3 (Very Very Quick Computation)
https://www.youtube.com/watch?v=m0_UMI2mnZs&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=2 ~North America Mathematic Contest Go Go Go
Video Solution 4 (Quick Computation)
https://youtu.be/C3n2hgBhyXc?t=37 ~ThePuzzlr
Video Solution 5 by OmegaLearn (Arithmetic Computation)
~ pi_is_3.14
Video Solution 6
~savannahsolver
Video Solution 7
~IceMatrix
Video Solution 8 (Problems 1-3)
~MathWithPi
Video Solution 9
~TheLearningRoyal
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
