Difference between revisions of "2022 AMC 8 Problems/Problem 8"
Hh99754539 (talk | contribs) (→Solution 2: Changed LaTeX error.) |
MRENTHUSIASM (talk | contribs) (Reformatted a little. Inserted the lead-in for expression.) |
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<math>\textbf{(A) } \frac{1}{462} \qquad \textbf{(B) } \frac{1}{231} \qquad \textbf{(C) } \frac{1}{132} \qquad \textbf{(D) } \frac{2}{213} \qquad \textbf{(E) } \frac{1}{22}</math> | <math>\textbf{(A) } \frac{1}{462} \qquad \textbf{(B) } \frac{1}{231} \qquad \textbf{(C) } \frac{1}{132} \qquad \textbf{(D) } \frac{2}{213} \qquad \textbf{(E) } \frac{1}{22}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| + | |||
Note that common factors (from <math>3</math> to <math>20,</math> inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes <cmath>\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.</cmath> | Note that common factors (from <math>3</math> to <math>20,</math> inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes <cmath>\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.</cmath> | ||
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==Solution 2== | ==Solution 2== | ||
| − | + | The original expression becomes <cmath>\frac{20!}{\frac{22!}{2}} = \frac{20! \cdot 2}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.</cmath> | |
| − | < | ||
~hh99754539 | ~hh99754539 | ||
Revision as of 19:48, 31 January 2022
Contents
Problem
What is the value of ![\[\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?\]](http://latex.artofproblemsolving.com/8/e/4/8e422ea624f4cc14204e47ec836017b740d90c3d.png)
Solution 1
Note that common factors (from 
to 
inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes ![\[\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.\]](http://latex.artofproblemsolving.com/2/2/3/223017c5281b22f6c831d21b4c389a60b0d57a19.png)
~MRENTHUSIASM
Solution 2
The original expression becomes ![\[\frac{20!}{\frac{22!}{2}} = \frac{20! \cdot 2}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.\]](http://latex.artofproblemsolving.com/3/b/4/3b4cd50d3b307220ce1d5d00591ff75793a25992.png)
~hh99754539
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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. 
