Difference between revisions of "2019 AMC 8 Problems/Problem 17"
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As you can easily see <cmath>\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)</cmath> | As you can easily see <cmath>\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)</cmath> | ||
− | + | If we remove all parentheses, we can easily see that the middle terms cancel, leaving us with | |
<cmath>\left(\frac{1\cdot100}{2\cdot99}\right)</cmath> = <math>\boxed{\textbf{(B)}\frac{50}{99}}</math> | <cmath>\left(\frac{1\cdot100}{2\cdot99}\right)</cmath> = <math>\boxed{\textbf{(B)}\frac{50}{99}}</math> | ||
− | ~phoenixfire | + | ~phoenixfire |
+ | -dreamr | ||
==See Also== | ==See Also== |
Revision as of 16:52, 20 November 2019
Problem 17
What is the value of the product
Solution 1
As you can easily see
If we remove all parentheses, we can easily see that the middle terms cancel, leaving us with
=
~phoenixfire -dreamr
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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.