Difference between revisions of "2019 AMC 8 Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
− | As you can easily see, we can write the product as <cmath>\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{ | + | As you can easily see, we can write the product as <cmath>\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}</cmath> |
If we remove all parentheses, we can easily see that the middle terms cancel, leaving us with | If we remove all parentheses, we can easily see that the middle terms cancel, leaving us with |
Revision as of 16:30, 21 November 2019
Problem 17
What is the value of the product
Solution 1
As you can easily see, we can write the product as
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.