Difference between revisions of "2019 AMC 8 Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
| + | 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> | ||
| + | And all middle terms get cancelled, all we are left with is | ||
| + | |||
| + | <cmath>\left(\frac{1\cdot100}{2\cdot99}\right)</cmath> = <math>\boxed{\textbf{(B)}\frac{50}{99}}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 13:29, 20 November 2019
Problem 17
What is the value of the product
![\[\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?\]](http://latex.artofproblemsolving.com/9/c/a/9ca55aef127eee6fc870173ae1b23e1abd56a02b.png)
Solution 1
As you can easily see ![\[\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\]](http://latex.artofproblemsolving.com/9/b/3/9b32c03a080e16757fdb6ccc0ba5c32cc17b79c2.png)
And all middle terms get cancelled, all we are left with is
![\[\left(\frac{1\cdot100}{2\cdot99}\right)\]](http://latex.artofproblemsolving.com/f/3/8/f3826858055ee43587920c290dc1da722bb59cb7.png)
= 
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. 
