Difference between revisions of "2019 AMC 8 Problems/Problem 17"
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<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> | ||
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==Solution 2== | ==Solution 2== | ||
If you calculate the first few values of the equation, all of the values tend to <math>\frac{1}{2}</math>, but are not equal to it. The answer closest to <math>\frac{1}{2}</math> but not equal to it is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.~heeeeeeeheeeee | If you calculate the first few values of the equation, all of the values tend to <math>\frac{1}{2}</math>, but are not equal to it. The answer closest to <math>\frac{1}{2}</math> but not equal to it is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.~heeeeeeeheeeee |
Revision as of 01:30, 25 November 2019
Contents
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
Solution 2
If you calculate the first few values of the equation, all of the values tend to , but are not equal to it. The answer closest to
but not equal to it is
.~heeeeeeeheeeee
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.