Difference between revisions of "2019 AMC 8 Problems/Problem 17"

(Solution 1)
(Solution 1)
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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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~phoenixfire
  
 
==See Also==
 
==See Also==

Revision as of 14:29, 20 November 2019

Problem 17

What is the value of the product \[\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?\]

$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50$

Solution 1

As you can easily see \[\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\]

And all middle terms get cancelled, all we are left with is

\[\left(\frac{1\cdot100}{2\cdot99}\right)\] = $\boxed{\textbf{(B)}\frac{50}{99}}$

~phoenixfire

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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

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