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

(Solution 1)
(Solution 1)
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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{99}{100}</cmath>
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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 \[\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?\]

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

Solution 1

As you can easily see, we can write the product as \[\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}\]

If we remove all parentheses, we can easily see that the middle terms cancel, leaving us with

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

~phoenixfire & dreamr

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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