Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

1988 AHSME Problems/Problem 14: Difference between revisions

Newer edit →
Timneh (talk | contribs)
2,460 edits
Created page with "==Problem== For any real number a and positive integer k, define <math>{a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}</math> What is <math>{-\frac..."
 
→Solution: Add one
Line 17: Line 17:
==Solution==
==Solution==


We expand both the numerator and the denominator.


<cmath>\begin{align*}
\binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100}
&= \frac{
  \dfrac{
    (-\frac{1}{2})
    (-\frac{1}{2} - 1)
    (-\frac{1}{2} - 2)
    \cdots
    (-\frac{1}{2} - (100 - 1))
  }{\cancel{(100)(99)\cdots(1)}}
}{
  \dfrac{
    (\frac{1}{2})
    (\frac{1}{2} - 1)
    (\frac{1}{2} - 2)
    \cdots
    (\frac{1}{2} - (100 - 1))
  }{\cancel{(100)(99)\cdots(1)}}
} \\
&= \frac{
    (-\frac{1}{2})
    (-\frac{1}{2} - 1)
    (-\frac{1}{2} - 2)
    \cdots
    (-\frac{1}{2} - 99)
}{
    (\frac{1}{2})
    (\frac{1}{2} - 1)
    (\frac{1}{2} - 2)
    \cdots
    (\frac{1}{2} - 99)
}
\end{align*}</cmath>
Now, note that <math>-\frac{1}{2}-1=\frac{1}{2}-2</math>, <math>-\frac{1}{2}-2=\frac{1}{2}-3</math>, etc.; in essence, <math>-\frac{1}{2}-n=\frac{1}{2}-(n+1)</math>. We can then simplify the numerator and cancel like terms.
<cmath>\begin{align*}
\frac{
    (-\frac{1}{2})
    (-\frac{1}{2} - 1)
    (-\frac{1}{2} - 2)
    \cdots
    (-\frac{1}{2} - 99)
}{
    (\frac{1}{2})
    (\frac{1}{2} - 1)
    (\frac{1}{2} - 2)
    \cdots
    (\frac{1}{2} - 99)
}
&= \frac{
    \cancel{(\frac{1}{2} - 1)}
    \cancel{(\frac{1}{2} - 2)}
    \cancel{(\frac{1}{2} - 3)}
    \cdots
    (\frac{1}{2} - 100)
}{
    (\frac{1}{2})
    \cancel{(\frac{1}{2} - 1)}
    \cancel{(\frac{1}{2} - 2)}
    \cdots
    \cancel{(\frac{1}{2} - 99)}
} \\
&= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\
&= \frac{-\frac{199}{2}}{\frac{1}{2}} \\
&= \boxed{\textbf{(E) } -199.}
\end{align*}</cmath>


== See also ==
== See also ==

Revision as of 15:00, 16 November 2017

Problem

For any real number a and positive integer k, define

${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$

What is

${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$?

$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$

Solution

We expand both the numerator and the denominator.

\begin{align*} \binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100} &= \frac{ \dfrac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} }{ \dfrac{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} } \\ &= \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } \end{align*}

Now, note that $-\frac{1}{2}-1=\frac{1}{2}-2$, $-\frac{1}{2}-2=\frac{1}{2}-3$, etc.; in essence, $-\frac{1}{2}-n=\frac{1}{2}-(n+1)$. We can then simplify the numerator and cancel like terms.

\begin{align*} \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } &= \frac{ \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cancel{(\frac{1}{2} - 3)} \cdots (\frac{1}{2} - 100) }{ (\frac{1}{2}) \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cdots \cancel{(\frac{1}{2} - 99)} } \\ &= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\ &= \frac{-\frac{199}{2}}{\frac{1}{2}} \\ &= \boxed{\textbf{(E) } -199.} \end{align*}

See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AHSME_Problems/Problem_14&oldid=88293"