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

Difference between revisions of "1985 AJHSME Problems/Problem 9"
 
(One intermediate revision by one other user not shown)
Line 14: Line 14:
  
 
Everything except for the first term is <math>1</math>, so the product is <math>\textbf{(A)}\frac{1}{10}</math>
 
Everything except for the first term is <math>1</math>, so the product is <math>\textbf{(A)}\frac{1}{10}</math>
 +
 +
==Video Solution==
 +
https://youtu.be/ZUZk6J2wCw8
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
Line 19: Line 24:
 
{{AJHSME box|year=1985|num-b=8|num-a=10}}
 
{{AJHSME box|year=1985|num-b=8|num-a=10}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
 +
 +
{{MAA Notice}}

Latest revision as of 08:24, 10 January 2023

Problem

The product of the 9 factors $\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =$

$\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}$

Solution

First doing the subtraction, we get \[\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10}\]

We notice a lot of terms cancel. In fact, every term in the numerator except for the $1$ and every term in the denominator except for the $10$ will cancel, so the answer is $\frac{1}{10}$, or $\boxed{\text{A}}$

If you don't believe this, then rearrange the factors in the denominator to get \[\frac{1}{10}\times\frac{2}{2}\times\frac{3}{3}\times\cdots\times\frac{9}{9}\]

Everything except for the first term is $1$, so the product is $\textbf{(A)}\frac{1}{10}$

Video Solution

https://youtu.be/ZUZk6J2wCw8

~savannahsolver

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_9&oldid=186210"