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 Problem 9"

(Created page with "== Problem == The product of the 9 factors <math>\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =</math> <math>\text{(A)}\ \...")
 
(→‎Solution 2 (Brute Force))
 
(6 intermediate revisions by 2 users not shown)
Line 3: Line 3:
  
 
<math>\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}</math>
 
<math>\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}</math>
 +
 +
== In-depth Solution by BoundlessBrain!==
 +
https://youtu.be/yBrLXnjasgg
  
 
== Solution ==
 
== Solution ==
Line 8: Line 11:
 
<cmath>\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.</cmath>
 
<cmath>\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.</cmath>
 
Numerators and denominators cancel to yield the answer: <math>\boxed{\text{(A)} \frac{1}{10}}.</math>
 
Numerators and denominators cancel to yield the answer: <math>\boxed{\text{(A)} \frac{1}{10}}.</math>
 +
 +
== Solution 2 (Brute Force) ==
 +
 +
Multiply the numerators and denominators
 +
 +
The expression is <math>\frac{362880}{3628800}</math> which is <math>\frac{1}{10}</math> or <math>\boxed{\textbf{(A)}\frac{1}{10}}</math>
 +
 +
 +
<math>\textbf{Note: Do not use unless no other method found}</math>

Latest revision as of 07:29, 6 October 2024

Problem

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

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

In-depth Solution by BoundlessBrain!

https://youtu.be/yBrLXnjasgg

Solution

This product simplifies to: \[\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.\] Numerators and denominators cancel to yield the answer: $\boxed{\text{(A)} \frac{1}{10}}.$

Solution 2 (Brute Force)

Multiply the numerators and denominators

The expression is $\frac{362880}{3628800}$ which is $\frac{1}{10}$ or $\boxed{\textbf{(A)}\frac{1}{10}}$


$\textbf{Note: Do not use unless no other method found}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problem_9&oldid=229085"