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 AIME Problems/Problem 1"

(→‎Solution)
(→‎Solution)
Line 4: Line 4:
 
==Solution==
 
==Solution==
 
Since <math>x_n=\frac{n}{x_{n-1}}</math>, <math>x_n \cdot x_{n - 1} = n</math>.  Setting <math>n = 2, 4, 6</math> and <math>8</math> in this equation gives us respectively <math>x_1x_2 = 2</math>, <math>x_3x_4 = 4</math>, <math>x_5x_6 = 6</math> and <math>x_7x_8 = 8</math> so <cmath>x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.</cmath> Notice that the value of <math>x_1</math> was completely unneeded!
 
Since <math>x_n=\frac{n}{x_{n-1}}</math>, <math>x_n \cdot x_{n - 1} = n</math>.  Setting <math>n = 2, 4, 6</math> and <math>8</math> in this equation gives us respectively <math>x_1x_2 = 2</math>, <math>x_3x_4 = 4</math>, <math>x_5x_6 = 6</math> and <math>x_7x_8 = 8</math> so <cmath>x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.</cmath> Notice that the value of <math>x_1</math> was completely unneeded!
 +
 +
==Solution 2==
 +
Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel:
 +
 +
<math>x_1x_2x_3x_4x_5x_6x_7x_8=x_1\cdot\dfrac{2}{x_1}\cdot\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}\cdot\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}\cdot\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}}=(x_1\cdot\dfrac{2}{x_1})\cdot(\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}})\cdot(\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}})\cdot(\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}})=2\cdot 4\cdot 6\cdot 8=\boxed{384}</math>
  
 
== See also ==
 
== See also ==

Revision as of 13:01, 30 May 2016

Problem

Let $x_1=97$, and for $n>1$, let $x_n=\frac{n}{x_{n-1}}$. Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$.

Solution

Since $x_n=\frac{n}{x_{n-1}}$, $x_n \cdot x_{n - 1} = n$. Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$, $x_3x_4 = 4$, $x_5x_6 = 6$ and $x_7x_8 = 8$ so \[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.\] Notice that the value of $x_1$ was completely unneeded!

Solution 2

Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel:

$x_1x_2x_3x_4x_5x_6x_7x_8=x_1\cdot\dfrac{2}{x_1}\cdot\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}\cdot\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}\cdot\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}}=(x_1\cdot\dfrac{2}{x_1})\cdot(\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}})\cdot(\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}})\cdot(\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}})=2\cdot 4\cdot 6\cdot 8=\boxed{384}$

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_1&oldid=78753"