Difference between revisions of "1985 AIME Problems/Problem 1"

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==Problem==
 
==Problem==
Let <math>x_1=97</math>, and for <math>n>1</math> let<math>x_n=\frac{n}{x_{n-1}}</math>. Calculate the product <math>x_1x_2x_3x_4x_5x_6x_7x_8</math>.
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Let <math>x_1=97</math>, and for <math>n>1</math> let<math>x_n=\frac{n}{x_{n-1}}</math>. Calculate the [[product]] <math>x_1x_2x_3x_4x_5x_6x_7x_8</math>.
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==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 <math>x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = 384</math>.
 
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 <math>x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = 384</math>.
==See Also==
 
*[[1985 AIME Problems/Problem 2|Next Problem]]
 
  
*[[1985 AIME Problems | Back to exam]]
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==See also==
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{{AIME box|year=1985|before=First Question|num-a=2}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 15:54, 6 March 2007

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 $x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = 384$.

See also

1985 AIME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AIME Problems and Solutions