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

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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>.
 
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>.
 
==Solution==
 
==Solution==
{{solution}}
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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==
 
==See Also==
 
*[[1985 AIME Problems/Problem 2|Next Problem]]
 
*[[1985 AIME Problems/Problem 2|Next Problem]]
  
*[[1985 AIME]]
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*[[1985 AIME Problems | Back to exam]]
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[[Category:Introductory Algebra Problems]]

Revision as of 13:33, 18 November 2006

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