# 1985 AIME Problems/Problem 1

## 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:

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