AIME 1985 · 第 1 题

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!

Another Way to think about the solution

Every time you multiply in the next term of the sequence, all the numbers before are flipped from the numerator to the denominator or the denominator to the numerator because they are divided. So, 97, the first term, will appear in the multiplied out form in this pattern: $NDNDNDND$ where $N$ is the numerator and $D$ is the denominator. The second term (2) will appear in the pattern $XNDNDNDN$ where $X$ means that the number is skipped the first term. All the pairs of $ND$ cross out and you find that only the even terms have an $N$ left over while all the values of the odd terms are crossed out.

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:

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