2013 BAMO-8 #4

 For a positive integer $n>2$, consider the $n-1$ fractions$$\dfrac21, \dfrac32, \cdots, \dfrac{n}{n-1}$$The product of these fractions equals $n$, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of $n$ for which this is possible and prove that you have found them all.



We now want to show that $n$ has to be a perfect square for all possible values of $n$. The answer is all perfect squares $n$ greater than $1$.To see this works, we can take $$\left[\frac{1}{2} \cdot \frac{2}{3} \dotsm\frac{\sqrt{n}-1}{\sqrt{n}}\right] \times \left[\frac{\sqrt{n}+1}{\sqrt{n}} \dotsm \frac{n}{n-1}\right].$$All of the first bracket simplifies to $\tfrac{1}{n}$ an the second to $n,$ and $\tfrac{1}{n} \times n = 1.$ To show this is the only solution, note the numerator $p$ has product $n!$ and the denominator $q$ has product $(n-1)!,$ so $pq = (n-1)!^2 \cdot n.$ Also note that for the value to be $1$, we must have $p=q$. Thus, because $pq= (n-1)!^2 \cdot n, $ $n$ must be a perfect square. $\square$

Comments

Popular posts from this blog

1995 IMO #2

Inequality problem I made