For a positive integer n > 2, consider the n-1 fractions 2/1; 3/2; ...; 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.

How can you make the product equal 1?
Find all values of n for which this is possible.

Any n that's a perfect square can have this property.

Each of the integers from 1..(n-1) appears exactly twice, and n appears once. For the fractions to multiple out to 1, the numerators and denominators must collectively have the same sets of factors, and so each factor must appear an even number of times between the two. Since all of the factors from 1..(n-1) appear an even number of times already, all of n's factors must also appear an even number of times, making n square.

As to *how* to accomplish the task, here's a recursive strategy:

For n = 2 the fractions are 2/1, 3/2, 4/3. Reverse the first one giving 1/2, 3/2, 4/3 whose product is 1

Assuming you have a product for n = k^2 that equals 1, consider the fractions from (k^2+1)/k^2 to (k+1)^2 / ((k+1)^2 -1). Flip the first k of these.

There will be two groups: the first k flipped and the remaining k+1 unflipped. In each case, the first numerator and the last denominator survive and the rest cancel out. In the first (flipped) batch, the surving numerator is k^2 (it was flipped) and the surviving denominator is (k^2 + k). In the second batch, the surving numerator is (k+1)^2 and the surviving denominator is (k^2 + k).

The two numerators' product is k^2 * (k+1)^2 and the denominators' is (k^2 + k)^2. But k^2 + k = k(k+1) so these two products are identical, and their ratio is 1. So, if you have a solution for n = k^2, then you can construct one for n = (k+1)^2. That combined with the explicit solution for n = 4 completes an inductive proof.

Posted by Paul on 2015-12-01 17:56:12