Skip to main content

Posts

Showing posts from August, 2014

Solving IMO 1989 #6 using Probability and Expectation

Learn One Weird Trick And Easily Solve The Product-Sum Problem

A tribute to Martin Gardner.

For which sets of positive integers does the sum equal the product? For example, when does \( x_1 + x_2 = x_1\cdot x_2\)? It's easy to see that this is only true when \(x_1 = x_2 = 2\).

In the general case our equality is \(\sum_{i=1}^{n} x_i= \prod_{i=1}^{n} x_i \). We can rearrange terms to give \[x_1+x_2+\sum_{i=3}^{n} x_i= x_1\cdot x_2\cdot \prod_{i=3}^{n} x_i,\] and this in turn factors nicely to give us \[\left( x_1\cdot \prod_{x=3}^{n} x_i - 1\right)\cdot \left( x_2\cdot \prod_{x=3}^{n} x_i - 1\right) = \left( \prod_{x=3}^{n} x_i \right)\cdot \left(\sum_{x=3}^{n} x_i \right) + 1.\] How does this help? Consider the case \(n=5\), and without loss of generality assume \(x_i \ge x_{i+1}\) for all \(i\). If \(x_3=x_4=x_5=1\) our factorized equation becomes \[(x_1-1)\cdot(x_2-1)=4,\] with the obvious solutions \( x_1=5, x_2=2; x_1=3, x_2=3\). The only remaining case to consider is \(x_3=2\), as any other case forces  \( \sum_{i=1}^{n} x_i < \prod_{…