Xi'an's Og

A riddle from The Riddler on how many different numbers one could at most produce from six initial values and the four basic operations. In other words, how many values could the terms in

a∅(b∅{c∅[d∅(e∅f)]})

and other positioning of the brackets could take? (With each ∅ being one of the four operations and a,…,f the initial values or a permutation of these.) A very crude evaluation leads to twenty million possible values, forgetting that addition and multiplication are commutative, while subtraction and division are anti-commutative. I tried a brute force approach in R, rather than exploring the tree of possible paths, but could not approach this figure by far, the number of different values still increasing for the largest manageable number of replicas I could try. Reducing the initial values at n=3, I could come closer to 123 with 95 different values and, for n=4, not too far from 1972 with 1687 values. Moving to n=6 by hopefully exhausting all (?) entries led to another guess of 50,524,809 values. I am however unsure that the R code is able to spot identical values, given the variability of the above figure…

This entry was posted on April 30, 2023 at 12:23 am and is filed under Books, R with tags arithmetics, brute force, evolution tree, mathematical puzzle, R, riddle, The Riddler. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.