Xi'an's Og

An arithmetic Le Monde mathematical puzzle of limited proportions (also found on Stack Exchange):

1. If x,y,z are distinct positive integers such that x+y+z=19 and xyz=p, what is the value of p that has several ordered antecedents?
2.  If the sum is now x+y+z=22, is there a value of p=xyz for which there are several ordered antecedents?
   
3. If the sum is now larger than 100, is there a value of p with this property?

The first question is dead easy to code

entz=NULL
for (y in 1:5) #y<z<x
for (z in (y+1):trunc((18-y)/2))
 if (19-y-z>z) entz=c(entz,y*z*(19-y-z))

and return p=144 as the only solution (with ordered antecedents 2 8 9 and 3 4 12). The second question shows no such case. And the last one requires more than brute force exploration! Or the direct argument that a multiple by κ of a non-unique triplet produces a sum multiplied by κ and a product multiplied by κ³. Hence leads to another non-unique triplet with an arbitrary large sum.

This entry was posted on March 19, 2018 at 12:18 am and is filed under Books, Kids with tags Le Monde, mathematical puzzle, R, Stack Exchange. 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.