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

*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?**If the sum is now x+y+z=22, is there a value of p=xyz for which there are several ordered antecedents?*

*If the sum is now larger than 100, is there a value of p with this property?*

**T**he 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.