**A**nother arithmetics Le Monde mathematical puzzle:

From the set of the integers between 1 and 15, is it possible to partition it in such a way that the product of the terms in the first set is equal to the sum of the members of the second set?can this be generalised to an arbitrary set {1,2,..,n}?What happens if instead we only consider the odd integers in those sets?.

I used brute force by looking at random for a solution,

pb <- txtProgressBar(min = 0, max = 100, style = 3) for (N in 5:100){ sol=FALSE while (!sol){ k=sample(1:N,1,prob=(1:N)*(N-(1:N))) pro=sample(1:N,k) sol=(prod(pro)==sum((1:N)[-pro])) } setTxtProgressBar(pb, N)} close(pb)

and while it took a while to run the R code, it eventually got out of the loop, meaning there was at least one solution for all n’s between 5 and 100. (It does not work for n=1,2,3,4, for obvious reasons.) For instance, when n=15, the integers in the product part are either 3,5,7, 1,7,14, or 1,9,11. Jean-Louis Fouley sent me an explanation: when n is odd, n=2p+1, one solution is (1,p,2p), while when n is even, n=2p, one solution is (1,p-1,2p).

A side remark on the R code: thanks to a Cross Validated question by Paulo Marques, on which I thought I had commented on this blog, I learned about the progress bar function in R, *setTxtProgressBar()*, which makes running R code with loops much nicer!

For the second question, I just adapted the R code to exclude even integers:

while (!sol){ k=1+trunc(sample(1:N,1)/2) pro=sample(seq(1,N,by=2),k) cum=(1:N)[-pro] sol=(prod(pro)==sum(cum[cum%%2==1])) }

and found a solution for n=15, namely 1,3,15 versus 5,7,9,11,13. However, there does not seem to be a solution for all n’s: I found solutions for n=15,21,23,31,39,41,47,49,55,59,63,71,75,79,87,95…