Xi'an's Og

The weekly puzzle from Le Monde is quite similar to #1157:

Is it possible to break the ten first integers, 1,…,10, into two groups such that the sum over the first group is equal to the product over the second? Is it possible that the second group is of cardinal 4? of cardinal 3?

An exhaustive R search returns 3 solutions by

library(R.utils)
bitz<-function(i)
  c(rev(as.binary(i)),rep(0,10))[d<-1:10]
for (i in 1:2^10)
  if (sum(d[!!bitz(i)])==prod(b<-d[!bitz(i)])) print(b)

[1]  1  4 10 #40
[1] 6 7 #42
[1] 1 2 3 7 #42

which brings a positive reply to the question. Moving from N=10 to N=19 produces similar results

[1]  1  9 18 #162
[1]  2  6 14 #168
[1]  1  3  4 14 #168
[1]  1  2  7 12 #168

with this interesting pattern of only two acceptable products, but I am obviously unable to run the same code for N=49, which is the subsidiary question to the puzzle. Turning to a more conceptual (!) approach, over a long insomnia bout (!!) and a subsequent run, I realised that if there are three terms, x¹,x² and x³, in the second group, they need satisfy

x¹x²x³+x¹+x²+x³=½N(N+1)

and if in addition one of them is equal to 1, x¹ say, this equation simplifies into

(x²+1)(x³+1)=½N(N+1)

which always leads to a solution, as e.g. for N=49,

x¹=1, x²=24 and x³=48.

A brute-force search also led to a four term solution in that case

x¹=1, x²=7, x³=10 and x⁴=17.

This entry was posted on October 6, 2020 at 12:20 am and is filed under Books, Kids, R with tags brute-force solution, Gare d'Austerlitz, Le Monde, mathematical puzzle, Paris, partition, R. 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.