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 solution 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 4 10 #40  6 7 #42  1 2 3 7 #42
which brings a positive reply to the question. Moving from N=10 to N=19 produced similar results
 1 9 18 #162  2 6 14 #168  1 3 4 14 #168  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 intellectual (!) approach, over a long insomnia bout, I realised that if there are three terms, x¹,x² and x³, in the second group, they need satisfy
and if in addition one of them is equal to 1, x¹ say, this equation simplifies into
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.