**T**he last and final Le Monde puzzle is a bit of a disappointment, to wit:

*A 4×4 table is filled with positive and different integers. A 3×3 table is then deduced by adding four adjacent [i.e. sharing a common corner] entries of the original table. Similarly with a 2×2 table, summing up to a unique integer. What is the minimal value of this integer? And by how much does it increase if all 29 integers in the tables are different?*

**F**or the first question, the resulting integer writes down as the sum of the corner values, plus 3 times the sum of the side values, plus 9 times the sum of the 4 inner values [of the 4×4 table]. Hence, minimising the overall sum means taking the inner values as 1,2,3,4, the side values as 5,…,12, and the corner values as 13,…,16. Resulting in a total sum of 352. As checked in this computer code in APL by Jean-Louis:

This configuration does not produce 29 distinct values, but moving one value higher in one corner does: I experimented with different upper bounds on the numbers and 17 always provided with the smallest overall sum, 365.

firz=matrix(0,3,3)#second level
thirz=matrix(0,2,2)#third level
for (t in 1:1e8){
flor=matrix(sample(1:17,16),4,4)
for (i in 1:3) for (j in 1:3)
firz[i,j]=sum(flor[i:(i+1),j:(j+1)])
for (i in 1:2) for (j in 1:2)
thirz[i,j]=sum(firz[i:(i+1),j:(j+1)])
#last
if (length(unique(c(flor,firz,thirz)))==29)
solz=min(solz,sum(thirz))}

and a further simulated annealing attempt did not get me anywhere close to this solution.

### Like this:

Like Loading...