https://www.researchgate.net/publication/316655944_Solution_to_Le_Monde_Puzzle_1006 ]]>

Merci, Jean-Louis! Some bits of the arguments did not make it through though…

]]>Let us assume that positive and negative values alternate ; start with a negative one and take the positive value being constant and equal to one; The 12 values are

-a 1 -b 1 -c 1 -d 1 -e f -a 1

One has to verify

-a>1-b ie b-1>a

which means that a minima b=a+2. Next 1>-b+1 means b>0 and

-b>1-c ie b<c-1

and a minima in c, c=b+2=a+4

Continuing this way,

d=c+2=a+6 and e=d+2=a_8

and since e<f-a

a+8< a-f

and -f=8

The last condition is

f>-a+1 or -f<a-1

and combined with the previous one

8<-f<a-1,

which means that a minima

a=11 and f=-9

so that the final solution is

-11 1 -13 1 -15 1 -17 1 -19 -9 -11 1