An easily phrased (and solved?) Le Monde mathematical puzzle that does not [really] require an R code:

The five triplets A,B,C,D,E are such that

and

Given that

find the five triplets.

Adding up both sets of equations shows everything solely depends upon E₁… So running an R code that checks for all possible values of E₁ is a brute-force solution. However, one must first find what to check. Given that the sums of the triplets are of the form (16s,4s,s), the possible choices for E₁ are necessarily restricted to

> S0=193+187+185+175
> ceiling(S0/16)
[1] 47
> floor((S0+175)/16)
[1] 57
> (47:57)*16-S0 #E1=S1-S0
 [1]  12  28  44  60  76  92 108 124 140 156 172

The first two values correspond to a second sum S₂ equal to 188 and 192, respectively, which is incompatible with A₁ being 193. Furthermore, the corresponding values for E₂ and E₃ are then given by

> S2==(49:57)*4
> E1=(49:57)*16-S0
> E2=S2-E1
> S3=S2/4
> S3-E2
[1] -103  -90  -77  -64  -51  -38  -25  -12    1

which excludes all values but E₁=172. No brute-force in the end…