A purely arithmetic Le Monde mathematical puzzle:

Find the four integers w, x, y, z such that the four smallest pairwise sums among the six pairwise sums are 59, 65, 66, and 69. Similarly, find the four smallest of the five integers v, x, y, z such that the five smallest pairwise sums among the ten pairwise sums are 56, 64 , 66, 69 and 70.

The first question is rather straightforward since there are only two possible orderings when x≤y≤z≤w :

x+y≤x+z≤x+w≤y+z≤y+w≤z+w and x+y≤x+z≤y+z≤x+w≤y+w≤z+w

which means

x+y=59, x+z=65, x+w=66, y+z=69, and x+y=59, x+z=65, y+z=66, x+w=69

but since the first system does not allow for an integer solution, the only possibility is the second system, with solution (x,y,z,w)=(29,30,36,40). And the second question is of the same complexity, with, when x≤y≤z≤w≤v :

x+y=56, x+z=64, y+z=66, x+w=69, x+v=70 or x+y=56, x+z=64, x+w=66, x+v=69, y+z=70

with solutions (x,y,z,w,v)=(27,29,37,42,43) and (x,y,z,w,v)=(25,31,39,41,44).