19 numbers are written on a circumference of a circle, in any order.

Their sum is 203 and the biggest number is X.

Any 3 adjacent numbers sum up to 31 or more.

What is the maximal possible value of X ?

If this number is selected, what can be said about the other 18?

So many people asked - I have to clarify few points.

A number is a number is a number .

Nothing in the text implies the numbers being positive, integer, rational or distinct.

6 triplets sum up to 6*31=186 at least , thus forcing the leftover number to be at most 17.

17 is therefore the correct answer,

provided there is no possibility of existence in a triplet (a,b.c) of a number bigger than 17.

I leave it to you to prove it.

If a=b=c than all the numbers ,except one, are 31/3 and the biggest one is 17.

Any triplet (a,b.c) qualifies as a valid answer ,iff a+b+c=31 and a+b,14=< b+c4=< and a+c14=< .

Examples; ( 0,17,14); ( 7,17,7); ( 9,10,12); (x,17,14-x)**, (31/3,31/3,31/3).

**Show that x cannot be negative, so 17 remains the biggest number.