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Nineteen triplets (Posted on 2010-09-23) Difficulty: 3 of 5
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?

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.0000 (2 votes)

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Hints/Tips I clarify few points....A SPOILER | Comment 7 of 10 |
 

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.


  Posted by Ady TZIDON on 2010-09-24 16:09:53
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