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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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Solution solution, I think | Comment 2 of 10 |

Eighteen of the numbers can be broken down into six triplets, summing to at least 186, meaning the one we left out can be at most 203-186 = 17. No other single number could be higher than this with the total sum adding to only 203, while keeping the triplets of the other 18 numbers at least 31 each.

The numbers on either side of the 17 must of course add up do at least 14, as do the two numbers to the 17's left or the two numbers to the 17's right.

In fact, each group of 3 among those 18 can be only exactly 31, as the average must be 31, and none can go below 31.

Whatever total is on the 17's left and right, must continue around the circle. Furthermore, the one that's on the 17's left is the beginning of a cycle when viewed in a certain direction (CW or CCW), but that on the right is the end of the cycle. But these must be equal

The only triplets that exceed a total of 31 are those that begin or end with the 17. In fact, they'l both begin and end with a 17: 17-7-17 for a total of 41, but that certainly fits in with the "31 or more".


  Posted by Charlie on 2010-09-23 17:57:18
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