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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Largest element ..... | Comment 5 of 10 |
(In reply to re: Largest element .........Please edit by Ady TZIDON)

Is this your point, Ady?

As Charlie observes, the configuration must be CCWCCW... or CWCWCW... If we maximise x to 17, then with the first configuration we have 17,17,x; but then 12*17 is 204, and we are 'bust'; or we have 17,x,x, with either 6 or 7 '17's'. But 6*17+13x makes x a fraction. So the number 17 appears exactly 7 times and the number 7 appears exactly 12 times. Now with the CWCW.. configuration, we obtain 10*17+9x which again gives a fraction, or 9*17+10x, but in that case x = 5 and adjacent numbers sum only to 27.

So there is one and only one configuration that works and that is 17,7,7.... with the 19th number being a 17.

  Posted by broll on 2010-09-24 05:04:19
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