Consider a sequence a 1 , a 2 , . . . , a 2 0 1 8 of positive integers. Extend it periodically to an infinite sequence a 1 , a 2 , . . . by defining a i + 2 0 1 8 = a i for all i ≥ 1 . If
a 1 ≤ a 2 ≤ . . . ≤ a 2 0 1 8 ≤ a 1 + 2 0 1 8
and
a a i ≤ i + 2 0 1 7 ,
for i = 1 , 2 , 3 , . . . , 2 0 1 8 .
Maximise a 1 + a 2 + . . . + a 2 0 1 8 .
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aofaof1≤2018.If we take a=2018 then a2018 is also equals to 2018.Taking all ai(i=1,2,....2018)=2018 we get the summation to be 2018x2018=4072324