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Algebra Level 5

Consider a sequence a 1 , a 2 , . . . , a 2018 a_1, a_2, . . . , a_{2018} of positive integers. Extend it periodically to an infinite sequence a 1 , a 2 , . . . a_1, a_2, . . . by defining a i + 2018 = a i a_{i+2018} = a_i for all i 1 i \ge 1 . If

a 1 a 2 . . . a 2018 a 1 + 2018 \large\ { a }_{ 1 } \le { a }_{ 2 } \le ...\le a_{ 2018 } \le a_1 + 2018

and

a a i i + 2017 \large\ { a }_{ { a }_{ i } } \le i + 2017 ,

for i = 1 , 2 , 3 , . . . , 2018 i = 1, 2, 3, ... , 2018 .

Maximise a 1 + a 2 + . . . + a 2018 { a }_{ 1 } + { a }_{ 2 } + ...+ a_{ 2018 } .


The answer is 4072324.

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1 solution

Alapan Das
Dec 19, 2018

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

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