A sequence is obtained by deleting all perfect squares from set of natural numbers . What is the remainder when the 2003^rd term of new sequence is divided by 2048?​

Dear student

There are 44 perfect squares between 1 and 2003 and after removing all the perfect squares from 12to 442 we see tat 2003 rd term of this new sequence becomes 2047(i.e 2003 +44) But since 2025 is a perfect square (which is 452) so it should also be removed . So the after removing 2025 from the  sequence the 2003rd terms becomes 2048.

hence remainder 0

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