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Q.

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

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Detailed Solution

Since $[\sqrt{2046}] = [\sqrt{2047}] = [\sqrt{2048}] = [\sqrt{2049}] = 45$

$∴$ 2003^rd term is 2003 + 45 = 2048
Hence remainder is 0

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