Question

A sequence is obtained by deleting all perfect squares from the set of all natural numbers starting from 1, then

Answer 1

Remainder is 0 when the 2003 rd term of  sequence obtained by deleting all perfect squares from the set of all natural numbers starting from 1 is divided by  2048

Step-by-step explanation:

Complete Question

What is the remainder when the 2003 rd term of new sequence is divided by

2048

√2003 > 44

Hence atleast 44 perfect Square has been deleted ( 1² , 2² to ........44²)

Now checking

45² = 2025

2003 + 44 = 2047  > 2025

Hence 2025 also deleted

46² = 2116 > 2003 + 45 = 2048 Hence 46² does not come till 2003rd term

Hence 45 terms had been deleted

so total terms before deleting perfect square = 2003 + 45 = 2048

so 2003rs term after deleting perfect square = 2048

2048 = 2048 * 1 + 0

Remainder = 0

Learn more:

how many different set of four consecutive consecutive numbers can ...

https://brainly.in/question/10719429