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goformit100

  • 2 years ago

Consider the sequence 2;3;5;6;7;8;10;::: of all positive integers that are not perfect squares. Determine the 2011th term of this sequence.

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  1. goformit100
    • 2 years ago
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    @oldrin.bataku

  2. oldrin.bataku
    • 2 years ago
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    Find the number of perfect squares under 2011 -- this is the difference between 2011 and our number.

  3. oldrin.bataku
    • 2 years ago
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    Well, we know \(\sqrt{2011}\approx45\) with rounding (since \(44^2=1936\), \(45^2=2025\)). This means we have skipped \(45\) square numbers at this point, so our 2011th term is \(2011+45=2056\)

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