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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

@oldrin.bataku

2. oldrin.bataku

Find the number of perfect squares under 2011 -- this is the difference between 2011 and our number.

3. oldrin.bataku

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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