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Expressed as a binary number, we have <math> S = a\times10_b^{2n}</math> where <math> 1\leq a<100_b</math>, an estimate for <math>\sqrt{S} = \sqrt{a}\times2^n</math> is <math>\sqrt{S} \approx 2^n</math>, since the [[geometric mean]] of the lowest and highest possible values with the given number of digits is <math>\sqrt{\sqrt{1} \cdot \sqrt{100_b}} = \sqrt[4]{4} = \sqrt{2} \approx 1.4</math>, which is close to one.
Slightly improved estimates can be obtained by expressing <math>S</math> in normalized scientific notation as <math> S = a\times10^b</math> where <math> 1\leq a<10</math>, and treating the cases <math>b = 2n</math> (even) and <math>b = 2n + 1</math> (odd) differently. Then the estimates are
:<math> \sqrt{S} \approx \begin{cases}
2 \cdot 10^n & \text{if } b = 2n, \\
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