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Hi to all I'm new this forum . I want to know easiest way to find squre root of large number with in a seconds without using calculator. Suppose I have 2601 and 4096 values. How to find the these value's of the root's with in seconds

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  • $\begingroup$ Say, if you are a programme, it is unthinkable that you would not know that $4096=2^{12}$, so $\sqrt{4096}=2^6=64$. With $2601$ I would start with the known value $\sqrt{2500}=50$. Because the derivative of $\sqrt{x}$ is $1/2\sqrt x$, the derivative at the point $x=2500$ is $1/100$. Therefore $\sqrt{2600}\approx 51$ (and I know that $\sqrt{2601}=51$ exactly - just trying to describe a mental arithmetic method here). Anyway, anchoring such an approximation to "known values" gives reasonably accurate approximations. $\endgroup$ – Jyrki Lahtonen Mar 8 '14 at 8:02
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There is no universal, simple formula.

If you know that the number has an exact (integral) square root, there are lots of mental tricks that apply to different situations.

For 2601, I first noted that 2+6+0+1 = 9, which means it is divisible by 9, so its square root must be divisible by 3.

2601 is a bit bigger than 2500, and I know its square root is 50. The first number larger than 50 divisible by 3 is 51. As 51*51 = 2601, 51 is the correct answer. If it had not been, I would have tried 54 then 57 (the next multiples of 3).

For 4096, unfortunately I know it is 2^12 (I'm a computer scientist) so the square root is 2^6 = 64. Had I not known that, I would have known the last digit was either a 4 or a 6 (because the square ends in 6), it is bigger than 60 (because 60^2 = 3600) and less than 70 (because 70^2 = 4900). This leaves only two possibilities, 64 and 66. There is a simple formula for calculating the square of a number ending in 5 in your head (I will spare you this one) from which you can determine the answer is less than 65 which means it must be 64.

There are lots of simple rules (like these) which you can use to calculate exact square roots in your head. The problem is that there are lots and lots of such rules and techniques, and different numbers require different rules. I am afraid there is not one (or even a couple of) simple rules that work for all numbers.

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  • $\begingroup$ U know 4096 = ` 2^12` but I don't that value, at that we can do? . I'm preparing for competitive exams $\endgroup$ – user3021349 Mar 8 '14 at 8:16
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Wikipedia has a list of methods of computing square roots. Perhaps the most relevant here is the base 10 digit-by-digit calculation (assuming you have access to a pen and paper) or the rough estimation.

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  • $\begingroup$ I'm asking shortcut way @Joel Reyes Noche $\endgroup$ – user3021349 Mar 8 '14 at 7:57
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So, you want to solve $x=\sqrt a$, $a$ being a large number. This means that you look for the solution of $$f(x)=x^2-a=0$$ and suppose that you have a rought estimate of the root (let us call it $x_0$). So, perform one iteration of Newton method; this will give you as a better estimates $$x_1=\frac{a+x_0^2}{2 x_0}$$ Let us try with $a=2601$; we know that $x_0=50$ is reasonable; so the previous formula gives $x_1=51.01$.

Do the same with $a=4096$; we know that $x_0=60$ is reasonable; so the previous formula gives $x_1=64.13$. Suppose that, instead of $x_0=60$, we use either $x_0=50$ or $x_0=70$; the above formula would respectively give either $x_1=65.96$ or $x_1=64.26$.

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There's a little known trick to calculate squares up to $75^2$, provided you know the squares up to $25^2$, exploiting $$a^2=|a-50|^2+100(a-25).$$ Here's how it works. To compute $37^2$, determine the distance from $37$ to $50$, that's $13$, and square it to get 169. Now add as much hundreds as $37$ is above $25$, namely $1200$, to get $37^2=169+1200=1369$.

For $51^2$ we get $1+2600=2601$. As $\sqrt{4096}$ might be $64$ or $66$ we compute $64^2=14^2+3900=4096$ and are done.

Another example: what's $\sqrt{2209}$? Candidates are $43^2=7^2+1800=1849$ and $47^2=3^2+2200=2209$. Bingo. These calculation are done in some seconds.

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