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The $50^{th}$ digit from the left in the expansion of $(\sqrt{50}+7)^{50}$ after the decimal point.

$\underline{\bf{My\; Try}}::$ Let $\left(\sqrt{50}+7\right)^{50} = I+f$, where $I = $Integer part and $f = $ fractional part. and $0\leq f<1$

Now Let $\left(\sqrt{50}-7\right)^{50} = g$ and here $0\leq g<1$

Now $\displaystyle \left(\sqrt{50}+7\right)^{50}+\left(\sqrt{50}-7\right)^{50} = 2\left(\binom{50}{0}\cdot (\sqrt{50})^{50}+\binom{50}{2}\cdot (\sqrt{50})^{48}\cdot 7^2+..............+\binom{50}{50}\cdot 7^{50}\right)$

$\displaystyle \left(\sqrt{50}+7\right)^{50}+\left(\sqrt{50}-7\right)^{50} = $Integer quantity.

$\displaystyle I+f+g = $ Integer Quantity. and $0\leq f+g<2$ So $(f+g) = 0$ or $(f+g) = 1$

So $I=$Integer quantity or $I = $ Integer quantity$\;\; -1$.

Now I did not understand how can i get it.

Help required.



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It's a bit easier than that: you know that $g\lt 1$ and that $I+f+g$ is some other integer, $j$ (your 'integer quantity'). Since $g\lt 1$, then it must be the case that $j$ (which is $(\sqrt{50}+7)^{50}+(\sqrt{50}-7)^{50}$) $=I+1$ and so $I$ is one less than this quantity. – Steven Stadnicki Oct 12 '13 at 3:56
Which digit are you looking for? $(\sqrt{50}+7)^{50} \approx 2.6\cdot 10^{57}$, so the fiftieth from the left is the hundred millions place. It that the one you want, or the fiftieth after the decimal point. The second is the more common question. – Ross Millikan Oct 12 '13 at 3:58
oh sorry actually it is fiftieth after the decimal point. Thanks for pointing out Ross Millikan. please explain me how can i get it. – juantheron Oct 12 '13 at 4:23
An answer was posted, then unfortunately the author deleted it when he thought he misread the problem. But in fact he got it right. I have voted to undelete. – Gerry Myerson Oct 12 '13 at 5:03
up vote 5 down vote accepted

Consider the number $N=(\sqrt{50}+7)^{50}+(\sqrt{50}-7)^{50}$.

Using the Binomial Theorem, or otherwise, we can show that $N$ is an integer.

Our number $(\sqrt{50}+7)^{50}$ is a little below $N$. How much below? Courtesy of the calculator, $(\sqrt{50}-7)^{50}\approx 4\times 10^{-58}$. So the $50$-th digit after the decimal point of $(\sqrt{50}+7)^{50}$ is $9$.

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You could replace the calculator by using $$\sqrt{50}-7=\frac{1}{\sqrt{50}+7} < \frac{1}{14}$$ so $(\sqrt{50}-7)^{50} < 10^{-50}$. – WimC Oct 12 '13 at 7:07
In the old days I would have used a log table. No batteries. – André Nicolas Oct 12 '13 at 7:17
Thanks André Nicolas and WimC, but i did not understand the line So the $50-$th digit after the decimal point of $(\sqrt{50}+7)^{50}$ is $ = 9$ – juantheron Oct 13 '13 at 8:32
@juantheron: Think about taking away say $0.000000005$ from $17$. The first $8$ digits after the decimal point are $9$'s. In the case of our roblem, the first $57$ digits after the decimal oint are $9$'s. – André Nicolas Oct 13 '13 at 15:06

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