Does this pattern continue $\lfloor\sqrt{44}\rfloor=6, \lfloor\sqrt{4444}\rfloor=66,\dots$?

By observing the following I have a feeling that the pattern continues.

$$\lfloor \sqrt{44} \rfloor=6$$ $$\lfloor \sqrt{4444} \rfloor=66$$ $$\lfloor \sqrt{444444} \rfloor=666$$ $$\lfloor \sqrt{44444444} \rfloor=6666$$

But I'm unable to prove it. Your help will be appreciated.

• I'd start by showing that the sequence $a_n=\frac{\sqrt{\sum_{i=0}^n44\cdot 100^i}}{\sum_{i=0}^n 6\cdot 10^i}$ converges to $1$. This might give you a clue where the answer comes from. Jul 13, 2018 at 12:21
• Nice observation (+1) Jul 13, 2018 at 12:59
• Well, $4=2^2$, so your pattern is the same as, \begin{align}\lfloor\sqrt{11}\rfloor&=6\div 2 = 3 \\ \lfloor\sqrt{1111}\rfloor&=33 \\ \lfloor\sqrt{111111}\rfloor&=333 \\ &\vdots\end{align} Jul 16, 2018 at 23:14
• This is a (very elegant!) variation of a well known puzzle: show that the pattern$$7^2=49\ ,\quad 67^2=4489\ ,\quad 667^2=444889\ ,\quad 6667^2=44448889$$continues indefinitely, proved for example here. Jul 17, 2018 at 23:49
• Actually with decimals, you have $\sqrt{0.444444444444...}=0.66666666666...$. Jul 20, 2018 at 1:15

Hint :

We have $$\left(\frac{6\cdot (10^n-1)}{9}\right)^2=\frac{4\cdot (10^{2n}-1)}{9}-\frac{8\cdot (10^n-1)}{9}$$

Try to find out why this proves that the pattern continues forever.

• @user571036 The equation has the form $a^2=b-c$. It is easy to see that $c<2a$. Now , we have $a^2=b-c<b$ , hence $a<\sqrt{b}$ and $(a+1)^2=a^2+2a+1=b-c+2a+1=b+1+2a-c>b+1>b$, hence $a+1>\sqrt{b}$. Hence, $\lfloor \sqrt{b} \rfloor=a$ Jul 13, 2018 at 12:48
• Then how did you get to know? Jul 13, 2018 at 12:53
• To find $\lfloor \sqrt{b} \rfloor$, we have to find the (unique) integer $a$ with $a\le \sqrt{b}<a+1$ that is simply the deifinition. Here the left inequality is also strict. The main problem was to establish the equation. I tried some cases and guessed it. Jul 13, 2018 at 12:55
• Thank you, Peter. You were so nice throughout the conversation. Jul 13, 2018 at 12:57
• How do you learn to solve such things? The solution comes from finding patterns in the problem. Most people just learn to find such patterns over the years without explicit training in methods of solving. There's nothing in the equation beyond say early high school. There are techniques though if you want to study them. For example, the book "How to Solve It", by renowned mathematician G. Polya but aimed for general audiences, is a good source. Jul 13, 2018 at 20:34

This can be proved a bit more simple.

The general term can be written as $4 \frac{10^{2n} - 1}{9}$. Taking square root will give you $\frac{2}{3} \sqrt{10^{2n} - 1}$. As $\frac{2}{3} \approx 0.66666666666$ and $\sqrt{10^{2n} - 1} \approx \sqrt{10^{2n}} = 10^n$, this explains why the result is $6;66;666;...$ etc. To prove rigorously, observe that: $66...66 = \frac{2}{3} 10^n - \frac{2}{3} < \frac{2}{3} \sqrt{10^{2n} - 1} < \frac{2}{3} 10^n = 66...66.67$.

• I think the rigorous part is not rigorous enough. The two bounds you give are not sufficient in explaining why flooring the value gives a $6$ as the last digit. Jul 18, 2018 at 8:17
• @M.Winter if you have a number in the interval $[665.67 ; 666.67]$, then its floor value must be $666$, isn't it? Jul 18, 2018 at 8:47
• Flooring means rounding to the smaller integer, so flooring 665.67 gives 665. Jul 18, 2018 at 8:53
• nice squeeze there Jul 18, 2018 at 8:56
• @SiXUlm sorry, consider 665.68. I made a typo. Jul 18, 2018 at 8:57

As a somewhat inferior alternative to Peter's algebraic approach, you could work out a square root in long-division style as shown here:

\begin{array}{r} 6 \; \phantom{0}6 \;\phantom{0}6 \;\phantom{0}6 \\[-3pt] \sqrt{44 \; 44 \; 44 \; 44} \\[-3pt] \underline{36}\; \phantom{66 \; 00 \; 00} \\[-3pt] 12\underline6|\phantom{0}8 \; 44 \phantom{\; 00 \; 00}\\[-3pt] \underline{7 \; 56} \phantom{\; 00 \; 00}\\[-3pt] 132\underline{6} |\phantom{0} \; 88 \; 44 \phantom{\; 00}\\[-3pt] \underline{79 \; 56} \phantom{\; 00}\\[-3pt] 1332\underline{6} | \phantom{0}8 \; 88 \;44\\[-3pt] \underline{7 \; 99 \; 56} \end{array}

From here you may be able to extrapolate some provable patterns that occur as you append more pairs of $4$s to the digits under the square root sign.