$\lfloor \underbrace{\sqrt{44...4}}_{2n} \rfloor = \underbrace{66..6}_{n} $

Question

Prove: $\lfloor \sqrt{44} \rfloor = 6 $, $\lfloor \sqrt{4444} \rfloor = 66 $, $\lfloor \sqrt{444444} \rfloor = 666$, ...

$$ \lfloor \underbrace{\sqrt{44...4}}_{2n} \rfloor = \underbrace{66..6}_{n} $$

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Attempt:

The approach is to prove $$ \underbrace{(66...6)^{2}}_{n} < \underbrace{44...4}_{2n} < (\underbrace{66...6}_{n} + 1)^{2}$$

Now notice that $$(66 + 1)^{2} = 66^{2} + 2(66) + 1 = 66^{2} +133$$ $$(666 + 1)^{2} = 666^{2} + 2(666) + 1 = 666^{2} +1333$$ $$ \vdots $$ $$(\underbrace{66...6}_{n} + 1)^{2} = \underbrace{(66...6)^{2}}_{n} + 1\underbrace{33...3}_{n}$$

Next, notice

$$ 66^{2} = 36(11^{2}) = 36(121) = 3600 + 720 + 36 = 4356 = 4444 - 2(44) $$

$$ 666^{2} = 36(111^{2}) = 36(12321) = 360000 + 72000 + 10800 + 720 + 36 = 443556 = 444444 - 2(444) $$

and after several trials it seems convincing that

$$ (\underbrace{66...6}_{n})^{2} = \underbrace{44...4}_{2n} - 2(\underbrace{44...4}_{n}) = \underbrace{44...4}_{2n} - \underbrace{88...8}_{n} $$

and since $\underbrace{88...8}_{n} < 1\underbrace{33...3}_{n}$, we are done.

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Q:

1.) But how to prove the: $$ (\underbrace{66...6}_{n})^{2} = \underbrace{44...4}_{2n} - \underbrace{88...8}_{n} $$ nicely?

2.) What other better ways?

Answer 1

Call $a_n$ the number with $n$ digits all equal to $6$ and $b_n$ the number with $2n$ digits all equal to $4$. We want to prove that $a_n^2 < b_n < (a_n+1)^2$. Observe that $$a_n = \frac{2}{3}(10^n-1) \text{ so } a_n+1 = \frac{2}{3}(10^n+\frac{1}{2}).$$ Hence $$a_n^2 = \frac{4}{9}(10^{2n} -2 \times 10^n + 1) \text{ so } (a_n+1)^2 = \frac{4}{9}(10^{2n} + 10^n + \frac{1}{4}),$$ whereas $$b_n = \frac{4}{9}(10^{2n}-1).$$ The desired inequalities follow.

Answer 2

Trick is to realize that $\underbrace{aaaa......a}_n = a\times \underbrace{11111.....1}_n=a\times \frac{\underbrace{99999...9}}9= a\times\frac {10^n - 1}9=\frac a9\times (10^n-1)$.

Then it's easy to calculate that ${\underbrace{66666....6}_n}^2 =$

$(\frac 69(10^n-1))^2 = \frac 49(10^{2n} - 2\times 10^n + 1)$

A bit of clever manipulation and

$\frac 49(10^{2n} - 2\times 10^{n} + 1) = \frac 49([10^{2n}-1] - 2\times[10^n-1])=$

$4\times \frac {10^{2n}-1}9 - 8\times \frac {10^n-1}9$

Now $\underbrace{4444....4}_{2n} = 4\times \frac{10^2n-1}9$ so

${\underbrace{66666....6}_n}^2 = 4\times \frac {10^{2n}-1}9 - 8\times \frac {10^n-1}9 < 4\times \frac{10^2n-1}9 = \underbrace{4444....4}_{2n}$.

.......

Likewise ${(\underbrace{66666....6}_n+1)}^2=$

$(\underbrace{66666....6}_n+1)^2 + 2\times (\underbrace{66666....6}_n) + 1=$

$4\times \frac {10^{2n}-1}9 - 8\times \frac {10^n-1}9 + 2\times 6\times \frac{10^n-1}9 + 1=$

$4\times \frac {10^{2n}-1}9 + 4\times \frac {10^n-1}9 + 1 >$

$4\times \frac {10^{2n}-1}9 = \underbrace{4444...4}_{2n}$

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So we have $(\underbrace{66666....6}_n)^2< \underbrace{4444...4}_{2n} < (\underbrace{66666....6}_n+1)^2$ so

$\underbrace{66666....6}_n< \sqrt{\underbrace{4444...4}_{2n}} < \underbrace{66666....6}_n+1$.

So $\lfloor \sqrt{\underbrace{4444...4}_{2n}}\rfloor=\underbrace{66666....6}_n$

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Or we can simply do induction. $6^2 = 36 = 44-8$ and $66^2 = 4444 - 88$.

So if we assume $(\underbrace{6666...6}_n)^2 = \underbrace{44.......4}_2n -\underbrace{8...8}_n$ then we can

show that $(\underbrace{666...6}_{n+1})^2 = $

$(10\times \underbrace{666...6}_{n} + 6)^2 = $

$100\times( \underbrace{6666...6}_n)^2 + 120\times \underbrace{666...6}_{n} + 36=$

$100\times (\underbrace{44.......4}_{2n} -\underbrace{8...8}_n) + 120\times \underbrace{666...6}_{n} + 36=$

$\underbrace{44.......4}_{2n}00 + 36 - \underbrace{8...8}_n00 + 720\times\underbrace{1111...1}_{n}=$

$\underbrace{44......4}_{2n}36 -\underbrace{8....8}_n00 + \underbrace{777...7}00 + \underbrace{2222...22}_n0 =$

$\underbrace{4444....4}_{2n}36 - \underbrace{111....1}_n00 +\underbrace{222...22}_n0=$

$\underbrace{4444....4}_{2n}36 - 1\underbrace{000....0}_{n-1}00 - \underbrace{1111...1}_{n-1}00 + \underbrace{222.....2}_{n-1}00 + 20=$

$\underbrace{4444....4}_{2n}36 - 1\underbrace{000....0}_{n-1}00 + \underbrace{111...1}_{n-1}00 + 20=$

$\underbrace{44444....4}_{2n}36 - \underbrace{888...8}_{n-2}900+20=$

$\underbrace{44444....4}_{2n}36 - \underbrace{888...8}_{n-2}000-900+20=$

$\underbrace{44444....4}_{2n}36 - \underbrace{888...8}_{n-2}00 -880=$

$\underbrace{44444....4}_{2n}44 -8 -\underbrace{888...8}_{n-2}880=$

$\underbrace{44444....4}_{2n}44-\underbrace{888...8}_{n-2}888 =$

$\underbrace{4444...4}_{2n+2} -\underbrace{8888...8}_{n+1}$

Answer 3

Using subscripts to indicate repeated digits. Your approach is fine. The main thing is to use the inductive assumption as a hint for how to split up the $6\dots$ and $4\dots$:

Inductive assumption: $$\color{red}{(6_n)^2 }\le \color{green}{4_{2n}} < \color{blue}{(6_n + 1)^2}$$

To prove: $$(6 \cdot 10^n + 6_n)^2 \le 44 \cdot 10^{2n} + 4_{2n} < (6 \cdot 10^n + (6_n + 1))^2$$ Expand: $$36 \cdot 10^{2n} + 12 \cdot 10^n \cdot 6_n+ (6_n)^2 \le 44 \cdot 10^{2n} + 4_{2n} < 36 \cdot 10^{2n} + 12 \cdot 10^n \cdot (6_n + 1) + (6_n + 1)^2$$

Left side, apply inductive assumption to the colored parts:

$$36 \cdot 10^{2n} + 12 \cdot 10^n \cdot 6_n+ \color{red}{(6_n)^2} \le 44 \cdot 10^{2n} + \color{green}{4_{2n}}$$ $$36 + 12 \cdot \frac{6_n}{10^n} \le 44$$ $$36 + 12 \cdot \left(\frac{2}{3} - \epsilon\right) \le 44$$

Right side:

$$44 \cdot 10^{2n} + \color{green}{4_{2n}} < 36 \cdot 10^{2n} + 12 \cdot 10^n \cdot (6_n + 1) + \color{blue}{(6_n + 1)^2}$$ $$44 < 36 + 12 \cdot \frac{6_n + 1}{10^n}$$ $$44 < 36 + 12 \cdot \left(\frac{2}{3} + \epsilon\right)$$