These Numbers Are Stupendous!

Relevant wiki: Algebraic Manipulation Problem Solving - Basic

First we have $\overline{\underbrace{dddddd\ldots d}_\text{n d's}}=\dfrac{d(10^n-1)}{9}$ . Then, we want to calculate:

$x=\left(\dfrac{6(10^{100}-1)}{9}\right)^2+\dfrac{8(10^{100}-1)}{9}=\dfrac{4(10^{100}-1)^2}{9}+\dfrac{8(10^{100}-1)}{9}$

$x=\dfrac{10^{100}-1}{9}\left(4(10^{100}-1)+8\right)=\dfrac{10^{100}-1}{9}(4\cdot 10^{100}+4)$

$x=\dfrac{4}{9}(10^{100}-1)(10^{100}+1)=\dfrac{4(10^{200}-1)}{9}$

$\boxed{x=\underbrace{444444\ldots 4}_\text{200 4's}}$

Notice that

$\begin{aligned}6^2 &= 36\\ 66^2&=4356\\ 666^2&=443556\\ 6666^2&=44435556\\ 66666^2&=4444355556\end{aligned}$

Therefore, we can generalize the pattern (a more rigorous proof available below):

$\large\left(\underbrace{666\ldots6}_{n\text{ 6's}}\right)^2 = \underbrace{444\ldots4}_{(n-1)\text{ 4's}}3 \underbrace{555\ldots5}_{(n-1)\text{ 5's}} 6$

Therefore,

$\left(\underbrace{666\ldots6}_{100\text{ 6's}}\right)^2+\underbrace{888\ldots8}_{100\text{ 8's}}\\ =\underbrace{444\ldots4}_{99\text{ 4's}}3 \underbrace{555\ldots5}_{99\text{ 5's}} 6+\underbrace{888\ldots8}_{100\text{ 8's}}\\ =\underbrace{444\ldots4}_{99\text{ 4's}}\underbrace{000\ldots0}_{101\text{ 0's}} + 3 \underbrace{555\ldots5}_{99\text{ 5's}} 6+\underbrace{888\ldots8}_{100\text{ 8's}}\\ =\underbrace{444\ldots4}_{99\text{ 4's}}\underbrace{000\ldots0}_{101\text{ 0's}} +\underbrace{444\ldots4}_{101\text{ 4's}}\\ =\boxed{\underbrace {444444\ldots4}_{200\text{ 4's}}}$

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Proof of my claim above by induction (as requested by @Pi Han Goh )

Statement: Let $P_n$ be the proposition that

$\large\left(\underbrace{666\ldots6}_{n\text{ 6's}}\right)^2 = \underbrace{444\ldots4}_{(n-1)\text{ 4's}}3 \underbrace{555\ldots5}_{(n-1)\text{ 5's}} 6$

for all positive integers $n$

Base Case: When $n=1$ ,

LHS $=6^2 = 36$ , RHS $=36$ , LHS = RHS

Induction Hypothesis: Assume $P_k$ is true for some positive integer $k$ . Consider $P_{k+1}$

LHS $P_{k+1}$

$=\left(\underbrace{666\ldots6}_{(k+1)\text{ 6's}}\right)^2\\ =\left(6\underbrace{000\ldots0}_{k \text{ 0's}}+ \underbrace{666\ldots6}_{k\text{ 6's}}\right)^2\\ =36\underbrace{000\ldots0}_{2k \text{ 0's}}+2\left(6\underbrace{000\ldots0}_{k \text{ 0's}}\right)\left( \underbrace{666\ldots6}_{k\text{ 6's}}\right) + \underbrace{444\ldots4}_{(k-1)\text{ 4's}}3 \underbrace{555\ldots5}_{(k-1)\text{ 5's}} 6\\ =36\underbrace{000\ldots0}_{2k \text{ 0's}}+12\left( \underbrace{666\ldots6}_{k\text{ 6's}}\underbrace{000\ldots0}_{k \text{ 0's}}\right) + \underbrace{444\ldots4}_{(k-1)\text{ 4's}}3 \underbrace{555\ldots5}_{(k-1)\text{ 5's}} 6\\ =36\underbrace{000\ldots0}_{2k \text{ 0's}}+ \underbrace{666\ldots6}_{k\text{ 6's}}\underbrace{000\ldots0}_{(k+1) \text{ 0's}}+ 1\underbrace{333\ldots3}_{(k-1)\text{ 3's}}2\underbrace{000\ldots0}_{k \text{ 0's}}+\underbrace{444\ldots4}_{(k-1)\text{ 4's}}3 \underbrace{555\ldots5}_{(k-1)\text{ 5's}} 6\\ =36\underbrace{000\ldots0}_{2k \text{ 0's}}+ \underbrace{666\ldots6}_{k\text{ 6's}}\underbrace{000\ldots0}_{(k+1) \text{ 0's}}+ 1\underbrace{777\ldots7}_{(k-1)\text{ 7's}} \underbrace{555\ldots5}_{k\text{ 5's}} 6\\ =36\underbrace{000\ldots0}_{2k \text{ 0's}}+ 8\underbrace{444\ldots4}_{(k-2)\text{ 4's}} 3\underbrace{555\ldots5}_{k\text{ 5's}} 6\\ =\underbrace{444\ldots4}_{k\text{ 4's}}3 \underbrace{555\ldots5}_{k\text{ 5's}} 6$

$=$ RHS $P_{k+1}$

Conclusion: By Mathematical Induction, since $P_1$ is true and $P_k$ is true implies $P_{k+1}$ is true, thus $P_n$ is true for all positive integers $n$

The quick and dirty way:

Firstly, the sum has easily more than 100 digits. We now only have 2 choices remaining. Notice how both addends are divisible by 4. Since 22 is not divisible by 4 but 44 is, by the divisibility test for 4, the only possible choice which could satisfy is $\underbrace{444444\ldots 4}_\text{200 4's}$ .

We can prove $dddddd......d = \frac{d(10^n-1)}{9}$ by following method:- $dddd...d = d(1111...1) = d(10^n+...+1000+100+10+1)$ Now we know that $10^n+...+100+10+1$ is a geometric series. So, $d(10^n+...100+10+1) = \frac{d(10^n-1)}{10-1} = \frac{d(10^n-1)}{9}$