Question

If ${(666...ntimes)}^2+(8888...ntimes)=(4444...Ktimes)$then $K$ is

Answer 1

Step 1. Expand $(666...ntimes)$ and solve it:

$\begin{array}{rcl}666\dots ntimes)& =& 6+60+600+..nterms\\ & =& 6(1+10+100+\dots .n)\\ & =& \frac{6({10}^n-1)}{9}\\ & =& \left(\frac{2}{3}\right)({10}^n-1)\end{array}$

Step 2. Expand $\mathbf{(}\mathbf{8888}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathit{n}\mathbf{}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathit{s}\mathbf{)}$ and solve it:

$\begin{array}{rcl}(8888\dots ntimes)& =& 8+80+800+..nterms\\ & =& 8(1+10+100+\dots .n)\\ & =& \frac{8({10}^n-1)}{9}\\ & =& \left(\frac{8}{9}\right)({10}^n-1)\end{array}$

Step 3. Put the values of $\mathbf{(}\mathbf{666}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathit{n}\mathbf{}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathit{s}\mathbf{)}$ and $\mathbf{(}\mathbf{8888}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathit{n}\mathbf{}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathit{s}\mathbf{)}$ in given equation, we get

$\begin{array}{rcl}{(666\dots ntimes)}^2+(8888\dots ntimes)& =& {\left(\frac{2}{3}\right)}^2{({10}^n-1)}^2+\left(\frac{8}{9}\right)({10}^n-1)\\ & =& \left(\frac{4}{9}\right){({10}^n-1)}^2+\left(\frac{8}{9}\right)({10}^n-1)\\ & =& \left(\frac{4}{9}\right)({10}^n-1)\left[\right({10}^n-1)+2]\\ & =& \left(\frac{4}{9}\right)({10}^n-1)\left[\right({10}^n+1]\\ & =& \left(\frac{4}{9}\right)({10}^{2n}-1)\\ & =& 4(1+10+100+\dots 2nterms)\\ & =& 4+40+400+\dots 2nterms\\ & =& 444\dots .ktimes(\mathrm{Given})\end{array}$

$\mathbf{\therefore }\mathit{K}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{n}$