\[\large (\ \underbrace{ 66666\ldots6 }_{n \text{ number of 6's}} \ )^2+ \underbrace{88888\ldots8}_{n \text{ number of 8's}}= \underbrace{44444\ldots4}_{2n \text{ number of 4's}} \]

Prove the above equation of positive integer \(n\).

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## Comments

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TopNewest\[ \underbrace{ 6666\ldots6 }_{n \text{ numbers of 6's}} = \dfrac{6}{9}(10^n - 1) = \dfrac{2}{3}(10^n - 1)\] \[ \underbrace{8888\ldots8 }_{n \text{ numbers of 8's}} = \dfrac{8}{9}(10^n - 1)\] \[ \underbrace{ 4444\ldots4 }_{2n \text{ numbers of 4's}} = \dfrac{4}{9}(10^{2n} - 1)\]

So, \[ (\underbrace{ 6666\ldots6 }_{n \text{ numbers of 6's}})^2 + \underbrace{8888 \ldots8}_{n \text{number of 8's}} = (\dfrac{2}{3}(10^n - 1))^{2} + \dfrac{8}{9}(10^n - 1)\] \[=\dfrac{4}{9}(10^{2n} - 2 \times 10^n +1 + 2 \times 10^n -2)\] \[=\dfrac{4}{9}(10^{2n} - 1)\] \[=\underbrace{4444 \ldots 4}_{2n \text{ number of 4's}}\]

Hence, proved.

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Great! Well done!

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Thank u

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@Pi Han Goh , I have even posted the solution using Induction. By the way can u please add me as your friend in Facebook. It's request from me.

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Observe that \( 6^2 + 8 = 44\). So, it is true for \(n=1\).

Let us suppose that for some \(k\),

\[(\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + \underbrace{8888\ldots8 }_{k \text{ numbers of 8's}} = \underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}\]

So, Consider

\[(\underbrace{ 6666\ldots6 }_{k+1 \text{ numbers of 6's}})^{2} + \underbrace{8888\ldots8 }_{k+1 \text{ numbers of 8's}}\] \[=(6 \times 10^{k} +\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + 8 \times 10^{k} +\underbrace{8888\ldots8 }_{k \text{ numbers of 8's}} \] \[=36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}} +(\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + 8 \times 10^{k} + \underbrace{8888\ldots8 }_{k \text{ numbers of 8's}}\] \[= 36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}} + 8 \times 10^{k} +\underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}\] \[= 36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \dfrac{6}{9}(10^{k} - 1) + 8 \times 10^{k} +\underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}\] \[= 44 \times 10^{2k} + \underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}\] \[= \underbrace{ 4444\ldots4 }_{2k+2 \text{ numbers of 4's}}\]

So, by \(\text{Principle of Finite Mathematical Induction}\). The given statement is true.

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