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How to prove that, if $n$ is a positive integer, then $$ (\underbrace{666 \ldots 6}_{n \text{ copies of } 6})^2 + \underbrace{888 \ldots 8}_{n \text{ copies of } 8} = \underbrace{444 \ldots 4}_{2n \text{ copies of } 4}? $$

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Hint :

I will show it for 3-digit and you will know the general trend.

$(666)^2 + 888 = 36(111)^2 + 8(111) = 111(36(111) + 8) = 444(9(111) + 2) = 444(1001) = 444444$

Basically you will always get $44\ldots4$ ($n$ times) common times $100\ldots1$ ($n-1$ zeroes) which will help repeating the digits.

You can write a formal proof using this idea very easily. Let $e_n$ be the integer with $n$ $1s$. Substitute $e_n$ for $111$ in proof above and note that $9e_n+2=10^n+1$.

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  • $\begingroup$ what should i do after This because I did exactly what you did here ..How do I prove it from here $\endgroup$ – MathEnthusiast Sep 20 '15 at 13:17
  • $\begingroup$ When you multiply 444 by 1001 you get 444444. You multiply 44 by 101 you get 4444... Basically you are multiplying 44...4 by $10^n + 1$ $\endgroup$ – Shailesh Sep 20 '15 at 13:20
  • $\begingroup$ Is that enough? It's actually 10^(n+1) $\endgroup$ – MathEnthusiast Sep 20 '15 at 13:20
  • $\begingroup$ How do I prove you statement mathematically $\endgroup$ – MathEnthusiast Sep 20 '15 at 13:22
  • $\begingroup$ Yes. Depends on how much rigour/formal is required. You can accept the answer if you are satisfied, otherwise I can help you on chat. $\endgroup$ – Shailesh Sep 20 '15 at 13:22
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Given $$\underbrace{(666666666666\ldots)^2}_{n~\text{times}}+\underbrace{(888888888888\ldots)}_{n~\text{times}}$$

Now we can write $$\underbrace{666666666666\ldots}_{n~\text{times}} = (6+6\cdot 10+6\cdot 10^2+\cdots+6\cdot 10^{n-1})$$

$$\displaystyle =6\left[\frac{10^n-1}{10-1}\right] = \frac{2}{3}\left[10^n-1\right]$$

Similarly we can write $$\underbrace{88888888888\ldots}_{n~\text{times}} = (8+8\cdot 10+8\cdot 10^2+\cdots+8\cdot 10^{n-1})$$

$$\displaystyle =8\left[\frac{10^n-1}{10-1}\right] = \frac{8}{9}\left[10^n-1\right]$$

So we get $$\displaystyle \left\{\frac{2}{3}\left[10^n-1\right]\right\}^2+\frac{8}{9}\left[10^n-1\right] = \frac{4}{9}(10^n-1)^2+\frac{8}{9}(10^n-1)$$

So we get $$\displaystyle = \frac{4}{9}(10^n-1)\cdot \left[10^n-1+2\right] = \frac{4}{9}(10^{2n}-1)$$

So we can write $$\displaystyle \frac{4}{9}(10^{2n}-1) = \underbrace{(444444444444\ldots)}_{2n~\text{times}}$$

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