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Someone had posted a question on this site as to what would be sum of digits of $999999999999^3$ (twelve $9s$ ) equal to?

I did some computation and found the pattern that sum of digits of $9^3 = 18$, $99^3 = 36$, $999^3 = 54$ and so on. So, I had replied that sum of digits of $999999999999^3 = 12 \cdot 18 = 216$.

Can anybody help me prove this, that sum of digits of $(\underbrace{999\dots9}_{n\text{ times}})^3 = 18n.$

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    $\begingroup$ Here is a link to the original question. As I have suggested in a comment to that question, it can be answered easily using the fact that $\underbrace{9\dots9}_{n\text{ times}}=10^n-1$. $\endgroup$ – barak manos Sep 5 '16 at 10:01
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    $\begingroup$ Thanks for adding that link. $\endgroup$ – user356774 Sep 5 '16 at 10:50
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    $\begingroup$ @the question: Similarly one can show that sum of digits for (99999.....9)^2 = 9n and for (99999.....9)^5 = 27n where (99999.....9) denotes n number of 9s $\endgroup$ – user356774 Sep 6 '16 at 13:06
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    $\begingroup$ See the pattern : $729, 970299, 997002999, 999700029999$ etc. Can you finish it from here? $\endgroup$ – Teresa Lisbon Oct 6 '16 at 12:23
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    $\begingroup$ did you try to expand $(10^{12}-1)^3$ using (a-b)^3=a^3+3ab^2-3ba^2-b^3$ $\endgroup$ – hamam_Abdallah Oct 6 '16 at 12:39
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Observe that:

$\left(\underbrace{9\dots9}_{n\text{ times}}\right)^3=(10^n-1)^3=10^{3n}-3\cdot10^{2n}+3\cdot10^{n}-1=\underbrace{9\dots9}_{n-1\text{ times}}7\underbrace{0\dots0}_{n-1\text{ times}}2\underbrace{9\dots9}_{n\text{ times}}$


Therefore, the sum of digits is $9(n-1)+7+2+9n=18n$.

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Well, all you have to do is to observe the pattern of the cubes themselves:

$9^3 = 729$.

$99^3=970299$

$999^3=997002999$

$9999^3=999700029999$

Now, you are in a position to make a conjecture:

Let $(k)_n$ mean the number in base $10$ represented by $k$ repeated $n$ times. Then, $((9)_n)^3 = (9)_{n-1}7(0)_{n-1}2(9)_n$.

I want you to go out and prove this conjecture yourself, use the fact that $(9)_n = 10^{n}-1$, and $(10^{n} -1 )^3 = 10^{3n} - 3\cdot 10^{2n} + 3 \cdot 10^n - 1$.

Now, use the fact that the sum of digits of $(k)_n$ is $nk$. Putting this formula above, the sum of digits of $((9)_n)^3$ is $9(n-1) + 7 + 2 + 9(n) = 18n$. Hence, for $12$ digits, your formula gives a sum of $216$, which is correct.

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With $n\geq 1$:

$(10^n-1)^3=10^{3n}-3\cdot 10^{2n}+3\cdot 10^n -1$ $=(10^{n-1}-1)10^{2n+1}+7\cdot 10^{2n}+2\cdot 10^n+10^n-1$.

Therefore $9(n-1)+7+2+9n=18n$.

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    $\begingroup$ True also for $n=1$ (our answers are literally almost identical I've noticed)... $\endgroup$ – barak manos Sep 5 '16 at 9:58
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    $\begingroup$ @barak manos: Thanks. - Yes, nearly the same.:-) $\endgroup$ – user90369 Sep 5 '16 at 10:18
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The cube of '9 is '72'9, where 'x is a digit before the base just before 10, so '7 = 10 -3 etc.

The sum of digits in the general base is 1'8 (ie 2,0 - 2)

When taken for the general case of 9(n)9³ = 9(n)7 0(n)2 9(n)9, it is 18n+18, but because we mean n to be the number of digits in the lead, 18(n).

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    $\begingroup$ Can anybody understand this? $\endgroup$ – TonyK Oct 6 '16 at 12:50
  • $\begingroup$ trying to figure it out.... $\endgroup$ – user356774 Nov 2 '16 at 8:26
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    $\begingroup$ The notation a(b) means here a repeated b times, so 9(4)9 means 9999 9. The cube of 9999 9 is 9999 7 0000 2 9999 9 or 9(4)7 0(4)2 9(9)9. Each column adds up to 18, so 9(5) the digits add to 90. $\endgroup$ – wendy.krieger Nov 2 '16 at 11:14

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