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Find the sum of digits in decimal form of the number (999...9)^3.

(There are 12 nines)

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    $\begingroup$ HINT: $999999999999=1000000000000-1$. $\endgroup$
    – barak manos
    Sep 4, 2016 at 7:19
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    $\begingroup$ Couldn't even start off $\endgroup$
    – S Aditya
    Sep 4, 2016 at 7:19

1 Answer 1

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HINT:

$999999999999^3=$

$(1000000000000-1)^3=$

$1000000000000^3-3\cdot1000000000000^2+3\cdot1000000000000-1$

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  • $\begingroup$ I don't think that would work. Consider $9=10-1$ . Let $f(x)$ denote sum of digits of $x$. $f(9) = 9$ and this is not equal to $f(10)-f(1)$ which is $0$. $\endgroup$
    – maverick
    Sep 4, 2016 at 7:28
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    $\begingroup$ @maverick: What's not to work here? I just gave a different way of representing $999999999999$, which allows you to calculate its 3rd power easily. And from there, calculating the digit sum is eminent (i.e., I did not suggest that you should calculate the digit sum of each term separately, and then combine them according to the sign of each term). $\endgroup$
    – barak manos
    Sep 4, 2016 at 7:33

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