Find the sum of digits in decimal form of the given number

Question

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

(There are 12 nines)

HINT: $999999999999=1000000000000-1$. – barak manos Sep 04 '16 at 07:19 — barak manos, Sep 04 '16 at 07:19

score 3 · Answer 1 · answered Sep 04 '16 at 07:22

3

HINT:

$999999999999^3=$

$(1000000000000-1)^3=$

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

answered Sep 04 '16 at 07:22

barak manos

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 $0$. – maverick Sep 04 '16 at 07:28
2

@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). – barak manos Sep 04 '16 at 07:33

1 Answers1