The question is question no $6$ from 2016 IWYMIC olympiad individual round (It can be found here.)
Rephrased in my words , in says :
Let $A =\,\underbrace{ 6666...66666} _{2016 \, times}\,\, $ and $\,\,B =\,\underbrace{ 99999...99999} _{2016 \, times}$
Let $N = A\times B.$ Find the sum of digits of $N$.
My Approach :
Using simple observation , we can see that :
$$6\times 9 = 54 \quad \quad \rightarrow S(N) = 9$$ $$66\times 99 = 6534 \quad \quad \rightarrow S(N) = 18$$ $$666\times 999 = 665334 \quad \quad \rightarrow S(N) = 27$$ Similarly we can conclude that:
$$\,\underbrace{ 6666...66666} _{2016 \, times}\,\, \times\,\,\,\underbrace{ 99999...99999} _{2016 \, times}\quad \quad \rightarrow S(N) = 2016\times 9 = 18144$$
Which is indeed the correct answer. But how can we mathematically prove this is always correct ?
My Take :
$$\underbrace{ 6666...66666} _{2016 \, times} = 6\, (\underbrace{ 1111...11111} _{2016 \, times})$$
And
$$\underbrace{ 9999...99999} _{2016 \, times} = 9\, (\underbrace{ 1111...11111} _{2016 \, times})$$
Therefore
$$N = 54 \,\,(\underbrace{ 1111...11111} _{2016 \, times})^2$$
And then Find A pattern in $(\underbrace{ 1111...11111} _{2016 \, times})^2$, Although the obvious pattern is only for small values and I lost my way quickly.
Can anyone suggest me a hint to prove this?