# Prove that the decimal representation consists of exactly $k(m-k) + 1$ digits

For a positive integer $n$, let $6^{(n)}$ be the natural number whose decimal representation consists of $n$ digits $6$. Let us define, for all natural numbers $m$, $k$ with $1 \leq k \leq m$ $$\left[\begin{array}{ccc}m\\ k\end{array}\right] =\frac{ 6^{(m)} \cdot 6^{(m-1)}\cdots 6^{(m-k+1)}}{6^{(1)} \cdot 6^{(2)}\cdots 6^{(k)}} .$$ Prove that for all $m, k$, $\left[\begin{array}{ccc}m\\ k\end{array}\right]$ is a natural number whose decimal representation consists of exactly $k(m-k) + 1$ digits.

We deduce that $$\left[\begin{array}{ccc}m\\ k\end{array}\right] = \prod_{i=0}^{k-1}\dfrac {10^{m-i}-1}{10^{i+1}-1}.$$ Where do I go from here?

• If I get the notation, $6^{(n)}=6 \cdot 10^{n-1},$ so maybe plugging that in will give a start, since we can apply the sum of consecutive integers formula. – coffeemath Jun 24 '16 at 20:11
• @coffeemath It should be $6^{(n)} = 6 \cdot 111\ldots 1$ where there are $n$ ones. – John Ryan Jun 24 '16 at 20:13
• A notation for the repunits that I have seen a few times is $R_n$. So your numbers are $6.R_n$ and in fact the $6$s are purely distraction; they all cancel out. Note that $R_n = (10^{n}-1)/(10-1)$ – Joffan Jun 24 '16 at 20:29
• John Ryan: So $6^{(2)}=6 \cdot 11,=66,$ not $6 \cdot 10^{2-1}=60,$ the way I looked at it. I guess Joffan's comment is best concerning what they are in terms of repunits. (my interpretation was off, sorry). – coffeemath Jun 24 '16 at 20:44