Challenge Level

$987654321$ | $=$ | $8$ | $\times$ | $123456789$ | $+$ | $9$ |

$98765432$ | $=$ | $8$ | $\times$ | $12345678$ | $+$ | $8$ |

$9876543$ | $=$ | $8$ | $\times$ | $1234567$ | $+$ | $7$ |

$987654$ | $=$ | $8$ | $\times$ | $123456$ | $+$ | $6$ |

... | ||||||

$9$ | $=$ | $8$ | $\times$ | $1$ | $+$ | $1$ |

Saul Foresta explained as follows why this pattern holds in the decimal system and in other number systems using bases other than base $10$:

I generalized the problem for any base $n$ and any number of digits $r$ where $r$ can be anywhere from $1$ to $(n - 1)$.

Then after rewriting both sides of the equality given in the problem using sigma notation I arrived at the following:

$$ {\sum_{k=1}^r (n-k)n^{r-k}} = {{(n-2)\sum_{k=1}^r kn^{r-k} + r}} $$

In each summation $k$ stands for the $k$th digit of the number we're dealing with, reading from left to right. For example, in the number $9876$, $k$ ranges from $1$-$4$, where $9$ is $k=1$, $8$ is $k=2$, and so on.

So all I need to do in order to prove that this pattern holds is show that the left side of this equality does indeed equal the right side. Taking the terms like $8 \times123456789$, that is

$$ (n-2)\sum_{k=1}^r kn^{r-k} $$

over to the left hand side, we will prove that this expression is equal to $r$.

[(n-1)n ^{r-1} + (n-2)n ^{r-2} + (n-3)n
^{r-3} + ... + (n-r)] - (n-2)[n ^{r-1} + 2n
^{r-2} + 3n ^{r-3} + ... + r] =

[n ^{r} - n ^{r-1} + n ^{r-1} - 2n
^{r-2} + n ^{r-2} - 3n ^{r-3} + ...+ n - r]
- [n ^{r} - 2n ^{r-1} + 2n ^{r-1} - 2.2n
^{r-2} + ... + (n-2)r]

The coefficient of $n^{r-k}$ on this left hand side is $[1-k] - [k+1-2k] = 0$ for

$ ( 1 \le k \le r-1) $

and the coefficient of $n^r$ is also $0$.The coefficient of $n^0$ is $[-r] - [-2(r)] = r$ and hence this expression is equal to $r$ as required.