### Code to Zero

Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.

### Binary Squares

If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?

We are used to writing numbers in base ten, using 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Eg. 75 means 7 tens and five units. This article explains how numbers can be written in any number base.

# Basic Rhythms

##### Stage: 5 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.