Whole Number Dynamics IV

Age 14 to 18
Article by Alan Beardon

Published 1997 Revised 2011

This is the fourth in a series of five articles and in this article we look at a dynamic system which ends in zero, or a cycle of four numbers, and investigate why this is the case.

Every whole number has a remainder $R$ when divided by $10$, for example, $53$ has remainder $3$, $67$ has remainder $7$ and so on.

Of course a whole number can be negative and we have to agree what we mean by the remainder in this case; for example, what is the remainder when $-12$ is divided by $10$? There are two possible answers here and both are equally valid:

First we could say that $-12 = (-1 \times10) + (-2)$ so the remainder is $-2$ or second, we could say that $-12 = (-2 \times 10) + (8)$ so that the remainder is $8$.

In this article the remainder will always be positive (or, of course, zero) so that, for example, the remainder of $-137$ is $3$, and the remainder of $-58$ is $2$. The remainder of $120$, and of $-120$, is $0$.

The general rule, then, is that any whole number can be written as a multiple of $10$ plus the remainder, where the remainder is always one of the numbers: $0$, $1$, $2$, $\ldots$, $9$.

Now let us consider the following rule. Given a whole number $N$, let us write $N$ as a multiple of $10$, say $M \times 10$, plus a remainder $R$; where (as always) $0 \leq R \leq 9$.

Starting with $N$ let's produce a new whole number $N\prime$ which is given by $N\prime = 10 R - M$

Be careful here, for this is not saying that $-M$ is the remainder of $N\prime$ (indeed, it cannot be if it is negative, and even if it is positive it might still be greater than $9$). An example will make this clearer.

Starting with $N =58$, we have $M =5$ and $R =8$ so that $N\prime= (8 \times 10) - 5 = 75$.

As another example, if $N = -123$ then $M = -13$ and $R =7$ so that in this case, $N\prime = (7 \times 10) - (-13) = 83$

You should now start with the whole number $68$ and keep on applying the rule: $N \to N\prime$. For example, starting with $68$ (which is $(6 \times 10) + 8$) and applying the rule gives $74$; now apply the rule to $74$, and so on until you notice something special.

When you have done this, repeat the whole process again, but this time starting with $28$; what do you notice now? Again, repeat the whole process, this time starting with $49$; what do you notice now?

Now start with $3164$.

What is (1) different, and (2) the same, about carrying out the process with this starting point? Now choose your own starting point, and keep on repeating the process until you again notice something special; then try again several times, each with different starting points. I suggest that you keep a careful record of your results and then try and write down what you have discovered. Ask your teacher, or a friend (who likes arithmetic!) to check your results for you.


You should have discovered by now that whatever number you start with, you will either end up at the number zero (and stay there thereafter), or find yourself circulating repeatedly around a cycle of four numbers. For example, one such cycle is:

$$13 \to (3 \times 10) - 1 = 29 \to (9 \times 10) - 2 = 88 \to (8 \times 10) - 8 = 72 \to (2 \times10) - 7 = 13$$

We shall now begin to try to analyse what is happening here.

First we notice that given any whole number $K$ there are exactly $10$ whole numbers which go to $K$ when we apply the rule; for example, the numbers $-530$, $-429$, $-328$, $-227$, $-126$, $-25$, $76$, $177$, $278$, $379$ are all the numbers that map to $53$.

To see this, we start with any whole number $N$, which we may write as $N = 10M+R$, and note that this ends up at $K$ if $10R - M = K$ or, equivalently, if $M= 10R- K$.

This means to say that if $N$ goes to $K$, then $$N = 10 \times M + R = 10(10R - K) = 101 R - 10 K$$ As $R$ can take any value between $0$ and $9$ inclusive (and only these values), the numbers which go to $K$ when we apply the rule once are precisely the numbers: \begin{eqnarray} -10 K, 101 - 10 K, 202 - 10 K, 303 - 10 K, \ldots, 909 - 10K \end{eqnarray} Let us now look at a special case of this general result.

Obviously $0$ goes to $0$ when we apply the rule (for if $N =0$ then $M =0$ and $R =0$).

What else goes to $0$ in one step?

According to the list (1) above, there are exactly ten numbers which go to $0$ in one step, and putting $K =0$ in the list, we see that these are: $$0, 101, 202, 303, \ldots,909$$ Notice that these are all multiples of $101$. Pick one of these, say $303$ and now ask what goes to $303$ in one step. Of course, if $N$ goes to $303$ in one step, then it goes to $0$ in two steps. Again using the list above, the set of numbers that go to $303$ in one step is $$-10 \times 303, 101-(10 \times 303), 202-(10 \times 303), \ldots,808-(10 \times 303), 909-(10 \times 303)$$ and these too are all multiples of $101$.

This argument shows much more than this. It is clear from (1) that if $K$ is a multiple of $101$, then all ten numbers that go to $K$ in one step are also multiples of $101$. As $0$ is a multiple of $101 (0 - 0 \times 101)$, we see that every number that goes to $0$ in any number of applications of the rule must also be a multiple of $101$.

Here is a question: does the number $123456$ end up at $0$ or in a cycle of four numbers? Of course you can carry out the process many times, but is obviously much less work to see whether or not $123456$ is a multiple of $101$. Try it. What about the number $987654321$?

We have seen that every number that goes (after a number of steps) to $0$ is a multiple of $101$. This is not the same thing as saying that every multiple of $101$ does go to $0$ (after a certain number of steps) but this is true and in the next article we shall show this is so.