Whole Number Dynamics V

Age 14 to 18
Article by Alan Beardon

Published 1998 Revised 2011

The last article Whole Number Dynamics IV was about the rule which takes $N$ to $N^{\prime}$ where $N$ was written in the form $N=10M+R$ and the remainder $R$ lies between $0$ and $9$ inclusive, and $N^{\prime}=10R-M$.

We suggested there that if we start with any whole number and then apply the rule $N \to N^{\prime}$ repeatedly, we will eventually reach $0$, or end in a cycle of four numbers. If you check many cases you will find this is so, but of course, this does not prove that it is so.

We also saw that for any whole number $K$ there are exactly ten numbers which go to $K$ in one step, and these are the numbers: $$-10K, 101-10K, 202-10K, 303-10K, ......, 808-10K, 909-10K $$ As a result of this, we then showed that if $N$ eventually reaches $0$ (after applying the rule sufficiently many times) then $N$ must be a multiple of $101$.

Now it is important to understand that the statement

(1) if $N$ is a multiple of $101$, then $N$ eventually reaches $0$,

is not the same as the statement:

(2) if $N$ eventually reaches $0$, then $N$ is a multiple of $101$.

A good way to see that these are different statements is to consider the following two (very similar) statements:

(3) if a whole number is a squared number, then it is positive or zero,


(4) if a whole number is positive or zero, then it is a squared number.

Obviously, (3) is true and (4) is false so that these two forms of a statement must be different. Returning to consider the statements (1) and (2), we recall that we have proved (2) in the last article; we shall now prove (1). Note that when combined together, (1) and (2) describe exactly those numbers which will eventually arrive at 0.

To show that (1) is true, let us consider any whole number $N$ that is a multiple of $101$; we want to show that this eventually reaches $0$. We can write $N= 101 P$, say, where $P$ is a whole number, and we can always write $P$ in the form $P =10 M+ R$, where $R$ is the remainder of $P$ when we divide $P$ by $10$. This means that: $$N = 101P = 101(10M + R) = 1010M + 100R + R = 10 X (101M + 10R) +R$$ so that $R$ is also the remainder of $N$.

Applying the rule to $N$, we see that $N$ goes to $N^{\prime}$, where $$N^{\prime} = 10 R - (101 M+ 10 R) = -101 M .$$ This shows, for example, that multiples of $101$ always go to multiples of $101$ (and we also know from last time that they can only come from multiples of $101$), so clearly the number $101$ plays an important role here!

Let us look at what we have just proved a little more closely.

In fact, we have just seen that if $N = 101(10 M+R)$ then $N^{\prime}= 101 \times(-M)$ , and roughly speaking,this says that when we apply the rule we just `drop' the units figure and change the sign). This shows, for example, that:

\begin{eqnarray} 101 \times 38792 & \to & 101 \times (-3879) \\ & \to & 101 \times 387 \\ & \to & 101 \times (-38) \\ & \to & 101 \times 3 = 303 \\ & \to & 0 \end{eqnarray}

A similar argument works with $38792$ replaced by any integer and this shows that if a whole number $N$ is a multiple of $101$, then it eventually reaches $0$.

Let us look again at the problem posed at the end of the last article. We asked there whether or not the number $123456$ ends up at $0$ or in a cycle of four numbers? Using a calculator, we see that $123456$ is not a multiple of $101$ so that (by what we have just shown) it cannot end up at $0$.

What about the number $12345678987654321$? This number is too big to put on a calculator so we need to find another approach for it would clearly be a very long task indeed to keep on applying the rule to this number! How do we handle this number?

We want to decide whether or not the number $12345678987654321$ is a multiple of $101$. Of course if a whole number $X$ is a multiple of $101$ then the number $12345678987654321 - X $ is also a multiple of $101$ and conversely, and using this we see that it is enough to subtract multiples of $101$ from $12345678987654321$ and then check whether or not the answer is a multiple of $101$. Better still, if P is any whole number then:

$$10000P = 9999P + P = (99 \times101)P + P$$ so that $10000 P$ is a multiple of $101$ plus $P$.

Thus, $$12345678987654321 = (1234567898765 x 10000) + 4321 = 1234567898765 + 101Q + 4321$$ for some whole number $Q$. Applying this again, we get $$1234567898765 = (123456789 x 10000) + 8765 = 123456789 + 101S + 8765$$ for some whole number $S$, and again, $$123456789 = (12345 x 10000) + 6789 = 12345 + 101T + 6789$$ for some whole number $T$.

Putting all these together, we find that the two numbers $12345678987654321$ and $( 4321 + 8765 + 6789 + 12345)$ differ by a multiple of $101$. It is enough, therefore, to check whether $(4321+8765+6789+12345)$ is, or is not, a multiple of $101$, and we have now reduced the problem to one that we can do on a calculator.

You can now answer the question : does $12345678987654321$ eventually reach $0$ or not ?

A final question : does $8765432123456789$ eventually reach $0$ or not?