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:

${\displaystyle -10K,101-10K,202-10K,303-10K,......,808-10K,909-10K}$

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

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 $N=101P$, say, where $P$ is a whole number, and we can always write $P$ in the form $P=10M+R$, where $R$ is the remainder of $P$ when we divide $P$ by $10$. This means that:

${\displaystyle N=101P=101(10M+R)=1010M+100R+R=10X(101M+10R)+R}$

so that $R$ is also the remainder of $N$.

Applying the rule to $N$, we see that $N$ goes to $N\text{'}$, where

${\displaystyle N\text{'}=10R-(101M+10R)=-101M.}$

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 $N=101(10M+R)$ then

101 x 38792		101 x (-3879)
		101 x 387
		101 x (-38)
		101 x 3 = 303
		0

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:

Thus,

${\displaystyle 12345678987654321=(1234567898765x10000)+4321=1234567898765+101Q+4321}$

for some whole number $Q$. Applying this again, we get

${\displaystyle 1234567898765=(123456789x10000)+8765=123456789+101S+8765}$

for some whole number $S$, and again,

${\displaystyle 123456789=(12345x10000)+6789=12345+101T+6789}$

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?

Whole Number Dynamics I
Whole Number Dynamics II
Whole Number Dynamics III
Whole Number Dynamics IV .