# I Interesting property of idempotent 10-adic number

1. Jul 6, 2016

### Happiness

We build a special number from $5$, by squaring it, appending the next digit of the square to it and repeating the steps.

$5^2=25.$ The next digit is 2, which is added to 5 to give 25.

$25^2=625.$ The next digit is 6, which is added to 25 to give 625.

$625^2=390625.$ The next digit is 0, which is added to 625 to give 0625.

$0625^2=390625.$ The next digit is 9, which is added to 0625 to give 90625.

$90625^2=8212890625.$ The next digit is 8, which is added to 90625 to give 890625.

We end up with the special number $F=\,...\,4106619977392256259918212890625.$

Why is it that the number in every step shares the same digits as its square? In other words, prove the existence of this number.

(The article mentioned that the digit 6 can be used. But I don't think so because $36\times36=1296$, which doesn't end with 36.)

Last edited: Jul 6, 2016
2. Jul 6, 2016

### micromass

Staff Emeritus
The number seems to be the limit $L=\lim_{n\rightarrow +\infty} 5^{2^n}$. We see that
$$L^2 = \lim_{n\rightarrow +\infty} (5^{2^n})^2 = \lim_{n\rightarrow +\infty} 5^{2^{n+1}} = L$$
So $L^2 = L$ meaning that your number is $L=0$ or $L = 1$.

We see that
$$|5^{2^n}|_{10} = 1~\text{and}~|5^{2^n} - 1|_{10} = 1$$
so the sequence doesn't converge. It does converge in the $5$-adic numbers:

$$|5^{2^n}|_5 = 2^{-n}$$
So your number equals $0$ in the $5$-adic numbers.

EDIT: the number does exist in the $10$-adic numbers. The $10$-adics are not an integral domain so $L^2 = L~\Rightarrow~L\in \{0,1\}$ is not valid. The number doesn't seem to be a rational number though

3. Jul 6, 2016

### Happiness

I am not familiar with p-adic numbers. Could this be proved without using p-adic numbers?

Last edited: Jul 6, 2016
4. Jul 6, 2016

### Happiness

$625^2=390625$ but $F$ ends with $890625$. So I think $F$ may not be the limit $L$.

5. Jul 6, 2016

### micromass

Staff Emeritus
Yes, you're right. I misunderstood the OP.

6. Jul 6, 2016

### micromass

Staff Emeritus
OK, so let $a_n$ be the number formed by the first $n$ digits of $F$. We know by construction that the $nth$ digit of $a_n$ is the $n$th digit of $a_{n-1}^2$. We prove by induction that the first $n$ digits of $a_n^2$ equal $a_n$.

We can obviously write $a_n = x 10^n + a_{n-1}$. Squaring gives us
$$a_n^2 = x^2 10^{2n} + 2x10^n a_{n-1} + a_{n-1}^2$$
We are interested in the first $n$ digits of this number. The number $x^2 10^{2n}$ has a zero on its first $2n$ digits, so it plays no role. The first $n-1$ digits of $2x10^n a_{n-1}$ are zero. The $n$th digit of this number is the first digit of $2xa_{n-1}$. But since the first digit of $a_{n-1}$ is $5$, we get that this first digit is the first digit of $2\cdot 5\cdot x$, that is: the first digit is $0$.

So the first $n$ digits of $a_n^2$ are the first $n$ digits of $a_{n-1}^2$, like we wanted.

7. Jul 6, 2016

### Happiness

$a_2=25=2(10)+5=2(10^1)+a_1$.

So $a_n = x 10^{n-1} + a_{n-1}$.

Right?

8. Jul 6, 2016

### micromass

Staff Emeritus
Right sorry, I should have proof read it. But the proof still goes through modulo these notational errors.

9. Jul 6, 2016

### micromass

Staff Emeritus
Using the same proof, I discovered a similar number:

$$2^5 = \mathbf{32},~32^5 = 33554\mathbf{432},~432^5 = 1504591950\mathbf{6432}, ...$$

10. Jul 6, 2016

### Staff: Mentor

The proof needs a slight modification, and relies on the ending number being 5 to make 2x*5 a multiple of 10.
That also explains why it works with 5 but not with 6.

The fifth power, starting with 2, also works because then 5x*2 is a multiple of 10. Apart from higher powers (^10, ^15, ... with 2, or ^4, ^6, ... with 5) that should be all we have in the decimal system. Other bases lead to similar results. Start with a divisor in this system, then use a power such that starting number * power is a multiple of the base.