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# Divisors of 999...9999 in the "right way"

I studied something after making that crazy thing. This time I will use the Euler's theorem which states that, if two positive integers $$a$$ and $$b$$ are relatively prime to themselves, then:

$$a^{\phi(b)}\equiv1\bmod {b}$$

Where $$\phi(b)$$ is the Euler's totient function, which returns the size of the set $$\mathbb{S}$$, where $$\mathbb{S}=\{x\in\mathbb{N}|\text{gcd}(b, x)=1\text{ and }x<b\}$$

This is true for $$a=10$$ and $$b$$ is a number that it is not a multiple of 2, nor 5. Therefore:

$$b\mid 10^{\phi(b)}-1=999....999$$

I should alert anyone reading this that $$\phi(b)$$ does not need to be the smallest positive integer that will make this work, let $$b=3$$:

$$3|10^{\phi(3)}-1=10^2-1=99$$

But:

$$3|9=10^1-1$$

However, we can expand this concept to sequences of other numbers; Writing this in another way, calling $$\phi(b)=n$$ because I already showed that a "$$n$$" exists under what conditions:

$$kb=10^n-1, k\in\mathbb{N}^*$$

I will multiply this by $$s=1+10^{n}+10^{2n}+...+10^{8n}=\sum^8_{k=0}10^{kn}$$, notice that s is a multiple of 9; This can be proved this way:

$$\sum^8_{k=0}10^{kn}\equiv\sum^8_{k=0}(10-9)^{kn}\equiv\sum^8_{k=0}1^{kn}\equiv\sum^8_{k=0}1\equiv9\equiv1\bmod{9}$$

Therefore $$s=9t, t\in\mathbb{N}^*$$

Multiplying both the two equations and using telescopic sums:

$$skb=(10^n-1)\times\sum^8_{k=0}10^{kn}$$

$$skb=10^n\times\sum^8_{k=0}10^{kn}-\sum^8_{k=0}10^{kn}$$

$$skb=\sum^9_{k=1}10^{kn}-\sum^8_{k=0}10^{kn}$$

$$skb=\sum^9_{k=1}10^{kn}-\sum^8_{k=0}10^{kn}$$

$$skb=10^{9n}-1=999...999$$

Remeber that $$s=9t$$

$$9tkb=999...999$$

$$tkb=111...111$$

Since $$t, k \in\mathbb{N}^*$$, $$tk\in\mathbb{N}^*$$; Therefore $$(\forall b\in\{x\in\mathbb{N})(2\nmid b \wedge 5\nmid b\}\exists f,n\in\mathbb{N}^* | fb=\sum^{n}_{k=0}10^k)$$

Just to make this complete, if $$2\mid b$$ or $$5\mid b$$, let $$b=2^\alpha5^\beta c, 2\nmid c, 5\nmid c, c\in\mathbb{N}^*$$, $$\gamma=\max(\alpha,\beta)$$

Multiply everything by $$z=10^\gamma2^{-\alpha}5^{-\beta}$$, note that this number is natural;

$$bz=10^{\gamma}c$$

Note that $$c$$ meets the conditions that allow me do this:

$$kc=\sum^n_{k=0}10^k$$

Multiplying by k the first equation:

$$bzk=10^\gamma\sum^n_{k=0}10^k=\sum^n_{k=0}10^{k+\gamma}=\sum^{n+\gamma}_{k=\gamma}10^k$$

This means I can generalize my result:

$$\forall b\in\mathbb{N}^*\exists l, m, n\in\mathbb{N}|bl=\sum^{m}_{k=n}10^k$$

I can generalize even more, using another base system:

Remember the Euler's theorem if a and b are co-prime :

$$a^{\phi(b)}\equiv1\bmod {b}$$

Rewriting, with $$\phi(b)=n$$:

$$bk=a^n-1= \overline{(a-1)(a-1)(a-1)....(a-1)(a-1)(a-1)}$$

Imagine that each (a-1) is a digit;

I will multiply everything by $$c=\sum^{a-2}_{k=0}a^{nk}$$, but first I will prove that this a multiple of $$a-1$$

$$\sum^{a-2}_{k=0}a^{nk}\equiv\sum^{a-2}_{k=0}(a-(a-1))^{nk}\equiv\sum^{a-2}_{k=0}1^{nk}\equiv\sum^{a-2}_{k=0}1\equiv a-1\equiv 0\bmod {a-1}$$

$$bck=(a^n-1)\sum^{a-2}_{k=0}a^{nk}$$

$$bck=a^n\sum^{a-2}_{k=0}a^{nk}-\sum^{a-2}_{k=0}a^{nk}$$

$$bck=\sum^{a-1}_{k=1}a^{nk}-\sum^{a-2}_{k=0}a^{nk}$$

$$bck=a^{n(a-1)}-1=\overline{(a-1)(a-1)(a-1)....(a-1)(a-1)}$$

As I proved before $$c=d(a-1), d\in\mathbb{N}^*$$

$$(a-1)bdk=\overline{(a-1)(a-1)(a-1)....(a-1)(a-1)}$$

$$bdk=(111...111)_a$$

Since $$dk\in\mathbb{N}^*$$, I can say:

$$(\forall b, a\in\mathbb{N}^*)(\text{gcd}(a,b)=1\Leftrightarrow\exists n, f|bf=\sum^n_{k=0} a^k)$$

The other way can be proven by observing:

$$bf\equiv1\mod a$$

This means $$bf-ka=1, k\in\mathbb{N}$$, with this in mind, I will prove that $$\text{gcd}(a,b)=L$$ must be 1:

$$b'L=b, a'L=a$$

$$a',b'\in\mathbb{N}$$

Doing a substitution

$$b'Lf-a'Lk=1$$

$$L(b'f-a'k)=1$$

As both of the factors are positive integers, both of them are 1

Now I will study the case where $$\text{gcd}(a,b)\neq1$$

I will consider the prime factorization of $$a$$:

$$a=\prod^z_{j=1}p_j^{a_j}$$

$$p_j$$ is the $$j$$-th prime, $$a_j$$ is its exponent, $$z$$ is the largest number that makes $$a_z\neq0$$

I will make $$b=c\prod^n_{j=1}p_j^{b_j}$$, this time $$(\forall j)(a_j\neq0\Leftrightarrow p_j\nmid c)\wedge(a_j=0\Rightarrow b_j=0)$$

Let $$\gamma=\max(\{x\in\mathbb{N}^*|x=\left\lceil\frac{b_j}{a_j}\right\rceil j\in\mathbb{N}\wedge a_j\neq0\})$$

I will multiply everything by $$d=a^\gamma\prod^n_{j=1}p_j^{-b_j}$$, notice that this number is a natural number

$$bd=a^\gamma c$$

Since $$\text{gcd}(a,c)=1$$, I can say that

$$ck=\sum^n_{j=0}a^j$$

Multiplying k on every side of the other equation:

$$bdk=a^\gamma ck$$

Substituting $$ck$$:

$$bdk=a^\gamma \sum^n_{j=0}a^j$$

$$bdk=a^\gamma \sum^n_{j=0}a^{j+\gamma}$$

$$bdk=\sum^{n+\gamma}_{j=\gamma}a^j$$

Since $$dk\in\mathbb{N}^*$$ I can say that(finally):

$$(\forall a,b\in\mathbb{N}^*)(\exists k,m,n\in\mathbb{N}^*)(bk=\sum^m_{j=n}a^j)$$

Note by Matheus Jahnke
9 months, 4 weeks ago

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