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):

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

## Comments

There are no comments in this discussion.