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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 \(b=3\):




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:


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

Multiplying both the two equations and using telescopic sums:






Remeber that \(s=9t\)



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;


Note that \(c\) meets the conditions that allow me do this:


Multiplying by k the first equation:


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 \(a-1\)

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





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



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:

\(b'L=b, a'L=a\)


Doing a substitution



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 \(a\):


\(p_j\) 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


Multiplying k on every side of the other equation:

\(bdk=a^\gamma ck\)

Substituting \(ck\):

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

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


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
4 months, 3 weeks ago

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