Divisors of 999...999 using division in a unusual way

Which integers can be multiplied by another integer so the product is in the form of 999....999?

To answer this question, will use some concepts, though I don't have a rigorous proof of these(I will update this note if I find something), a irreducible fraction:

1) Will only have terminating decimals if the denominator can be written as product of powers of two and five;

"Proof": n2a×5b=n2a×5b×10c10c=n×2ca×5cb10c\frac{n}{2^a\times5^b}=\frac{n}{2^a\times5^b}\times \frac{10^c}{10^c}=\frac{n\times2^{c-a}\times5^{c-b}}{10^c}, with c=max(a,b)c=max(a,b)

2) Will only have non-terminating decimals if the denominator is not a multiple of two and five, but is a multiple of another prime number;

3) Will have both terminating(non-repeating) and non-terminating decimals if the denominator is a multiple of 2 or 5, and is a multiple of another prime;

I will use the second case; It is possible to revert it back to a fraction:

0.(part that repeats)=part that repeatsa sequence of 9 with the same number of digits as the part that repeats0.\overline{\text{(part that repeats)}}=\frac{\text{part that repeats}}{\text{a sequence of 9 with the same number of digits as the part that repeats}}

So there is a way to transform a(irreducible) fraction whose denominator is not a multiple of two and five to a fraction whose denominator is a sequence of nines

1n=a99...999\frac{1}{n}=\frac{a}{99...999} if n is not a multiple of 2 nor 5(if n=1n=1(not a multiple of another prime) then it is trivial)

n=99...999an=\frac{99...999}{a}

n×a=99...999n\times a=99...999

With this, I can, for instance, find the number that, multiplied by 17, will return a sequence of nines:

117=0.0588235294117647=5882352941176479999999999999999\frac{1}{17}=0.\overline{0588235294117647}=\frac{588235294117647}{9999999999999999}

17=999999999999999958823529411764717=\frac{9999999999999999}{588235294117647}

17×588235294117647=999999999999999917\times588235294117647=9999999999999999

Note by Matheus Jahnke
2 years, 6 months ago

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Every integers except those divisible by 2 and 5, can be multiplied in such a way.

For any integer q q ,

Consider order of 10(modq)10 (mod q) ,

denote it k k .

Then 10k199999.....90(modq) 10^k \equiv 1 \Rightarrow 99999.....9 \equiv 0 (mod q)

But we can generalise the result, to For any integer qq ,

Let α,β \alpha, \beta be the maximum power of 2,5 dividing qq respectively.

Then there always exists a number 999...9990000.....000 999...9990000.....000 with number of zero =max(α,β) = max (\alpha , \beta ) .

A Brilliant Member - 2 years, 5 months ago

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Well, actually, you can generalize even more, when you consider another base system, here I used another way Euler Theorem to make this more rigorous: https://brilliant.org/discussions/thread/divisors-of-9999999-in-the-right-way/?ref_id=1339441

Matheus Jahnke - 2 years, 2 months ago

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