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# 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": $$\frac{n}{2^a\times5^b}=\frac{n}{2^a\times5^b}\times \frac{10^c}{10^c}=\frac{n\times2^{c-a}\times5^{c-b}}{10^c}$$, with $$c=max(a,b)$$

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:

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

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

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

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

$$n\times a=99...999$$

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

$$\frac{1}{17}=0.\overline{0588235294117647}=\frac{588235294117647}{9999999999999999}$$

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

$$17\times588235294117647=9999999999999999$$

Note by Matheus Jahnke
6 months, 1 week ago

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

For any integer $$q$$ ,

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

denote it $$k$$ .

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

But we can generalise the result, to For any integer $$q$$,

Let $$\alpha, \beta$$ be the maximum power of 2,5 dividing $$q$$ respectively.

Then there always exists a number $$999...9990000.....000$$ with number of zero $$= max (\alpha , \beta )$$. · 5 months, 1 week ago