Waste less time on Facebook — follow Brilliant.
×

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

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 \( = max (\alpha , \beta ) \). L Km · 5 months, 1 week ago

Log in to reply

@L Km 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 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...