## 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
japiga
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### 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Find the least natural number n, so that anyone bellow divisions may not be shorten:

7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

I found that such number n = 17, but how to prove it?

stapel_eliz
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What is meant by "shortening" a division?

japiga
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Yes, sorry. "shorten" means "divided not to be further divided". Or generally, not to be divided at all. Is it now more clear?

stapel_eliz
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Yes, sorry. "shorten" means "divided not to be further divided". Or generally, not to be divided at all. Is it now more clear?
Sorry; no.

Martingale
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Find the least natural number n, so that anyone bellow divisions may not be shorten:

7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

I found that such number n = 17, but how to prove it?
If I understand you then 17 doesn't work...

$\frac{19}{17+21}=\frac{19}{38}=\frac{1}{2}$

Martingale
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Yes, sorry. "shorten" means "divided not to be further divided". Or generally, not to be divided at all. Is it now more clear?
Sorry; no.

I believe that japiga wants the smallest $n\in\mathbb{N}$

such that

$\gcd(k,n+(k+2))=1$

$\forall k\in\{7,8,\ldots,31\}$

Though I could be wrong

japiga
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Yes, I am wrong, sorry. Thanks Martingale you are correct, but What is solution? What is the least number for n, such as that each division is not to be further divided for all line: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33) ?
And, to add, solution for "n" should be any integer number, but we are looking for the least one ("the smallest one").

Martingale
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Yes, I am wrong, sorry. Thanks Martingale you are correct, but What is solution? What is the least number for n, such as that each division is not to be further divided for all line: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33) ?
And, to add, solution for "n" should be any integer number, but we are looking for the least one ("the smallest one").
Note:

for $\frac{k}{n+k+2}$

if n=k-2 we get

$\frac{k}{n+k+2}=\frac{k}{k-2+k+2}=\frac{k}{2k}=\frac{1}{2}$

that should eliminate a lot of possibilities.

or similarly...

if k=n+2

$\frac{k}{n+k+2}=\frac{n+2}{n+n+2+2}=\frac{n+2}{2(n+2)}=\frac{1}{2}$

japiga
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Thanks! It make sense if final result is 1/2, but I am little confused with original assignment: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33). Why he have line only with those numbers, not for all, like 32/(n + 34), 33/(n+35).... It seems that this rule is valid for any k/(n+k+2). Do you think so?

Martingale
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Thanks! It make sense if final result is 1/2, but I am little confused with original assignment: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33). Why he have line only with those numbers, not for all, like 32/(n + 34), 33/(n+35).... It seems that this rule is valid for any k/(n+k+2). Do you think so?

if you were allowed to use all the numbers (ie get stuff like 32/(n + 34), 33/(n+35)...) then there would be no value of n that worked (by what I showed), but here we are only allowed to go to 31/(n+33). This restricts things ...making it so there is a solution.