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?

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?

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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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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japiga wrote: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
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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japiga wrote: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...

- Martingale
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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stapel_eliz wrote:japiga wrote: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

such that

Though I could be wrong

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").

And, to add, solution for "n" should be any integer number, but we are looking for the least one ("the smallest one").

- Martingale
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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japiga wrote: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

if n=k-2 we get

that should eliminate a lot of possibilities.

or similarly...

if k=n+2

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
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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japiga wrote: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.