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

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

Postby japiga on Thu Sep 23, 2010 8:28 am

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?
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Postby stapel_eliz on Thu Sep 23, 2010 10:23 am

What is meant by "shortening" a division? :confused:
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Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Postby japiga on Thu Sep 23, 2010 11:44 am

Yes, sorry. "shorten" means "divided not to be further divided". Or generally, not to be divided at all. Is it now more clear?
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Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Postby stapel_eliz on Thu Sep 23, 2010 7:25 pm

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

Postby Martingale on Thu Sep 23, 2010 9:38 pm

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...

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

Postby Martingale on Thu Sep 23, 2010 9:54 pm

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. :oops:



I believe that japiga wants the smallest

such that





Though I could be wrong :wink:
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Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Postby japiga on Fri Sep 24, 2010 7:57 am

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

Postby Martingale on Fri Sep 24, 2010 1:26 pm

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

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

Postby japiga on Fri Sep 24, 2010 2:14 pm

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

Postby Martingale on Fri Sep 24, 2010 2:18 pm

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.
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