11 consecutive compounded numbers  TOPIC_SOLVED

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11 consecutive compounded numbers

Postby japiga on Sat Oct 09, 2010 2:17 pm

Please help in this assignment:
Prove the existence of 11 consecutive compounded numbers.
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Postby stapel_eliz on Sat Oct 09, 2010 10:10 pm

japiga wrote:Prove the existence of 11 consecutive compounded numbers.

What is a "compounded" number? How does your book define this term?

Thank you! :wink:
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Re: 11 consecutive compounded numbers

Postby japiga on Mon Oct 11, 2010 1:32 pm

compunde number is number which is possible to bi divided as natural number, oppoite from the simople number such as 1, 3, 5, 7, 11.
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Postby stapel_eliz on Mon Oct 11, 2010 8:11 pm

Ah. So you mean "compound" versus "prime". ("Compounded" and "simple" generally refer to interest earned on investments.)

What results do you have that you can apply to this proof? What have you tried so far?

Please be complete. Thank you! :wink:
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Re: 11 consecutive compounded numbers

Postby japiga on Tue Oct 12, 2010 12:55 pm

Possible I have solution, but I would like to be confirmed from others.
(n+2)!+2, (n+3)!+3, (n+4)!+4, (n+5)!+5, (n+6)!+6, (n+7)!+7, (n+8)!+8, (n+9)!+9, (n+10)!+10, (n+11)!+11 and (n+12)!+12 , since each number could be extracted by a number in the brackets and divided what it should be a prove that number is not prime, but opposite of the prime (I coined it as compounded number). Do you think it is true?
For n=1
1*2*3+2=2*(1*3+1)=2*4, the result is compounded number since it is divided by 2;
1*2*3*4+3=3*(1*2*3*4+1)=3*25, the result is compounded number since it is divided by 3;
1*2*3*4*5+4=4*(1*2*3*4*5+1)=4*81, the result is compounded number since it is divided by 4, ……
and the last one
1*2*3*4*5*6*7*8*9*10*11*12*13+12=12*(1*2*3*4*5*6*7*8*9*10*11*13+12)= some number but divided by 12.
This should be proved!
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Re: 11 consecutive compounded numbers

Postby Martingale on Tue Oct 12, 2010 12:58 pm

japiga wrote:Possible I have solution, but I would like to be confirmed from others.
(n+2)!+2, (n+3)!+3, (n+4)!+4, (n+5)!+5, (n+6)!+6, (n+7)!+7, (n+8)!+8, (n+9)!+9, (n+10)!+10, (n+11)!+11 and (n+12)!+12 , since each number could be extracted by a number in the brackets and divided what it should be a prove that number is not prime, but opposite of the prime (I coined it as compounded number). Do you think it is true?
For n=1
1*2*3+2=2*(1*3+1)=2*4, the result is compounded number since it is divided by 2;
1*2*3*4+3=3*(1*2*3*4+1)=3*25, the result is compounded number since it is divided by 3;
1*2*3*4*5+4=4*(1*2*3*4*5+1)=4*81, the result is compounded number since it is divided by 4, ……
and the last one
1*2*3*4*5*6*7*8*9*10*11*12*13+12=12*(1*2*3*4*5*6*7*8*9*10*11*13+12)= some number but divided by 12.
This should be proved!


Are those numbers consecutive?
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Re: 11 consecutive compounded numbers

Postby japiga on Tue Oct 12, 2010 2:31 pm

Yes, they are consecutive numbers, it is provided by the second part of the "compounded" number, e.g. (n+2)!+2 by ..+2, ..+3, ..+4, ..+5,.... ..+12.
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Re: 11 consecutive compounded numbers

Postby Martingale on Tue Oct 12, 2010 3:28 pm

japiga wrote:Yes, they are consecutive numbers, it is provided by the second part of the "compounded" number, e.g. (n+2)!+2 by ..+2, ..+3, ..+4, ..+5,.... ..+12.



japiga wrote:(n+2)!+2, (n+3)!+3, (n+4)!+4, (n+5)!+5, (n+6)!+6, (n+7)!+7, (n+8)!+8, (n+9)!+9, (n+10)!+10, (n+11)!+11 and (n+12)!+12



Are not consecutive numbers.
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Re: 11 consecutive compounded numbers

Postby japiga on Wed Oct 13, 2010 8:06 am

Or, possible this should be solution:

12!+2, 12!+3, 12!+4, 12!+5, 12!+6, 12!+7, 12!+8, 12!+9, 12!+10, 12!+11 and 12!+12 , (total 11 consecutive numbers) since each number could be extracted by a number behind the “+” (e.g. +2, +3, +4….+12) and divided with the same number 2, 3, 4, 5,…. 12 respectively what it should be a prove that number is not prime, but opposite of the prime (I coined it as compounded number). Do you think that now it is true?
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Re: 11 consecutive compounded numbers  TOPIC_SOLVED

Postby Martingale on Wed Oct 13, 2010 10:40 am

japiga wrote:Or, possible this should be solution:

12!+2, 12!+3, 12!+4, 12!+5, 12!+6, 12!+7, 12!+8, 12!+9, 12!+10, 12!+11 and 12!+12 , (total 11 consecutive numbers) since each number could be extracted by a number behind the “+” (e.g. +2, +3, +4….+12) and divided with the same number 2, 3, 4, 5,…. 12 respectively what it should be a prove that number is not prime, but opposite of the prime (I coined it as compounded number). Do you think that now it is true?


yes
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