## 11 consecutive compounded numbers

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
japiga
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### 11 consecutive compounded numbers

Prove the existence of 11 consecutive compounded numbers.

stapel_eliz
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Prove the existence of 11 consecutive compounded numbers.
What is a "compounded" number? How does your book define this term?

Thank you!

japiga
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### Re: 11 consecutive compounded numbers

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.

stapel_eliz
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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?

japiga
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### Re: 11 consecutive compounded numbers

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!

Martingale
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### Re: 11 consecutive compounded numbers

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?

japiga
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### Re: 11 consecutive compounded numbers

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.

Martingale
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### Re: 11 consecutive compounded numbers

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.
(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.

japiga
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### Re: 11 consecutive compounded numbers

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?

Martingale
Posts: 333
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### Re: 11 consecutive compounded numbers

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