Please help in this assignment:

Prove the existence of 11 consecutive compounded numbers.

Please help in this assignment:

Prove the existence of 11 consecutive compounded numbers.

Prove the existence of 11 consecutive compounded numbers.

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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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
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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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!

(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
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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Are those numbers consecutive?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!

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
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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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.

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?

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**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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yesOr, 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?