11 consecutive compounded numbers

[japiga]

Please help in this assignment:
Prove the existence of 11 consecutive compounded numbers.

[stapel_eliz]

Prove the existence of 11 consecutive compounded numbers.

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

Thank you!

[japiga]

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]

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!

[japiga]

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]

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]

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]

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]

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]

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