Please help in this assignment:
Prove the existence of 11 consecutive compounded numbers.
japiga wrote:Prove the existence of 11 consecutive compounded numbers.
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!
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
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?