Find 2000 consecutive composite numbers.
(Of course you can't do this by trial and error alone)
For those who do not know what a composite number is, it is any integer greater than 1 that is not prime. (4, 6, 8, 9, ...)
First I would define what is meant by (N)! (pronounced as N - Factorial), where N is positive integer.
(N)! is defined(mathematically) as the product of all positive natural numbers from 1 to N and is written as:
(N)! = (N).(N-1).(N-2).(N-3)....4.3.2.1 or,
(N)! = 1.2.3.4.....(N-3).(N-2).(N-1).(N)
So, we see that all the numbers from 1 to N are factors of (N)!. So, for any positive integer, say M, lying between 1 and N is a factor of
[(N)! + M]. Using this, we can say that 2 is a factor of (N)! + 2, 3 is a factor of (N)! + 3, 4 is a factor of (N)! + 4,....., and so on, provided the numbers 2,3,4,....all are less than N. Proceeding in a similar manner, we can write N is a factor of (N)! + N. Thus we see that in this case we are getting (N-1) consecutive composite numbers.
So, to find 2000 consecutive composite numbers, put N = 2001 and we get the resulting numbers as [(2001)! + 2],[(2001)! + 3],[(2001)! + 4],[(2001)! + 5],....,[(2001)! + 2000],[(2001)! +2001], which are nothing but 2000 consecutive composite numbers.
My Solution !!!!
