Ok, my question is: prove that all of the following numbers are composite: 1000! +2, 1000! +3, 1000+4, ....., 1000! + 1002
Now, I assume I have to start with proving that 1000! is composite or not and then prove the other numbers being added to the factorial?
Also, our class definition of composite is: A positive integer a is called composite provided there is an integer b such that 1< b < a and b|a.
Help is much appreciated!
Not sure of a formal way to prove this, but I'll try and get my point across in a clear(ish) way.
The number can be written as .
So your , let's call this equation *.
Now first we'll deal with .
For , from our definition * we know that 2 is a factor, therefore is a composite number. Use the same logic for 3-1000.
, this is the same as . We know from * that 7, 11 and 13 are factors of , therefore is a composite number.
Repeat this technique for
Not the most well written proof but hope it makes sense.
It should be obvious that n!, for n> 2, is composite, just from the definition. In fact, n! is divisible by every integer from n down.
And 1000!+ 2 is divisible by 2, 1000!+ 3 is divisible by 3, 1000!+ 4 is divisible by 4, etc. Once we get to 10001 we have to be a bit more careful: 1001 is 11*91 so 1000!+ 1001 is divisible by 11, and 1002 is divisible by 2 so 1000!+ 1002 is divisible by 2.
Now, I assume I have to start with proving that 1000! is composite or not and then prove the other numbers being added to the factorial?
Also, our class definition of composite is: A positive integer a is called composite provided there is an integer b such that 1< b < a and b|a.
Help is much appreciated!