so if you add any number less then 1000 we could factor it out and therefore it is composite.
for so you could factor out either 143 or 7
for then next one 1002 is even so...
I hope this helps
Good Luck.
B
How you one go about doing this:
Prove that all of the following numbers are composite: 1000! + 2, 1000! + 3, 1000! + 4, ... , 1000! + 1000, 1000! + 1001, 1000! + 1002. The point of this problem is to present a long list of consecuetive numbers, all of which are composite.
Thanks in advance for any advice.
1000! = (1)(2)(3)(4) ....... (1000).
So ..... do you see the common factor in each of the 1000! + 2, 1000! + 3, 1000! + 4, ... , 1000! + 1000 ......?
As for 1000! + 1001, 1000! + 1002 ...... Note that 1001 = (7)(11)(13) and 1002 = (2)(3)(167) ..... Again, can you see the common factor ......?
Here's a random one or two:
1000! + 4 = (1)(2)(3)(4)(5) .... (1000) + 4
Take out the common of four:
= 4 {(1)(2)(3)(5) ......(1000) + 1}
which is a product of two numbers therefore composite.
1000! + 1002 = (1)(2)(3)(4)(5) .... (1000) + (2)(3)(167)
Take out the common factors of 2 and 3:
= (2)(3) {(1)(4)(5) .... (1000) + 167}
which is a product of two numbers therefore composite.
Do you see it now?!