# Factorial problem! Help!

• Mar 24th 2008, 08:16 PM
jzellt
Factorial problem! Help!
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.

• Mar 24th 2008, 08:58 PM
TheEmptySet
$\displaystyle 1000!=1 \cdot 2 \cdot 3 \cdot ... \cdot 1000$

so if you add any number less then 1000 we could factor it out and therefore it is composite.

for $\displaystyle 1000! +1001=1000!+143*7$ so you could factor out either 143 or 7

for then next one 1002 is even so...

I hope this helps

Good Luck.

B
• Mar 24th 2008, 09:01 PM
mr fantastic
Quote:

Originally Posted by jzellt
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.

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 ......?
• Mar 24th 2008, 09:41 PM
jzellt
I appreciate all of the input here, but it's still not clicking as to how I show that all of these numbers are composite.
• Mar 24th 2008, 10:19 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
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?!