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Math Help - Factorial problem! Help!

  1. #1
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    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.

    Thanks in advance for any advice.
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  2. #2
    Behold, the power of SARDINES!
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    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 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
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  3. #3
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    Quote Originally Posted by jzellt View Post
    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 ......?
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  4. #4
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    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.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    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?!
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