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Math Help - Composite numbers

  1. #1
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    Composite numbers

    let n be an integer with n> or equal 2. Show that n! + 2 is even. Now show that m|n! + m, for all m such that 2(</=)m(</=)n and explain why n! + m is composite. Use this to find a sequence of n - 1 consecutive compositive integers. (the sequence is an a_2,a_3,...,a_n , with a_i+1 = a_i +1 and all a_i composite.
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  2. #2
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    Divisibility of factorials

    Hello mitch_nufc
    Quote Originally Posted by mitch_nufc View Post
    let n be an integer with n> or equal 2. Show that n! + 2 is even. Now show that m|n! + m, for all m such that 2(</=)m(</=)n and explain why n! + m is composite. Use this to find a sequence of n - 1 consecutive compositive integers. (the sequence is an a_2,a_3,...,a_n , with a_i+1 = a_i +1 and all a_i composite.
    Look at the definition of n! for a positive integer n. It consists of the product of all integers from 1 to n, inclusive. Can you see why this is divisible by m, provided 2 \le m \le n?

    Can you now see why n!+m is also divisible by m?

    So what about the numbers n!+2, n!+3, ..., n!+n?

    Grandad
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  3. #3
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    i still dont get it
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  4. #4
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    Divisibility of factorials

    Hello mitch_nufc

    n! = 1 \times 2 \times 3 \times \dots \times n

    So n! is divisible by all integers from 2 to n. OK so far?

    In particular n! is divisible by 2. So if you add 2 to n!, you get another number divisible by 2, because an even number plus an even number is always even.

    That was the first thing you were asked to prove.

    Next, note that if two integers, a and b, say, are each divisible by another integer m, then (a+b) is also divisible by m. (Can you see why?)

    Now we've already said that n! is divisible by all integers between 2 and n. In other words, it's divisible by m if 2 \le m \le n. And m is (obviously) also divisible by m. So n! + m is also ...?

    Now think again about the integers n! + 2, n! + 3, \dots, n! + n. What can you say about them?

    Grandad

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