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Thread: 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 $\displaystyle n!$ for a positive integer $\displaystyle n$. It consists of the product of all integers from $\displaystyle 1$ to $\displaystyle n$, inclusive. Can you see why this is divisible by $\displaystyle m$, provided $\displaystyle 2 \le m \le n$?

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

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

    Grandad
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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

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

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

    In particular $\displaystyle n!$ is divisible by $\displaystyle 2$. So if you add $\displaystyle 2$ to $\displaystyle n!$, you get another number divisible by $\displaystyle 2$, 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, $\displaystyle a$ and $\displaystyle b$, say, are each divisible by another integer $\displaystyle m$, then $\displaystyle (a+b)$ is also divisible by $\displaystyle m$. (Can you see why?)

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

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

    Grandad

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