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Thread: integers and divisibility

  1. #1
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    integers and divisibility

    Find the smallest positive integer n such that 2n −1 is divisible by 47.
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  2. #2
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    Hello, anncar!

    Find the smallest positive integer $\displaystyle n$ such that $\displaystyle 2n -1$ is divisible by 47.

    If $\displaystyle 2n-1$ is divisible by 47, then: .$\displaystyle 2n-1 \:=\:47a$ for some integer $\displaystyle a.$

    . . Then we have: .$\displaystyle n \:=\:\frac{47a+1}{2}$


    Since $\displaystyle n$ is a positive integer, $\displaystyle 47a+1$ is divisible by 2.

    . . The least value occurs when $\displaystyle a = 1$


    Therefore: .$\displaystyle n \;=\;\frac{47(1) + 1}{2}\quad\Rightarrow\quad\boxed{n \:=\:24}$

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