integers and divisibility

**anncar** · December 7th 2007, 09:26 AM

Find the smallest positive integer n such that 2n −1 is divisible by 47.

**Soroban** · December 7th 2007, 09:40 AM

Hello, anncar!

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

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

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

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

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

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

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