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

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- Dec 7th 2007, 09:26 AManncarintegers and divisibility
Find the smallest positive integer n such that 2n −1 is divisible by 47.

- Dec 7th 2007, 09:40 AMSoroban
Hello, anncar!

Quote:

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