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 May 2nd, 2017, 06:42 PM #1 Newbie   Joined: May 2017 From: Shanghai, China Posts: 1 Thanks: 0 Divisibility Rules? https://brilliant.org/discussions/th...ility-rules-4/ Prove that if x is divisible by 3, then $\displaystyle 2^x - 1$ is divisible by 7
May 2nd, 2017, 11:20 PM   #2
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Quote:
 Originally Posted by anush2209 https://brilliant.org/discussions/th...ility-rules-4/ Prove that if x is divisible by 3, then $\displaystyle 2^x - 1$ is divisible by 7
Let $\ \ \ x = 3y \ \ \$ for any positive integer $\ \ y$

Then

$2^x - 1 = 2^{3y} - 1$

$2^{3y} - 1 = (2^3)^y - 1$

Factor

$2^{3y} - 1 = (2^3 - 1)[(2^3)^{y- 1} + (2^3)^{y-2} + ... + 1)]$

$2^{3y} - 1 = 7k$

This shows $\ \ \ 2^x - 1 \ \$ (given your condition) is a multiple of $\ 7 \$ therefore divisible by $\ 7$

Last edited by agentredlum; May 2nd, 2017 at 11:22 PM.

 May 2nd, 2017, 11:22 PM #3 Global Moderator   Joined: Dec 2006 Posts: 18,845 Thanks: 1567 The polynomial $y^n - 1$ is divisible by $y - 1$ by the factor theorem. Hence if $y = a^b$ for positive integers $a$ and $b$, $a^b - 1$ divides $y^n - 1$, which is $a^{bn} - 1$. For the given problem, use $a = 2$ and $b = 3$ so that $a^b- 1 = 7$.

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