It's not always divisible by 11

K = 1111111 11 n number of 1’s 2 1111111 11 n number of 1’s K = \underbrace{1111111 \ldots 11}_{n \text{ number of 1's}}2 \underbrace{1111111 \ldots 11}_{n \text{ number of 1's}}

For positive integer n n , can K K ever be a prime number?

It's impossible to know. Yes No

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1 solution

Chris Galanis
Dec 28, 2015

K = 111 11 n digits 2 11 111 n digits = 111 11 ( n + 1 ) digits 000 00 n digits + 11 111 ( n + 1 ) digits = 111 11 ( n + 1 ) digits 1 0 n + 11 111 ( n + 1 ) digits = 111 11 ( n + 1 ) digits ( 1 0 n + 1 ) \begin{aligned} K & = \underbrace{111 \ldots 11}_{n \text{ digits}}2 \underbrace{11 \ldots 111}_{n \text{ digits}} \\ & = \underbrace{111 \ldots 11}_{(n+1) \text{ digits}}\underbrace{000 \ldots 00}_{n \text{ digits}} + \underbrace{11 \ldots 111}_{(n+1) \text{ digits}} \\ & = \underbrace{111 \ldots 11}_{(n+1) \text{ digits}}\cdot 10^n + \underbrace{11 \ldots 111}_{(n+1) \text{ digits}} \\ & = \underbrace{111 \ldots 11}_{(n+1) \text{ digits}}\cdot (10^n +1) \end{aligned}

Thus K K is a composite number for every n N n \in \mathbb{N^*}

Nicely done!

Adarsh Kumar - 5 years, 5 months ago

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