A number with $2016$ zeros that is written as $101010 \dots 0101$ is given, in which the zeros and ones alternate. Prove that this number is not prime.
Problem
Source: Germany - BWM Round 1 2016, #1
Tags: prime numbers, number theory, Germany, decimal representation
Diogene
13.11.2016 03:12
2017 is prime .
Tetrahedron49
01.02.2017 00:52
The sum of the digits is 1008, 1008 is divisible by 9 so the number is not prime.
hunter117
01.02.2017 01:00
Is it just me or is the sum $2017$?
mathguy5041
01.02.2017 01:04
Call that number $n$. $99n=10^{4034}-1=(10^{2017}-1)(10^{2017}+1)$. Let the first factor be $a$ and the second one be $b$. $n=\frac{a}{9}*\frac{b}{11}$, and it is obvious to check with divisibility rules that both of these numbers are integers that are greater than one, so $n$ is not prime.
tchebytchev
01.02.2017 01:07
Tetrahedron49 wrote: The sum of the digits is 1008, 1008 is divisible by 9 so the number is not prime. The sum is 2017 Let N=$101010 \dots 0101 =\sum_{i=0}^{2016} 10^{2i}=\sum_{i=0}^{2016} 100^{i}=\frac{100^{2017}-1}{99}$ is not a prime
Tetrahedron49
27.02.2017 10:14
I misread the problem sorry about that.
minusonetwelth
28.07.2024 15:47
Let \[ n = 101010\ldots 0101 = 10^0 + 10^2 + 10^4 + \ldots + 10^{2016 \cdot 2} \]Let $a = 10^2$, then \[ 101010\ldots 0101 = 1 + a + a^2 + \ldots + a^{2016} = \frac{a^{2017} - 1}{a - 1} = \frac{(10^{2017} - 1)(10^{2017} + 1)}{(10 - 1)(10 + 1)} \]Now, $10 \equiv 1 \pmod{10-1}$, so $10^{2017} \equiv 1 \pmod{10-1} \iff 10^{2017} - 1 \equiv 0 \pmod{10-1}$, thus $10-1 \mid 10^{2017} - 1$. Similarly, because $2017$ is odd, $10 + 1 \mid 10^{2017} + 1$. Therefore, the two factors $\frac{10^{2017} - 1}{10-1}$ and $\frac{10^{2017} + 1}{10+1}$ are positive integers. Since they are obviously greater than 1, $n$ cannot be a prime number.