String of 1s and Prime numbers

Can you prove that if 10n19\frac{10^{n}-1}{9} is a prime number then 'n' is also a prime number?

#NumberTheory #Mathematics

Note by Vikram Pandya
7 years, 4 months ago

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Comments

Umm.. Yes, I can :)

Let's prove the contrapositive. In other words, we're going to prove that if nn is not a prime number, then 10919\frac{10^9-1}{9} will never be a prime number.

Let nn be equal to abab where aa and bb are two positive integers greater than 11.

Now notice that

10ab19=10a19×(1+10a+102a+103a+10a(b1))\frac{10^{ab}-1}{9}=\frac{10^a-1}{9}\times (1+10^a+10^{2a}+10^{3a}+\cdots 10^{a(b-1)})

which is definitely not a prime number.

A fun fact: primes in the form of 10n19\frac{10^n-1}{9} are called repunit primes.

Mursalin Habib - 7 years, 4 months ago

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This is true for any base.A repunit is of the form bn1b1\dfrac{b^n-1}{b-1} in base bb.The above result holds in any base that a repunit is a prime if nn is a prime.

Rahul Saha - 7 years, 4 months ago

Here's a related problem:

How many numbers of the form 10100101 1010\ldots 0101 are prime?

Calvin Lin Staff - 7 years, 4 months ago

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  1. http://mathforum.org/library/drmath/view/61896.html http://askville.amazon.com/binary-numbers-form-101-10101-1010101-prime/AnswerViewer.do?requestId=30307452

Shamik Banerjee - 7 years, 4 months ago

Some initial thoughts:

101101 is prime.

1010110101 is a multiple of 33. (This holds for all other cases where the number of 11s are a multiple of 33)

10101011010101 is the product of at least two factors, namely 101101 and 1000110001. This holds for all greater cases where the number of 11s are even.

Haven't thought of when the number of 11s are odd though.

Michael Tong - 7 years, 4 months ago

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When the number of 1 1 s is odd, the number is divisible by the repunit you get by taking out all of the zeroes; e.g. 10101 10101 is divisible by 111 111 , 101010101 101010101 is divisible by 11111 11111 , and so on. This is because 1+102+104++102n=(1+10+102++10n)(110+102+10n). 1+10^2 + 10^4 + \cdots + 10^{2n} = (1+10+10^2+\cdots+10^n)(1-10+10^2-\cdots+10^n). Note that the last sign in that alternating factor is a + + only if n n is odd.

Patrick Corn - 7 years, 4 months ago

only 101101 is a prime

Sagnik Saha - 7 years ago

Yeah the one way of doing this. I was not sure how to put proof related questions on this forum. But this is good level 4 question otherwise :)

Vikram Pandya - 7 years, 4 months ago
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