Prove that there exists a prime with at least ## n ## of its digits.

[Math100]

Homework Statement

For any ## n\geq 1 ##, prove that there exists a prime with at least ## n ## of its digits equal to ## 0 ##.
[Hint: Consider the arithmetic progression ## 10^{n+1}k+1 ## for ## k=1, 2, ##....]

Relevant Equations

None.

Proof:

By Dirichlet's theorem, we have that if ## a ## and ## d ## are two positive coprime numbers,
then there are infinitely many primes of the form ## a+nd ## for some ## n\in\mathbb{N} ##.
Let ## n\geq 1 ## be a natural number.
Now we consider the arithmetic progression ## 10^{n+1}k+1 ## for some ## k\in\mathbb{N} ##.
Then ## a=10^{n+1} ## and ## d=1 ##.
This means ## gcd(a, d)=1 ## where ## a ## and ## d ## are coprime numbers.
Thus, ## 10^{n+1}k+1 ## has at least ## n ## consecutive zeros for every ## k ##.
Therefore, there exists a prime with at least ## n ## of its digits equal to ## 0 ## for any ## n\geq 1 ##.

 

[fresh_42]

Homework Statement:: For any ## n\geq 1 ##, prove that there exists a prime with at least ## n ## of its digits equal to ## 0 ##.
[Hint: Consider the arithmetic progression ## 10^{n+1}k+1 ## for ## k=1, 2, ##...]
Relevant Equations:: None.

Proof:

By Dirichlet's theorem, we have that if ## a ## and ## d ## are two positive coprime numbers,
then there are infinitely many primes of the form ## a+nd ## for some ## n\in\mathbb{N} ##.

Close. There are infinitely many primes in the set ##\{a+nd\, : \,n\in \mathbb{N}\}.## It is not for some ##n\in \mathbb{N}##. ##n## is actually a counter: ##a+1\cdot d\, , \,a+2\cdot d\, , \,a+3\cdot d\, , \,\ldots##

Let ## n\geq 1 ## be a natural number.
Now we consider the arithmetic progression ## 10^{n+1}k+1 ## for some ## k\in\mathbb{N} ##.
Then ## a=10^{n+1} ## and ## d=1 ##.

Better: Set ## d:=10^{n+1} ## and ## a:=1 ##. You need it the other way around. This uses ##n## as a fixed natural number, so we need another letter for the counter in the arithmetic progression. I will take ##k## below.

This means ## gcd(a, d)=1 ## where ## a ## and ## d ## are coprime numbers.
Thus, ## 10^{n+1}k+1 ## has at least ## n ## consecutive zeros for every ## k ##.
Therefore, there exists a prime with at least ## n ## of its digits equal to ## 0 ## for any ## n\geq 1 ##.

See? Here you use that ##\{a+nd\, : \,n\in \mathbb{N}\}=\{1+k\cdot 10^{n+1}\, : \,k\in \mathbb{N}\}## has infinitely many primes. O.k., we need only one prime in there.

 

[mathwonk]

this special case of Dirichlet's theorem is much easier than the general case. see e.g. p.2 of:
https://static1.squarespace.com/sta...a49636398ee2dce/1472161971095/primes1mod4.pdf