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