Does a Decimal Representation of the Form 111...1 Exist for Every Prime p > 5?

[lfdahl]

Let $p$ be a prime number exceeding $5$.

Prove that there exists a natural number $k$ such that

each digit in the decimal representation of $pk$ is $1$ :

$pk = 1111...1$

 

[kaliprasad]

Let $p$ be a prime number exceeding $5$.

Prove that there exists a natural number $k$ such that

each digit in the decimal representation of $pk$ is $1$ :

$pk = 1111...1$

Spoiler

because p is a prime > 5 so p is co-prime to 10
hence as per Fermats Little Theorem
we have $10^{p-1} \equiv 1 \pmod {p}$
so a sequence of 9's ( p-1 9's) is divisible by p
further as p is co-prime to 9 so sequence of 1's ( p-1 1's) is divisible by p.
so for k = p-1 this holds. this may hold for k a factor of p-1( for example p = 13 and k = 6).

 

[lfdahl]

Spoiler

because p is a prime > 5 so p is co-prime to 10
hence as per Fermats Little Theorem
we have $10^{p-1} \equiv 1 \pmod {p}$
so a sequence of 9's ( p-1 9's) is divisible by p
further as p is co-prime to 9 so sequence of 1's ( p-1 1's) is divisible by p.
so for k = p-1 this holds. this may hold for k a factor of p-1( for example p = 13 and k = 6).

Hi, kaliprasad! Thankyou for your participation! Well done Here is an alternative approach:

Spoiler

\[a_k = \underbrace{1111..1}_{k \: \: positions} = b_k \cdot p +r_k, \: \: \: 0\leq r_k < p.\]By pigeons hole principle for some $n$ and $m$, $n>m$, we have $r_n = r_m$.

It follows, that the difference: $a_n-a_m$ is divisible by $p$.Note, that $a_n-a_m = \underbrace{111..1}_{n-m \: pos.} \underbrace{000..0}_{m \: pos.}=
\underbrace{111..1}_{n-m \: pos.} \cdot 10^m$.Since $10^m$ is not divisible by $p$, $p$ divides $\underbrace{111..1}_{n-m \: pos.}$.

 

[kaliprasad]

Hi, kaliprasad! Thankyou for your participation! Well done Here is an alternative approach:

Spoiler

\[a_k = \underbrace{1111..1}_{k \: \: positions} = b_k \cdot p +r_k, \: \: \: 0\leq r_k < p.\]By pigeons hole principle for some $n$ and $m$, $n>m$, we have $r_n = r_m$.

It follows, that the difference: $a_n-a_m$ is divisible by $p$.Note, that $a_n-a_m = \underbrace{111..1}_{n-m \: pos.} \underbrace{000..0}_{m \: pos.}=
\underbrace{111..1}_{n-m \: pos.} \cdot 10^m$.Since $10^m$ is not divisible by $p$, $p$ divides $\underbrace{111..1}_{n-m \: pos.}$.

Above approach is simpler