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

• MHB
• lfdahl
In summary, we can prove that the decimal representation pk = 11111....1 exists for every prime, p > 5 by using the fact that any prime number greater than 5 can be expressed as 6k+1 or 6k-1 and using the formula pk = (10^k - 1)/9. This formula works for primes that can be expressed as 6k+1 and for primes that can be expressed as 6k-1, the decimal representation will be pk = (10^k + 1)/9. However, there are exceptions to this rule, such as the prime numbers 2 and 5. There is no specific significance to this decimal representation, but it is a
lfdahl
Gold Member
MHB
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 $1$ :

$pk = 1111...1$

lfdahl said:
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 $1$ :

$pk = 1111...1$

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).

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:

$a_k = \underbrace{1111..1}_{k \: \: positions} = b_k \cdot p +r_k, \: \: \: 0\leq r_k < p.$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 $\underbrace{111..1}_{n-m \: pos.}$.

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

$a_k = \underbrace{1111..1}_{k \: \: positions} = b_k \cdot p +r_k, \: \: \: 0\leq r_k < p.$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

## 1. How can you prove that the decimal representation pk = 11111....1 exists for every prime, p > 5?

To prove this, we can use the fact that any prime number greater than 5 can be expressed as 6k+1 or 6k-1 for some integer k. This means that pk will always be a multiple of 6, and therefore, pk will always end in either 1 or 9. Since we are looking for a decimal representation that ends in 1, we can focus on the case where pk = 6k+1. In this case, we can express pk as (10^k - 1)/9, which is a repeating decimal of 1's. Therefore, 11111....1 exists for every prime, p > 5.

## 2. Is there a specific formula or method for determining the decimal representation pk = 11111....1 for each prime, p > 5?

Yes, as mentioned in the previous answer, we can use the formula pk = (10^k - 1)/9, where k is an integer, to determine the decimal representation for each prime, p > 5. This formula works for primes that can be expressed as 6k+1. For primes that can be expressed as 6k-1, the decimal representation will be pk = (10^k + 1)/9.

## 3. Can you provide an example of a prime number, p > 5, with a decimal representation of pk = 11111....1?

One example is the prime number 37. Using the formula pk = (10^k - 1)/9, we can see that p = 37 can be expressed as pk = 1111111111111111111111111111111111111 (with 36 ones).

## 4. Are there any exceptions to the rule that pk = 11111....1 exists for every prime, p > 5?

Yes, there are a few exceptions. The first exception is the prime number 2, which does not follow the formula mentioned above. Another exception is the prime number 5, which ends in 5 and not 1. However, for all other primes greater than 5, the decimal representation of pk = 11111....1 will exist.

## 5. Is there any significance to the decimal representation pk = 11111....1 for primes, p > 5?

There is no specific significance to this decimal representation, but it is a useful tool in understanding the patterns and properties of prime numbers. It also highlights the fact that prime numbers can be expressed in different ways and that there are many interesting relationships between them.

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