What two prime numbers are never a factor of $n$ 9s

Let's use the Pigeonhole Theorem to solve this.

Assume $p\neq2,5$ is a prime, then consider a series:

$9,99,999,\dots,\underbrace{999...99}_{p+1\text{ times}}$

We can always find two numbers congruent under modulo $p$ , say that is $\underbrace{999...99}_{a\text{ times}}\equiv \underbrace{999...99}_{b\text{ times}}\pmod{p}$

WLOG, assume $a>b$ , then we have $\underbrace{999...99}_{a\text{ times}}-\underbrace{999...99}_{b\text{ times}}\equiv0 \pmod p$

So this number, which has $(a-b)$ $9$ 's in front and followed by $b$ $0$ 's, is divisible by $p$ , but since $p$ is not equal to $2,5$ , then we can cancel all of the zeros behind $9$ 's, so $\underbrace{999...99}_{(a-b)\text{ times}}=10^{a-b}-1$ will be the multiple of $p$ , hence proved.

By Fermat's Little Theorem , if $p$ is a prime that doesn't divide $10,$ then $p \text{ divides } 10^{p-1} - 1$

Remark: The problem has changed both in problem text and answer type since I posted this solution. The previous sentence is actually the complete solution for the current form (as a "featured problem"). I've left the rest of my solution below unedited, but it is unnecessary for the current form.

This covers all primes other than $2, 5.$ For these two primes $p \in \{2,5\},$ we note that the $p$ divides $10^n$ (since $n\geq 1$ ), so $p \text{ divides } 10^n-1 \iff p \text{ divides } -1$ but since no prime divides $-1,$ we conclude neither $2$ nor $5$ is a factor of $10^n - 1$ for any $n \geq 1.$

Their product, and therefore the answer to the problem, is $2\times 5 = \boxed{10}$

The last digit of any prime (other than $2$ or $5$ ) will cycle through all ten digits in some order repeatedly, with every tenth ending number being $0$ .

This means we can take any number $x_n$ , add to it $q\times p$ where $p$ is the prime in question and $q$ is a positive integer less than $10$ , and produce a number such that $x_n+pq=10x_{n+1} + 9$

For example, where $p=97$ : $679+97\times60=6499$ $6499+97\times500=54999$

In these cases, $x_1=67$ , $x_2=64$ , $x_3=54$ .

Let's assume there is some prime number which causes the value of $x$ to loop, and so would never produce a number $10^y-1$ with this method:

Such a loop would produce the situation $x+pQ=10^yx+10^y-1$ , where $Q$ is the sum of all $q$ within a loop, and $y$ is some amount of $9$ s. (e.g. $13+pQ=139...999$ )

However, as $x$ must be a multiple of $p$ , we know that $x+pQ$ is a multiple of $p$ , as is $10^yx$ . For the equation to balance, $10^y-1$ must also be a multiple of $p$ , and thus, even if a prime number loops using this method, it will still be a divisor of a number with the form $10^y-1$ .

It's much easier than it seems.

When you factorize 10 into prime numbers you get $5 \times 2$ . If you raise 10 to the nth power and you factorize it too, you get $5^n \times 2^n$ .

However, we are subtracting 1 from $10 ^n$ . This means, that we are only 1 unit away to be able to factorize the number by 2. Similarly, we are only 1 unit away to be able to factorize the number by 5. However, the number isn't away a fixed number from being able to be divisible by any other number different to 2 and 5.

So $10^n$ isn't divisible FOR SURE for both 2 and 5. Therefore, $10^n$ might be divisible by any other prime number.

Roses are red,

Violets are blue,

Let's say "yes" is false

And therefore "no" is true

:)

Consider $1/p$ for prime $p \neq 2, 5$ . The decimal expansion of this repeats after n digits for some integer n since all rational fractions have a repeating decimal expansion.

Thus $10^n * 1/p - 1/p$ is an integer so p divides $10^n - 1$ .

We can go further using Fermat's little theorem to show n = p-1 is one possible value but other people have discussed this already.

To be divisible by 2 last digit should be multiple of 2.

Hence, 2 cannot divide (10^n-1)

To be divisible by 5 last digit should be 5.

Hence, 5 cannot divide (10^n-1)

Product of 2 numbers = 2 " 5 = 10

Given a prime number $p$ different from 2 and 5. We want to find a solution to 10^{n-1} =0 /) (mod p) or \(10^n =1 (mod p). Since p is coprime to 10, we know that 10 is not zero mod p. By Fermats little theorem we have $10^{p-1}= 1$ (mod p).