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If $p$ is a prime coprime to $10$, show (or disprove with a counterexample) that at least one of the numbers $1,11,\cdots, \underbrace{11\cdots 1}_\text{p}$ is divisible by $p$.

Show (or disprove) that if you attach at most $p$ $3$'s to the end of $p$, you will at some point obtain a multiple of $p$ that ends in at least one $3$. That is, an element of the sequence $(a_n)_{n=1}^{p}, a_1 = 10p+3, a_n = 10a_{n-1}+3$ for $n\ge 2$ is a multiple of $p$.

I know that if $p$ is prime, then if $p | ab$ where $a,b\in\mathbb{Z}, p|a$ or $p|b$. Also, if $\gcd(10,p)=1$ there exist integers $s,t$ so that $10s+tp=1$. In addition, $10$ is an invertible element in the multiplicative group of integers modulo $p, $ which can be denoted by $\mathbb{Z}_p^*$. So the numbers $10, 20,\cdots (p-1)10$ all have different residues modulo $p$. I also know Lagrange's theorem, which states that the order of an element of a group divides the order of the group. I also know some properties of the Legendre function such as the fact that $(\frac{ab}p) = (\frac{a}p)\cdot (\frac{b}p)$ for integers $a,b$, if $a\equiv b\mod p, (\frac{a}p) = (\frac{b}p), (\frac{-1}p) = (-1)^{(p-1)/2},$ but I'm not sure if these are useful.

But I'm not sure how to come up with a counterexample or proof for the above two claims; they seem to work for small primes like $13$ and $11$.

Bill Dubuque
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user3472
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2 Answers2

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For $p = 3$ this is true.

Assume $p \neq 2,3,5$. Note that $$\underbrace{1...1}_{p-1\text{ times}} = \frac{10^{p-1}-1}{9} \equiv 0 \mod p.$$

acreativename
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You are correct in identifying that all of the listed values must have different residues modulo $p$. The other thing you need to do is consider - how many unique residues exist modulo $p$, and how many of them are in the list? What happens if none of them are congruent to 0?

ConMan
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