# Can $7$ be the smallest prime factor of a repunit?

Repunits are numbers whose digits are all $$1$$. In general, finding the full prime factorization of a repunit is nontrivial.

Sequence A067063 in the OEIS gives the smallest prime factor of repunits. There are no $$7$$'s (just the number $$7$$, not as a digit in a different prime) that I can see in the sequence.

$$7$$ does divide many repunits, but always does when $$3$$ is also a factor, so $$7$$ never shows up as the smallest prime factor. The table of the first $$508$$ smallest prime factors of repunits shows that $$7$$ is not the smallest prime factor of any of those repunits.

My question is can $$7$$ ever be the smallest prime factor of a repunit?

I tried to find a repunit with $$7$$ as the smallest factor but searched very far and $$3$$ is always a factor whenever $$7$$ is.

I tried to prove that it can't be but I have no idea how.

• Hint: think about $R(6) = 111111$. Also the fact that $$1001=7\cdot 11 \cdot 13$$ will help. Nov 28 '19 at 2:23
• yes $7$ is a factor of $111111$ but so is $3$ ... so it is not the smallest prime factor Nov 28 '19 at 2:25
• I don't think Zubin Mukerjee was trying to give a counter example. I think that was a hint as to why $7$ can never appear without $3$. If $3\not\mid 1111....$ but $7\mid 111....$ then $1111.....111\pmod {1001}$ is divisible by $7$ but $3$ isn't. Is that possible. Nov 28 '19 at 2:48
• Yes, $R(6)$ is what I used to show that $7$ cannot be the smallest prime factor, ever. The idea is that if $R(n)$ is divisible by $7$, then $n$ is divisible by $6$, which means $n$ is divisible by $3$, which means $R(n)$ is divisible by $3$, so $7$ isn't the smallest prime factor. See my answer for the full argument @VincentLuo Nov 28 '19 at 2:50

Can $$7$$ be the smallest prime factor of a repunit?

No, it cannot.

The main idea is as follows: If the $$n^\text{th}$$ repunit $$R(n)$$ is divisible by $$7$$, then $$n$$ is divisible by $$6$$, which means $$n$$ is divisible by $$3$$, which means $$R(n)$$ is divisible by $$3$$, so $$7$$ cannot be the smallest prime factor.

The full argument is fleshed out below.

Let $$R(n)$$ be the $$n^\text{th}$$ repunit, e.g. $$R(4) = 1111$$.

One interesting property about repunits is that if an integer $$m$$ divides $$R(n)$$, then $$m$$ divides $$R(2n)$$, and $$m$$ divides $$R(3n)$$, and so on. This is because, for any positive integer $$n$$, we have the factorization $$R(2n) = \left(10^{n}+1\right)R(n)$$ and, more generally, for any positive integers $$k$$ and $$n$$, we can factor as follows:

$$R(kn) = \left(\displaystyle\sum\limits_{j=0}^{k-1} 10^{jn}\right)R(n)$$

What this means, for your problem, is if a prime $$p$$ divides a repunit $$R(n)$$, then it will also divide $$R(kn)$$ for all $$k$$.

This tells us that:

• Every other repunit is divisible by $$11$$. $$11 \,|\, R(2) \implies\,\Big(\,\,\,n \equiv 0 \pmod{2} \iff 11 \,|\, R(n) \,\,\,\Big)$$ Here, the "$$\iff$$" bidirectionality follows from the fact that $$11 \not| \,\,R(1)$$.

• Every third repunit is divisible by $$3$$. $$3 \,|\, R(3) \implies \,\Big(\,\,\,n \equiv 0 \pmod{3} \iff 3 \, |\, R(n) \,\,\,\Big)$$ Here, the "$$\iff$$" bidirectionality follows from the fact that $$3 \not| \,\,R(1)$$ and $$3 \not| \,\,R(2)$$.

• Every third repunit is divisible by $$37$$. $$37 \,|\, R(3) \implies \,\Big(\,\,\,n \equiv 0 \pmod{3} \iff 37 \, |\, R(n) \,\,\,\Big)$$ Here, the "$$\iff$$" bidirectionality follows from the fact that $$37 \not| \,\,R(1)$$ and $$37 \not| \,\,R(2)$$.

• Every sixth repunit is divisible by $$13$$.$$13 \,|\, R(6) \implies \,\Big(\,\,\,n \equiv 0 \pmod{6} \iff 13 \, |\, R(n) \,\,\,\Big)$$ Here, the fact that each of $$1, 11, 111, 1111, 11111$$ are not divisible by $$13$$ implies the "$$\iff$$" bidirectionality.

• Every sixth repunit is divisible by $$7$$.$$7 \,|\, R(6) \implies \,\Big(\,\,\,n \equiv 0 \pmod{6} \iff 7 \, |\, R(n) \,\,\,\Big)$$ Here, the fact that each of $$1, 11, 111, 1111, 11111$$ are not divisible by $$7$$ implies the "$$\iff$$" bidirectionality.

This last fact implies that $$7$$ can never be the smallest prime factor of a repunit, because whenever $$n \equiv 0 \pmod{6}$$, it's also true that $$n \equiv 0 \pmod{3}$$, so whenever $$7$$ is a factor of a repunit, $$3$$ will be as well.

By a similar argument, $$13$$ and $$37$$ also cannot be the smallest prime factor of a repunit.

• Beautiful. +1.. Nov 28 '19 at 2:51

The repunits as a recurrence obey $$r_{n+1} = 10 r_n + 1$$ see https://en.wikipedia.org/wiki/Pisano_period

The code making the table of Pisano periods below. The simple outcome: the possible values of $$r_n \pmod{21}$$ are $$0, 1, 2, 6, 11, 19 .$$ In particular, the remainder after dividing by 21 can be $$6,$$ but is never $$7$$ or $$14 \pmod {21}.$$ Thus, $$r_n$$ can be divisible by $$3$$ alone, without any $$7.$$ Once $$r_n$$ is divisible by $$7,$$ the only possibility is $$0 \pmod {21},$$ meaning also divisible by $$3$$

   mpz_class rep = 0;
for(int n = 1; n <= 45; ++n){
rep = rep * 10 + 1;
cout << "  n: " << n  << "  rep: " << rep << " mod 21: " << rep % 21 << endl;
}


with no extra symbols. This is also a complete proof:

jagy@phobeusjunior:~$./mse Wed Nov 27 18:45:37 PST 2019 n: 1 rep: 1 mod 21: 1 n: 2 rep: 11 mod 21: 11 n: 3 rep: 111 mod 21: 6 n: 4 rep: 1111 mod 21: 19 n: 5 rep: 11111 mod 21: 2 n: 6 rep: 111111 mod 21: 0 n: 7 rep: 1111111 mod 21: 1 n: 8 rep: 11111111 mod 21: 11 n: 9 rep: 111111111 mod 21: 6 n: 10 rep: 1111111111 mod 21: 19 n: 11 rep: 11111111111 mod 21: 2 n: 12 rep: 111111111111 mod 21: 0 n: 13 rep: 1111111111111 mod 21: 1 n: 14 rep: 11111111111111 mod 21: 11 n: 15 rep: 111111111111111 mod 21: 6 n: 16 rep: 1111111111111111 mod 21: 19 n: 17 rep: 11111111111111111 mod 21: 2 n: 18 rep: 111111111111111111 mod 21: 0 n: 19 rep: 1111111111111111111 mod 21: 1 n: 20 rep: 11111111111111111111 mod 21: 11 n: 21 rep: 111111111111111111111 mod 21: 6 n: 22 rep: 1111111111111111111111 mod 21: 19 n: 23 rep: 11111111111111111111111 mod 21: 2 n: 24 rep: 111111111111111111111111 mod 21: 0 n: 25 rep: 1111111111111111111111111 mod 21: 1 n: 26 rep: 11111111111111111111111111 mod 21: 11 n: 27 rep: 111111111111111111111111111 mod 21: 6 n: 28 rep: 1111111111111111111111111111 mod 21: 19 n: 29 rep: 11111111111111111111111111111 mod 21: 2 n: 30 rep: 111111111111111111111111111111 mod 21: 0 n: 31 rep: 1111111111111111111111111111111 mod 21: 1 n: 32 rep: 11111111111111111111111111111111 mod 21: 11 n: 33 rep: 111111111111111111111111111111111 mod 21: 6 n: 34 rep: 1111111111111111111111111111111111 mod 21: 19 n: 35 rep: 11111111111111111111111111111111111 mod 21: 2 n: 36 rep: 111111111111111111111111111111111111 mod 21: 0 n: 37 rep: 1111111111111111111111111111111111111 mod 21: 1 n: 38 rep: 11111111111111111111111111111111111111 mod 21: 11 n: 39 rep: 111111111111111111111111111111111111111 mod 21: 6 n: 40 rep: 1111111111111111111111111111111111111111 mod 21: 19 n: 41 rep: 11111111111111111111111111111111111111111 mod 21: 2 n: 42 rep: 111111111111111111111111111111111111111111 mod 21: 0 n: 43 rep: 1111111111111111111111111111111111111111111 mod 21: 1 n: 44 rep: 11111111111111111111111111111111111111111111 mod 21: 11 n: 45 rep: 111111111111111111111111111111111111111111111 mod 21: 6 Wed Nov 27 18:45:37 PST 2019 jagy@phobeusjunior:~$

• Hi, this isn't related to this question at all, but I was reading this article on axioms for geometry and in the middle of it, there is a theorem named after you. I thought "Hey! I've seen that name before!" :) maa.org/sites/default/files/images/upload_library/22/Ford/… Dec 3 '19 at 6:05

$$7|1001 = 7*11*13$$

So $$7|1001*111 = 111,111$$ so if $$7|\underbrace{1111.....1}_n$$ then

$$7|\underbrace{11111.....1}_n - 111111$$

$$= \underbrace{1111111....111}_{n-6}000000$$

and as $$10$$ and $$7$$ are relatively prime. $$7|\underbrace{1111111....111}_{n-6}$$

So by induction if $$n \equiv a \pmod 6$$ then $$7|\underbrace{1111...1}_{a}$$.

So we just need to show when does $$7|1,11,111,1111,11111,111111$$

And that is only $$7|111111 = 111*1001 = (3*37)*(7*11*13)$$.[*]

So $$7|\underbrace{1111......1}_n$$ if and only if $$6|n$$ in which case $$3,11,13,37$$ will also divide the repunit.

Thanks to Zubin Mukerjee for the hint.

.....

[*]. $$7|\underbrace{111...1}_{a < 6}\iff 7|\underbrace{111...1}_{6-a}$$ but $$7\not \mid 1(a=1)$$ and $$7\not \mid 11 (a=2)$$ and so $$7\not \mid 11111 (6-1)$$ and $$7\not\mid 1111 (6-2)$$ while $$7\not \mid 111 =3*37$$