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Prove that $371\cdots 1$ is not prime.

I tried mathematical induction in order to prove this, but I am stuck.

My partial answer:

To be proved is that $37\underbrace{111\cdots 1}_{n\text{ ones}}$ is never prime for $n\geq 1$. Let $P(n)$ be the statement that $37\underbrace{111\cdots 1}_{n\text{ enen}}$ is not prime.

For $n=1$, we can write $371=7\cdot 53$, and therefore $P(1)$ is true. Let $P(k)$ be true for $k>1$. Then we now have to prove that $P(k+1)$ is true.

I found that $37\underbrace{111\cdots 1}_{k\text{ ones}}$ can be written as $37\cdot 10^k+10^{k-1}+10^{k-2}+\cdots +10^0$ and that $37\underbrace{1111\cdots 1}_{k+1\text{ ones}}$ can be written as $37\cdot 10^{k+1}+10^{k}+10^{k-1}+\cdots+10^0$.

I think I'm pretty close to the answer know, but I don't know how to proceed.

Bart Michels
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Dr. Heinz Doofenshmirtz
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    Is it 33...3 or is it 371...1? – naslundx Apr 28 '15 at 17:20
  • 371..1, thanks for editing :) – Dr. Heinz Doofenshmirtz Apr 28 '15 at 17:25
  • Try experimenting with factoring it for different values of $n$. You'll see a pattern with the small prime factors that repeats every time $n$ is increased by $6$. – aes Apr 28 '15 at 17:29
  • After doing some experimentation, I suspect this problem is not that trivial. Call $x_k$ the number with $k$ 1s. When $k \equiv 2$ mod 3, the situation is simple: the digit sum is divisible by $3$ so the number is divisible by $3$. But the other possible moduli don't seem so simple. In particular, it appears that you have to consider mod 6, not just mod 3, because $x_1$ and $x_4$ actually have no factors in common. – Ian Apr 28 '15 at 17:39
  • @CauchytheDog You can find a rigorous proof in the answer I linked to in my answer. If something is not clear there then please ask questions on that answer and I will be happy to elaborate. – Bill Dubuque Apr 28 '15 at 19:48
  • @Bill I am afraid I am not yet familiar with modular arithmetic.. – Dr. Heinz Doofenshmirtz Apr 28 '15 at 20:59

4 Answers4

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First note that $111111=3 \cdot 7 \cdot 11\cdot 13\cdot 37$.

If $371\cdots 1$ is divisible by any of these prime factors, so will be that number followed by six ones. It suffices to look at the factorizations of $371$, $3711$, $37111$, $371111$, $3711111$, and $37111111$.

$371 = \mathbf{7} \cdot 53$

$3711 = \mathbf{3} \cdot 1237$

$37111 = 17 \cdot \mathbf{37} \cdot 59$

$371111 = \mathbf{13} \cdot 28547$

$3711111 = \mathbf{3} \cdot 1237037$

$37111111 = \mathbf{37} \cdot 1003003$

Umberto P.
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  • I could not get it. In my opinion, it seems like a guess. It couldn't be a proof. When you want to prove "If 371⋯1 is divisible by any of these prime factors, so will be that number followed by six ones." It is not sufficient to take a look at few examples but ALL possible cases. – Hyoyoung Jung Apr 28 '15 at 17:50
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    @HyoyoungJung. That's because every number of the form $371...1$ is of the form $x * 10^k + n * 111111$ where $x$ is one of the above. – jxnh Apr 28 '15 at 17:56
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    @HyoyoungJung, if $37\underbrace{1\cdots1}{n}$ is divisible by one of the primes then so is $37\underbrace{1\cdots1}{n}000000$, thus so is $37\underbrace{1\cdots1}{n}111111=37\underbrace{1\cdots1}{n+6}$, because the difference $111111$ is divisible by all of them. – Yoni Rozenshein Apr 28 '15 at 17:56
  • @YoniRozenshein, JHance Thank you. Your answers are good supporting clues for me. – Hyoyoung Jung Apr 28 '15 at 18:04
  • I understand the explanation of Yoni, but could you elaborate a bit, because I am afraid I don't seem to understand it yet.. – Dr. Heinz Doofenshmirtz Apr 28 '15 at 18:06
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Hints not to give this away completely:

Notice $111 = 37 \times 3$ so since it starts with 37, what happens if you add a number of ones divisible by $3$, like $37,111$ or $37,111,111$, etc?

Also notice if you take $3711$, the sum of digits is divisible by 3, so this should take care of $3,711$ or $3,711,111$ or $3,711,111,111$ or the like.

This leaves only one sequence class for you to deal with. Can you?

gt6989b
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If $n$, the number of ones, is a multiple of $3$, then $N=37111\cdots111$ is a multiple of $37$ (because $111=37\cdot3$).

If $n=3k+2$ then $N$ is a multiple of three, because the sum of its digits is $3k+12$, a multipe of three.

For the remainding cases, consider that $371$ is a multiple of $7$ , $371111$ is a multiple of $13$ and $111111=3\cdot 7\cdot13\cdot 11\cdot37$

ajotatxe
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1

Hint $\ $ Do a covering congruence argument mod $\color{#c00}6$ using the divisors $3,7,13,37\,$ of $\,10^{\color{#c00}6}-1$

Community
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Bill Dubuque
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