Show that there is a number consisting only of 1’s that is divisible by 2001

Question

Show that there is a number consisting only of 1’s that is divisible by 2001.

ie, we have to prove 2001 divides $1+10+10^2+10^3+...+10^k$ for some $k$,

How can we proceed now?

https://math.stackexchange.com/questions/1009511/primes-dividing-11-111-1111 — lab bhattacharjee, Jun 23 '19 at 02:23

score 7 · Answer 1 · answered Jun 23 '19 at 02:36

Consider the set $S = \{1, 11, 111, 1111, \ldots \}$. By the Pigeonhole Principle, two of these numbers must be equal modulo $2001$. Subtracting yields a number divisible by $2001$ of the form $10^k s$, where $s \in S$. But $\gcd(10, 2001) = 1$, so $2001 \mid s$.

score 6 · Answer 2 · answered Jun 23 '19 at 02:16

6

By Euler's theorem, $2001|10^{\phi(2001)}-1$.

Since $2001=3\times23\times29, \phi(2001)=2\times22\times28=1232.$

Therefore $\dfrac{2001} 3=23\times29=667|\dfrac{10^{\phi(2001)}-1}9, $ so $2001=3\times667 |\dfrac{10^{3\phi(2001)}-1}9. $

answered Jun 23 '19 at 02:16

J. W. Tanner

60,406

so the number with $3696$ $1$s is divisible by $2001$ – J. W. Tanner Jun 23 '19 at 02:20
Could you explain the last step? – Jun 23 '19 at 03:06
@ChrisCuster: sorry if wasn't clear; does this help? If $a\equiv1\pmod9$ then also $a\equiv1\pmod3$ so $3|a^2+a+1$ so $27|a^3-1=(a-1)(a^2+a+1)$ ; use that with $a=10^{\phi(2001)}$ – J. W. Tanner Jun 23 '19 at 03:43

score 2 · Answer 3 · answered Jun 23 '19 at 04:23

First, we have

$$2001 = (3)(23)(29).$$

Now, let $R_{n} = \frac{10^{n} - 1}{9}$ denote the $nth$ repunit.

Since $3 | R_{3}, \, 23 | R_{22},$ and $29 | R_{28}$, and

Because $LCM(3, 22, 28) = 924$, it follows that

$$2001 \, | \, R_{924} = \frac{10^{924} - 1}{9}.$$

score 1 · Accepted Answer · answered Jun 23 '19 at 20:08

$2001=3\cdot 23\cdot 29$, and $23\cdot 29$ is a divisor of $10^{\text{lcm}(22,28)}-1=10^{308}-1$.
It follows that $2001$ is a divisor of $\frac{10^{924}-1}{9}$, which is a number made by $924$ digits equal to $9$.

As an alternative, the sequence defined by $$ a_0 = 0, \qquad a_{n+1} = 10 a_n+1 $$ is periodic $\!\!\pmod{m}$ for any $m$, see here. In particular there is some $k\in[1,2001]$ such that $a_k\equiv a_0\equiv 0\pmod{2001}$.

Show that there is a number consisting only of 1’s that is divisible by 2001

4 Answers4