Croatian Mathematical Competitions

Let $N$ be the number whose decimal representation consists of 2187 digits 1. Then
$$N = \underbrace{111\ldots 1}_{2187\ \text{digits}} = \frac{10^{2187} - 1}{9}.$$

We want to show that $2187$ divides $N$.

Note that $2187 = 3^7$.

It suffices to show that $3^7$ divides $N$.

Since $N = \frac{10^{2187} - 1}{9}$, we need to show that $3^7$ divides $10^{2187} - 1$.

Observe that $10^k \equiv 1 \pmod{9}$ for any $k$, but we need divisibility by $3^7$.

Let us use Euler's theorem:

Since $\gcd(10, 3^7) = 1$, Euler's theorem gives $10^{\varphi(3^7)} \equiv 1 \pmod{3^7}$, where $\varphi(3^7) = 3^7 - 3^6 = 2187 - 729 = 1458$.

So $10^{1458} \equiv 1 \pmod{2187}$.

But $2187 = 3^7$, and $2187$ divides $10^{2187} - 1$ if and only if the order of $10$ modulo $2187$ divides $2187$.

Alternatively, note that $N = \frac{10^{2187} - 1}{9}$, so $N$ is an integer, and $3^7$ divides $N$ if and only if $10^{2187} \equiv 1 \pmod{3^8}$, since $N = \frac{10^{2187} - 1}{9}$ and $3^7$ divides $N$ if $10^{2187} \equiv 1 \pmod{3^8}$.

But actually, since $N$ is a repunit, and $3^k$ divides $10^{3^k} - 1$ for $k \geq 1$, we can use induction or known results:

For any $k \geq 1$, $3^k$ divides $10^{3^k} - 1$.

Proof by induction:

Base case: $k = 1$, $10^3 - 1 = 999$, which is divisible by $3$.

Assume $3^k$ divides $10^{3^k} - 1$.

Consider $10^{3^{k+1}} - 1 = (10^{3^k})^3 - 1 = (10^{3^k} - 1)((10^{3^k})^2 + 10^{3^k} + 1)$.

By the induction hypothesis, $3^k$ divides $10^{3^k} - 1$, so $3^k$ divides $10^{3^{k+1}} - 1$.

But also, $10^{3^k} \equiv 1 \pmod{3^k}$, so $(10^{3^k})^2 + 10^{3^k} + 1 \equiv 1 + 1 + 1 = 3 \pmod{3^k}$, so $3$ divides this factor, so $3^{k+1}$ divides $10^{3^{k+1}} - 1$.

Therefore, $3^7$ divides $10^{3^7} - 1 = 10^{2187} - 1$.

Thus, $3^7$ divides $N = \frac{10^{2187} - 1}{9}$.

Therefore, $2187$ divides the number whose decimal representation consists of $2187$ digits 1.