Show that $\underbrace{1\ldots1}_{81\text{ times}}$ is divisible by $81$ but not by $243$?

Question

Question

How do I show that $$\underbrace{1\ldots1}_{81\text{ times}}$$ is divisible by $81$ but not by $243$?

My Attempt

Write $$\underbrace{1\ldots1}_{81\text{ times}}=\underbrace{111,111,111}_{9\text{ times}}$$ and note that $111,111,111=9 \times 12,345,679$. This also shows that $9 \mid 111,111,111$.

Now, the sum of digits of $$\underbrace{12,345,679}_{9\text{ times}}$$ is $37 \times 9$ which is clearly divisible by $9$, thereby showing that $9 \mid 12,345,679$.

This in turn shows that $$81 \mid \underbrace{111,111,111}_{9\text{ times}}$$

Next, note that $$111,111,111=243 \times 457247.37\overline{037}=243 \times \left(457247+\frac{370}{999}\right)$$

As a result, we have $$\underbrace{111,111,111}_{9\text{ times}}=243 \times \left(457247+\frac{370}{999}\right) \times \left(1+10^{9}+\cdots+10^{72}\right)$$

To show that $$243 \nmid \underbrace{111,111,111}_{9\text{ times}}$$ we need only show that $999 \nmid (1+10^9+\cdots+10^{72})$.

Indeed, as $10^3 \equiv 1 \pmod {999}$, so $10^9=(10^3)^3 \equiv 1^3 \equiv 1 \pmod {999}$ and in general $10^{9a} \equiv 1 \pmod {999}$ for $a \in \mathbb{N}$. Since $$(1+10^9+\cdots+10^{72}) \equiv \underbrace{1+\cdots+1}_{9\text{ times}} =9 \pmod {999}$$ so $999 \nmid (1+10^9+\cdots+10^{72})$ and we get the desired result.

Doubt

Is the above solution correct? If it is correct, is there a shorter and more elegant solution?

Answer 1

The problem is equivalent to show that $10^{81}-1$ is divisible by $3^6$ but not by $3^7$. This follows from the fact that, for $x=10$, $$\begin{align} x^{81}-1&=(x^{27}-1)(1+x^{27}+x^{54})\\ &=(x^{9}-1)(1+x^9+x^{18})(1+x^{27}+x^{54})\\ &=(x^{3}-1)(1+x^3+x^6)(1+x^9+x^{18})(1+x^{27}+x^{54})\\ &=\underbrace{(x-1)}_{=3^2}\underbrace{(1+x+x^2)}_{3k_1}\underbrace{(1+x^3+x^6)}_{3k_2}\underbrace{(1+x^9+x^{18})}_{3k_3}\underbrace{(1+x^{27}+x^{54})}_{3k_4} \end{align}$$ where $k_1,k_2,k_3,k_3$ are coprime with $3$.

Answer 2

Let us study $\nu_3(10^n-1)$. Since $10^n-1 = 3^2(10^{n-1}+\ldots+1)$ and $10^{n-1}+\ldots+1\equiv n\pmod{3}$, $$ 3\nmid n\rightarrow \nu_3(10^n-1)=2.$$ Let us assume $n=3m$ now: $10^{3m}-1 = (10^m-1)(10^{2m}+10^{m}+1)$ with $10^{2m}+10^{m}+1\equiv 3\pmod{9}$, so $$ 3\mid n \rightarrow \nu_3(10^{n}-1) = 1+ \nu_3(10^{n/3}-1)$$ and by induction $$ \boxed{\nu_3(10^n-1) = 2+\nu_3(n)}. $$ If we consider $n=81$ we have $\nu_3(n)=4$, hence the largest power of $3$ dividing $10^{81}-1$ is $3^6$ and the largest power of $3$ dividing the $81$th repunit is $3^4=81$.

Answer 3

An alternative solution is using binomial theorem. Notice $$\begin{align} X \stackrel{def}{=}\underbrace{1\ldots 1}_{81} &= \frac{10^{81}-1}{10-1} = \frac{(1+9)^{81} - 1}{9}\\ &= \binom{81}{1} + \binom{81}{2}9 + \binom{81}{3}9^2 + 9^3 (\cdots)\end{align}$$ where $(\cdots)$ is some integer we don't care. Since $243 = 3^5 | 9^3$, we get $$\begin{align} X & \equiv 81 + \frac{81\cdot 80}{2} 9 + \frac{81\cdot 80\cdot 79}{6} 9^2 \pmod {3^5}\\ & \equiv 3^4 + 40(3^6) + (40)(79) 3^7 \pmod {3^5}\\ & \equiv 3^4 \pmod {3^5}\end{align}$$ When we divide $X$ by $243$, the remainder is $81 = 3^4$. This implies $X$ divides $81$ but not $243$.