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I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.

So that means that $\frac{10^a-1}{9}$ must be divisible by 3.

$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.

When is that last statement true?

Bill Dubuque
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O.S.
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4 Answers4

1

You can use induction to prove

$$\frac{10^n-1}{9n} \tag{1}\label{eq1A}$$

is an integer when $n = 3^k$. The base case of $k = 0$ gives $3^k = 1$, with \eqref{eq1A} becoming $\frac{10-1}{9} = 1$. Next, assume \eqref{eq1A} is an integer for $k = m$ for some integer $m \ge 0$. Then, for $k = m + 1$,

$$\begin{equation}\begin{aligned} \frac{10^{3^{m+1}} - 1}{9(3^{m+1})} & = \frac{\left(10^{3^{m}}\right)^3 - 1}{\left(9(3^{m})\right)3} \\ & = \frac{(10^{3^{m}} - 1)(10^{3^{2m}} + 10^{3^{m}} + 1)}{\left(9(3^{m})\right)3} \\ & = \left(\frac{10^{3^{m}} - 1}{9(3^{m})}\right)\left(\frac{10^{3^{2m}} + 10^{3^{m}} + 1}{3}\right) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

The first factor above is an integer based on the induction hypothesis, i.e., \eqref{eq1A} being an integer for $n = 3^{m}$. For the second factor, note

$$10 \equiv 1 \pmod{3} \implies 10^{3^{2m}} \equiv 10^{3^{m}} \equiv 1 \pmod{3} \tag{3}\label{eq3A}$$

This means

$$10^{3^{2m}} + 10^{3^{m}} + 1 \equiv 3 \equiv 0 \pmod{3} \implies 3 \mid 10^{3^{2m}} + 10^{3^{m}} + 1 \tag{4}\label{eq4A}$$

i.e., the second factor is also an integer. With the product of $2$ integers being an integer, \eqref{eq2A} shows \eqref{eq1A} is an integer for $n = 3^{m+1}$. This proves by induction that \eqref{eq1A} is an integer for $n = 3^k$ for all integers $k \ge 0$.

John Omielan
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  • This essentially repeats the proof of LTE in the cubic case (cf. my answer). Better to prove the general LTE (just as simple) and invoke it as a Lemma rather than repeat its proof every time it is needed. – Bill Dubuque Mar 22 '21 at 10:24
0

Using induction,

If $10^r=1+3^sk$ where $3\nmid k$

$$10^{3r}=(10^r)^3=(1+3^sk)^3=1+3^{s+1}k+3^{2s+1}k^2+3^{3s}k^3\equiv1\pmod{3^{s+1}}$$

for $s+1\le2s+1\iff s\ge0,3s\ge s+1\iff 2s\ge1$

Now for the base case $r=1=3^0, s=2$

So, by using weak induction, $$10^{3^a}-1$$ is divisible by $3^{a+2}$ for $a\ge0$

lab bhattacharjee
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  • This essentially repeats the proof of LTE in the cubic case (cf. my answer). Better to prove the general LTE (just as simple) and invoke it as a Lemma rather than repeat its proof every time it is needed. – Bill Dubuque Mar 22 '21 at 10:24
0

The statement is equivalent to proving $$10^{3^k}-1\equiv_{3^{k+2}}0$$ Notice that $\phi(3^{k+2})=3^{k+2}-3^{k+1}=2\cdot 3^{k+1}$, so that for a unit $a$ we have $$a^{2\cdot 3^{k+1}}\equiv_{3^{k+2}}1$$ We also have

Claim: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}}$$

Proof: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}-2\cdot 3^{k}}\equiv_{3^{k+2}}10^{3^{k+1}}$$

Let $t=10^{3^k}$, then the above claim may be written as $$t^3\equiv_{3^{k+2}}t\iff t(t+1)(t-1)\equiv_{3^{k+2}}0$$ Both $t,t+1$ are units since $t,t+1\not\equiv_{3}0$, therefore we must have $t-1\equiv_{3^{k+2}}0$.

cansomeonehelpmeout
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  • This too is equivalent to reproving LTE in this case (cf. my prior answer). But the general proof is no more difficult (and the general proof can be reused much more generally). – Bill Dubuque Mar 22 '21 at 19:51
-1

Recall a simple form (with simple proof) of LTE = Lifting The Exponent is

$$a\equiv b\!\!\! \pmod{kn}\,\Rightarrow\, a^k\equiv b^k\!\!\!\! \pmod{k^2n}$$

Applied inductively for $\,k=3\,$ (i.e. repeatedly cubing) yields the claim as follows $$\begin{align} &10^{\large 3^\color{#c00}0}\! \equiv 1\!\!\! \pmod{3^{2+\color{#c00}0}}\\ \Rightarrow\ &10^{\large 3^\color{#c00}1}\! \equiv 1\!\!\! \pmod{3^{2+\color{#c00}1}}\ \ \rm by\ \ LTE\\ \Rightarrow\ &10^{\large 3^\color{#c00}2}\! \equiv 1\!\!\! \pmod{3^{2+\color{#c00}2}}\ \ \rm by\ \ LTE\\ &\ \ \ \ \ \ \ \ \ \vdots\\ \Rightarrow\ &10^{\large 3^\color{#c00}k}\! \equiv 1\!\!\! \pmod{3^{2+\color{#c00}k}}\ \ \rm by\ \ LTE\\[.1em] \Rightarrow\ &9\cdot 3^k\mid 10^{3^k}\!\!-1\\[.1em] \Rightarrow\ & 9\cdot n\ \mid 10^n-1 \end{align}\qquad$$

Bill Dubuque
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