MHB Are All Numbers in This Mathematical Sequence Exact Cubes?

[anemone]

Prove that all numbers of the sequence $\dfrac{107811}{3},\,\dfrac{110778111}{3},\,\dfrac{111077781111}{3},\,\cdots$ are exact cubes.

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[anemone]

Congratulations to the following members for their correct solutions:

1. kaliprasad
2. Opalg

Solution from kaliprasad:

Spoiler

Let us look at 1st few terms

1st term = $\dfrac{107811}{3} = \dfrac{1 * 10^5 + 77 * 10^2 + 111}{3}$

2nd term = $\dfrac{110778111}{3} = \dfrac{11 * 10^7 + 777 * 10* 3 + 1111}{3}$

3rd term = $\dfrac{111077781111}{3} = \dfrac{111 * 10^9 + 7777 * 10^4 + 11111}{3}$

so

$\begin{align*}\text{nth term}&= \dfrac{(10^n- 1) * 10^{2n+3} + 7 * 10^{n+2} -1)(10^{n+1} + (10^{n+2} - 1)}{3*9}\\&=\dfrac{ (10^n-1) * 1000 * 10^{2n} + 7 *( 100 * 10^n - 1)(10 * 10^n) + (100 * 10^n - 1)}{27}\\&=\dfrac{1000 * 10^{3n} - 1000 * 10^{2n} + 7 *1000 * 10^{2n} - 7 *10 * 10 ^n + 100 * 10^n - 1}{27}\\&=\dfrac{1000 * 10^3n - 300 * 10 ^2n + 30 * 10 ^n -1}{27}\\&=\dfrac{10^{3n+3} - 3 * 10^{2n+2} + 3 * ^10{n+1} -1}{27}\\&=\dfrac{(10^{n+1} - 1)^3}{27} \end{align*}$so nth term = $(\dfrac{10^{n+1} -1}{3})^3$, which is a perfect cube

Solution from Opalg:

Spoiler

The numerator, call it $N_k$, of the general term in this sequence consists of $3k$ digits which we can divide into three groups of $k$ digits. The first group consists of $k-1$ ones followed by a zero; the second group consists of $k-1$ sevens followed by an eight; and the third group consists of $k$ ones:
\[
N_k = \overbrace{11\cdots10}^k \overbrace{77\cdots78}^k \overbrace{11\cdots11}^k.
\]
Let $P_k = \frac13\!\cdot\!10^k(10^k - 1) = \overbrace{33\cdots33}^k \overbrace{00\cdots00}^k$, using the fact that $10^k - 1 = \overbrace{99\cdots99}^k.$ Then
\[
N_k + P_k = \overbrace{11\cdots10}^k \overbrace{77\cdots78}^k \overbrace{11\cdots11}^k + \overbrace{33\cdots33}^k \overbrace{00\cdots00}^k = \overbrace{11\cdots11}^k \overbrace{11\cdots11}^k \overbrace{11\cdots11}^k = \tfrac19(10^{3k} - 1).
\]
Therefore $N_k = \frac19(10^{3k} - 1) - P_k = \frac19(10^{3k} - 1) - \frac13\!\cdot\!10^k(10^k - 1) = \frac19(10^{3k} - 3\!\cdot\!10^{2k} + 3\!\cdot\!10^k - 1) = \frac19(10^k-1)^3.$But $10^k - 1 = 3R_k$, where $R_k = \overbrace{33\cdots33}^k$. It follows that $N_k = 3R_k^3$ and so $N_k/3$ is the cube of $R_k.$