Largest value of sequence

Question

What is the idea of this problem? I couldn't understand it

Any help would be much appreciated

EDIT: further details inside this post

Find $\max_{i \in \mathbb{N}} a_i$ where $a_i = \left\lfloor \frac{10^{i+1}}{7}\right\rfloor-100\left\lfloor\frac{10^{i-1}}{7}\right\rfloor$

Answer 1

The formula means "what is the largest digit in decimal expansion of $\dfrac 1{13}$" ?

Since $\dfrac 1{13}=0.\overline{076923}\cdots$ then this digit is $9$.

Multiply by $10^i$ shifts the sequence to the left for instance by $1000$ we get $76.923\overline{076923}\cdots$

The entire part is $76$. The other term is shifted one less place but multiplied by $10$ so it is $70$. The difference is $6$ which is the digit at this $3^{rd}$ place in the expansion.

You can write $\displaystyle \dfrac 1{13}=\sum_{k=1}^{\infty} d_k10^{-k}$ apply the formula above and realize you'll find $d_i$ as an answer.

I did it once in this post : Find $\max_{i \in \mathbb{N}} a_i$ where $a_i = \left\lfloor \frac{10^{i+1}}{7}\right\rfloor-100\left\lfloor\frac{10^{i-1}}{7}\right\rfloor$

This is a bit tedious to write, please refer to that post for more details.

Answer 2

The period is $$0,7,6,9,2,3,$$ which gives that $9$ is a maximum.

Answer 3

The correct choice is $(D)$

The maximum value of $9$ occurs at $i=10$

that is $a_{10} =9$ is the largest.