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Define a sequence $\{a_i\}_{1}^{\infty}$ as follows: $$a_i = \left\lfloor \dfrac{10^{i+1}}{7}\right\rfloor-100\left\lfloor\dfrac{10^{i-1}}{7}\right\rfloor.$$ Find $\max_{i \in \mathbb{N}} a_i$.

Let $10^{i-1} \equiv j \pmod{7}$, where $0 \leq j < 7$. We have $$\left\lfloor \dfrac{10^{i+1}}{7}\right\rfloor-100\left\lfloor\dfrac{10^{i-1}}{7}\right\rfloor = \left\lfloor \dfrac{10^{i+1}}{7}\right\rfloor-100 \cdot \dfrac{10^{i-1}-j}{7},$$ where $0 \leq j < 7$. I didn't see how to continue.

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$a_i$ is the last two digits before decimal point of $\frac{10^{i+1}}{7}$

But this digits can be only $1428571428571....$

The maximum of which is $85$

EDIT: More details at OP's request:

The decimal representation of your numbers is:

$$\frac{10^{i+1}}{7}=\overline{d_1d_2...d_{i+1}.d_{i+2}d_{i+3}...}$$ where $d_1,d_2\ldots$ cycle through $1,4,2,8,5,7$

So: $$\left\lfloor\frac{10^{i+1}}{7}\right\rfloor=\overline{d_1d_2...d_{i+1}}$$

The other number: $$\frac{10^{i-1}}{7}=\overline{d_1d_2...d_{i-1}.d_id_{i+1}...}$$ where $d_1,d_2\ldots$ cycle through $1,4,2,8,5,7$

So $$\left\lfloor\frac{10^{i-1}}{7}\right\rfloor=\overline{d_1d_2...d_{i-1}}$$ $$100\left\lfloor\frac{10^{i-1}}{7}\right\rfloor=\overline{d_1d_2...d_{i-1}00}$$

So $a_i=\overline{d_id_{i+1}}$

Which cycles through $14,42,28,85,57,71$, the maximum of which it is $85$

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  • $\begingroup$ How is $a_i$ the last two digits before decimal point of $\frac{10^{i+1}}{7}$? $\endgroup$ – Puzzled417 Jan 11 '17 at 3:22
  • $\begingroup$ I used the properties of floor function, specifically $\lfloor\lfloor 10^{i+1}/7\rfloor/100\rfloor=\lfloor 10^{i+1}/700\rfloor=\lfloor 10^{i-1}/7\rfloor$ $\endgroup$ – Momo Jan 11 '17 at 3:27
  • $\begingroup$ I don't understand. Can you elaborate? $\endgroup$ – Puzzled417 Jan 11 '17 at 3:30
  • $\begingroup$ ... and for the rest of it think at the $\lfloor 123.45\rfloor-100\cdot\lfloor 1.2345\rfloor= 23$. Can you see now how the last two digits before decimal point are extracted? $\endgroup$ – Momo Jan 11 '17 at 3:34
  • $\begingroup$ Yes, but I want to see the proof of this. $\endgroup$ – Puzzled417 Jan 11 '17 at 3:36
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Since every rational has a cyclic fractional part, let's explicit Momo's answer with full sum notation. It is a little tedious, and shows why the overline notation is so powerful, but anyway, let's do it!

$\frac{1}{7}=0.[142857]...=\sum\limits_{k=1}^{\infty}142857\times10^{-6k}$

$$\frac{1}{7}=\sum\limits_{k=1}^{\infty}c_{k-1}\,10^{-k}$$

With $c_0=1,\;c_1=4,\;c_2=2,\;c_3=8,\;c_4=5,\;c_5=7$

$c_{6q+r}=c_{r}$ for $r=0\,..\,5$

REM: I change a little bit $a_i$ definition to ease expressions

$$a_{i}=\lfloor\frac{10^{2+i}}{7}\rfloor-100\lfloor\frac{10^i}{7}\rfloor$$


Let's have a look at

$$a_i=\lfloor\sum\limits_{k=1}^{\infty}c_{k-1}10^{2+i-k}\rfloor-100\lfloor\sum\limits_{k=1}^{\infty}c_{k-1}10^{i-k}\rfloor$$

Let's note partial sums $S(k_0,k_1)=\lfloor\sum\limits_{k=k_0}^{k_1}c_{k-1}10^{2+i-k}\rfloor-100\lfloor\sum\limits_{k=k_0}^{k_1}c_{k-1}10^{i-k}\rfloor$

$a_i=S(1,\infty)=S(1,i)+S(i+1,i+2)+S(i+3,\infty)$


We notice that in $S(1,i)$ since $k\le i$ the powers of $10$ are positive so we can get rid of the floor function, these all are integers.

$S(1,i)=\sum\limits_{k=1}^{i}c_{k-1}\,10^{2+i-k}-100\sum\limits_{k=1}^{i}c_{k-1}\,10^{i-k}=0$ since both terms are equal.


We notice that $k\ge i+3$ then $(i-k)<0$ and $(2+i-k)<0$.

All powers of $10$ are strictly negative, thus the sum is $<1$ and floor value is $0$.

$S(i+3,\infty)=0-100\times 0=0$


It remains only one term :

$S(i+1,i+2)=\lfloor c_{i}10^1+c_{i+1}10^0\rfloor-100\lfloor c_{i}10^{-1}+c_{i+1}10^{-2}\rfloor$

The first term is an integer, and the second one is $<1$ thus

$S(i+1,i+2)=c_{i}10^1+c_{i+1}10^0-100\times 0=10c_{i}+c_{i+1}$


$$a_i=10\,c_i+c_{i+1}$$

So since the $c_i$ are cyclic, the values for $\{a_i\}_i$ are $\{14,42,28,85,57,71\}$ and the max is $85$.

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