$\color{green} \textbf{Additive Remainders.}$ Define $r_i$ differently now: not only does range of the slider $i$ is extended from $(0,a)$ to $(0,c-1)$, but also not by remainder: after defining $r_0 = 0$, define $r_{i+1} \equiv r_i+b \pmod c$. $\color{green} \rule{25cm}{0.4pt}$ $\color{green} \textbf{In All Its Glory.}$ If $(b,c) \ne 1$, it's impossible for $1$ to be a part of $r_i$, as it will result in all $r_i$ being a multiple of $\text{gcd}(b,c)$. So, assuming that the set $\{r_0,r_1,\ldots,r_{c-1}\}$ is complete, we get that $\{r_{a+1},\ldots,r_{c-1}\}$ equals to $\{a+1,a+2,\ldots\}$ (I purposefully emitted the existence of $c-1$ there --- those two numbers are the one we will consider.) $\color{green} \rule{25cm}{0.4pt}$ First, we will prove the obvious bound that $b \leq a$. This is because $r_1$ is $\textit{exactly}$ $b$, and if it exceeds $a$, then $r_1 \not\in \{0,1,2,\ldots,a\}$. $\color{green} \rule{25cm}{0.4pt}$ Then, since $b \ne 1$, the two numbers $a+1-b$ and $a+2-b$ modulo $c$ are greater than zero for $b \leq a$ reasons; and are at most $a$ due to $b$'s positivity. Then, they should be among $\{r_0,r_1,\ldots,r_a\}$ --- but only one of them (a.k.a. $r_a$), when added with $b$, should result in a member from the big set $\{r_{a+1},\ldots,r_{c-1}\}$.

Experimenting right of the bat, there are $c-2$ candidates for $a$ (from $1$ to $c-2$) and $c-2$ candidates for $b$ (from $2$ to $c-1$.) $\color{cyan} \rule{25cm}{2pt}$ $\color{green} \text{Sec 1: Preliminary Counting.}$ Excluding the blatant choices when $a = 0$ and $a = c-1$ (the problem is blatantly wrong if we pick only $r_0$ or all $r_0$ until $r_{c-1}$ --- and from here on I've set my eyes to consider all of $r_0$ until $r_{c-1}$ instead of ignoring the late $r_{a+1}$ until $r_{c-1}$), I tried to derive a contradiction from $a = 1$ or $a = c-2$ (the next bordered values.) Surprisingly (or better worded: $\textit{obviously}$), those two can contradict the problem statement only when $r_1 = 1$ or $r_{c-1} = c-1$; when the multiplier $b$ is equal to $1$. This also makes spotting easy patterns to contradict on $a$ small to be useless --- any middle schooler would instantly recognize the fallacy of extracting a pattern with no intricacy at all. Also, by the 5 MOHS argument of forcing gcd of $b$ and $c$ to be $1$, there's no point in analyzing the $``\text{degenerate}''$ cases when $b = 2, c= 4$; for example, so I directly skipped to $c = 5$ for less obvious examples. $\color{green} \rule{25cm}{0.4pt}$ For the still easy case of $c = 5$, there are two non-obvious $a$s to start: $a = 2$ and $a = 3$. Seeing that the condition \[ \{r_0,r_1,r_2\} = \{0,1,2\} \Rightarrow \{r_1,r_2\} = \{1,2\} \]quickly devolved into a senseless condition (after all, $r_0$ shouldn't be considered as a free variable/one thing to particularly zoom in, of course) I decided to stick to the remaining option: $a = 3$. Here I recalled choosing $a$ (the threshold) instead of $b$ (the multiplier) because the main problem seemed to be how big the threshold were, and I wanted the multiplier's weirdness to adapt to the threshold's (apparently fixed) size. Well, if any of that makes sense for an $a = 3$ anyway; $(a,b) = (3,2)$ gave $\{r_1,r_2,r_3\} = \{2,4,1\}$ with $3$ disappearing --- and $b = 4$ made a $4$ in $r_1$. At this point I realized that the condition is equivalent to $4$ missing from the first four resides, and I was back to square one. At least I got the impression that $1$ shouldn't be missing from any set, and that's apparentlya a big deal (from all attempts, I found that it's especially hard to make $1$ disappear unconditionally.) $\color{cyan} \rule{25cm}{2pt}$ $\color{green} \text{Sec 2: Some relatively useless progress.}$ Since small values of $c$'s behavior is easily manipulated and determined, I tried to jump towards high $c$, and a relatively large but comfortable prime to work on was $11$. Well, since this is (at least seems to be) a size problem, the instant loophole I spotted was that \[\begin{tabular}{p{12cm}} if 1 is in the front, then it's easy to find some small number which must lie at the back of the pack. \\ \end{tabular}\]and so I fixed $1$'s position in the back, say at the seventh position $r_7$. By that, I obtained the decoy $b = 8$. (By the way, the rigorous way to say the above observation was that if $b^{-1}$ is less than $c-a-1$, then we're done. Such an ugly statement but it usually works if $b^{-1}$ is indeed small.) $\color{green} \rule{25cm}{0.4pt}$ Now since I've represented the problem as following: \[ \left\{ c \left\{\dfrac{0}{c}\right\}, c \left\{\dfrac{b}{c}\right\},\ldots, c\left\{\dfrac{ab}{c}\right\} \right\} = \{0,1,\ldots,a\} \]I started to view $b$ not necessarily in the form of \[ b \times \{0,1,\ldots,a\} = \{0,1,\ldots,a\} \]as post $\#9$ views it; but in the additive sense in my Solution as they provide a means to bound $r_i$'s size. Speaking of that, a problem which (also) immediately surfaced was the guaranteed existence of $b$, a.k.a. $8$, in the first remainder $r_1$. This is already a very big number, which kind of forced $a$ to be $9$ or more. So I cut it there; and got the very nice bound (which acutely narrows the possibility of $a$ in relation to a big summand $b$) \[ a \geq b \]and here I stopped at the configuration \[ \{0,8,5,13 = 2,10,7,4,1,9,6\} = \{r_0,r_1,r_2,r_3,r_4,r_5,r_6,r_7,r_8,r_9\} \]with apparently an (almost) perfect config with a misplaced $3$ in the back and a $10$ in the front. (To add a bit of satisfaction: this kind of suits my initial goal of making a config $(a,b)$ looking kind of sus; apparently all almost-contradicting sets' misplaced numbers are very small or very big --- none in the middle.) $\color{cyan} \rule{25cm}{2pt}$ $\color{green} \text{Sec 3: The Lucky Break, Extended}$. Now I didn't really focus to the particular values of $r_0$ until $r_a$ (the method of precise output is usually put into a series/polynomial as in $\#4$ or $\#6$; I'll get to it towards the $\color{red} \text{end of this Motivation.}$ To begin, I picked another decoy prime $c = 17$ and set $a \geq b = 12 \geq 8$. What can I infer from blindly picking these numbers? Clearly, I would want the extreme numbers $1$ and $16$ in the front/back of the whole pack, obviously, but what can I get from separating the remainders $r_0$ to $r_16$ somewhere in the middle (to be exact, after the $a^{th}$ remainder) and having them naturally arranged based on size? Naturally, to contrast $\textbf{more}$ large/small numbers with each other! (The Lucky Break.) Going away from the apparently must-be-big set $\{r_0,r_1,\ldots,r_a\}$, I went to look at the smaller \[ \{r_{13},r_{14},r_{15},r_{16}\} = \{13,14,15,16\} \]sized set. If you sniff $\textit{very}$ closely, it would seem that something is forced inside this set, is it not? Not only that, it didn't seem like a $14$ when added with a $8$ could also big; could it be that not some, but $\textit{all}$ terms don't actually have their identity to be $\textit{that}$ big? Apparently, yes! While I initially considered terms where $r_i = c-1$ or $r_j = c-2$, since we're talking about the border of $r_a$ and $r_{a+1}$ here, it would seem best to make the gap insultingly small; just select $r_i$ and $r_j$ equal to $a+1$ to some $a+c$ ($c$ constant) since the transition can only happen once! I was overjoyed, but my mind was still filled with a lot of other contradictions; I'd say what's most difficult in this problem is not about $\textit{finding a contradiction}$; rather; it's more to $\textit{finding the contradiction that finishes the problem.}$ \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]Bonus: why I think this problem is very beautiful.After writing the above Solution and before writing this Motivation, I skimmed the above Solutions --- and found out that they're $\textit{very much related}$ to one another. $\color{red} \rule{25cm}{1pt}$ $\color{red} \text{Adding from the back, or divisibilities from size issues.}$ The most similar ones out of all above seem to be post $\#2$; Utilizing the transition from $\{r_0,r_1,\ldots,r_a\}$ to $\{r_{a+1},\ldots,r_{c-1}$ instead of mine, from $a+1$ and $a+2$ (elements of the back set) to the front; ruling out the case $a = 1$ instead of $a = c-2$ in my part; after pulling out the bound in $a+b-1$ and $c$, the finalization occured when comparing $r_{c-1}$ in relation to the set $\{a+1,\ldots,c-1\}$ instead of comparing $r_1 = b$ in relation to the set $\{0,1,\ldots,a\}$ with the main idea being $c-b$, presumably with $b$ small, is still fairly large when compared with $a+1$, the transiting term towards $\{a,a-1,\ldots,1,0\}$. $\color{red} \rule{25cm}{0.4pt}$ And with that, reviewing the next post $\#3$: it has some very fatal typos which rendered the Solution wrong, but it has some interesting methodology to it (being a true impressive one-liner, at that.) Now, the blatantly confusing pick of the number $r$ with \[ 0 < r < a, a < r+b < c \]was the groundbreaking idea: it highlights the $\color{green} \textbf{Additive Remainders}$ idea in an algebraic way, that is: assuming the set $\{r_0,r_1,\ldots,r_{a}\}$ is as the problem says, any remainder $r$ less than $a$ will happen, and here we'd like to specify one of them (say $r = r_k$) that will yield something greater than $a$, but not so big as to exceed the modulo class $c$, when added with the summand $b$. However, $kb$ were not to be divided with $a$, but with $c$ --- so the correct expression would be \[ kb = mc + [r] \Rightarrow (k+1)b = mc + [r+b]; \ \text{with} \ r+b > a \Rightarrow r \geq a-b+1\]and modifying the bound a bit by considering $r_1$, we can be sure that $\textbf{that number is positive.}$ Since there's the problem of incidentally selecting $r_k = r_a$, the problem is not immediately over, since $k+1$ may as well be $a+1$, indicating that \[ (a+1)b = mc + [r+b] \Rightarrow r_{a+1} = [r+b] = (a-b+1)+b=a+1 \]if we turned around from this statement to selecting a number $a-b+2$ as the new $r$, we can proceed as in my Solution: but we'd have to verify that $(a-b+2)+b = a+2$ does not cycle back to zero (or reach $c$); i.e. $a+2$ shouldn't be equal to $c$ or $a \ne c-2$. $\color{red} \rule{25cm}{0.4pt}$ However, the equation \[ r_{a+1} = a+1 \]seemed to be very interesting to zoom into: and from here I was quite excited to see the link between this approach towards this approach to the Solutions in $\#5$ and $\#8$. First of all, I deemed the divisibility \[ (a+1)(b-1) = mc \Rightarrow c \mid (a+1)(b-1) \]to be relatively useless on its own; as I didn't directly see a connection between $p \mid c$ implying $a \equiv -1 \pmod p$ and $b \equiv 1 \pmod p$. The latter even implies that when considering mod $p$ (prime) divisor of $c$, the remainders $r_1$ until $r_{c-1}$ rises in a way forbidden to what the problem dictates ($b \not\equiv 1$). So, I took a step back and analyzed the terms $\textbf{surrounding}$ $r_{a+1}$ as follows: \[ \begin{tabular}{c|c|c|c|c|c|c} $k$&$\dots$&$a-1$&$a$&$a+1$&$a+2$&$\dots$ \\ \hline $r_k$&$\dots$&$a-2b+1$&$a-b+1$&$a+1$&$a+b+1$&$\dots$ \\ \end{tabular} \]and assuming we've already known that $a$ is greater than $b$, it would seem that before it reaches $k = 1$ (going down the ladder from $a$ to $1$), some small negative with magnitude less than $b$ will pop up and causing a term of magnitude at least $c-b+1$ (the term creating the negative would have to be $1$ or greater, for it cannot be zero) --- somewhat similar to the \[ r_{b^{-1}-1} \equiv 1-b \pmod c = c-b+1 \]argument in post $\#8$. But we're not done yet. Since the identity $c \mid (a+1)(b-1)$ promises that if $c \nmid b-1$ (which should happen due to $b < c$, $a+1$ will have to share some prime divisor with $c$, rendering its gcd to $\textbf{not be}$ 1. Why should that be? If its gcd is equal to one, then $b = 1$, which results in the statement \[ r_k = k \ \text{for every} \ k, \ 1 \leq k \leq c-1 \]So what should happen if $\text{gcd}(a+1,c) = g$? Take $a = 5$ and $c = 15$, for example, yielding $\text{gcd}(6,15) = 3$. However, by Bezout (or Euclidean Algorithm in general), we get that adding $6$ to $6$ will yield \[ r_{12} = 12, r_{18} \equiv 18 \ (\text{mod} \ 15), r_{9} = 9, r_{15} = 15 \ (\text{or} \ r_0 = 0) \]which fulfilled our promise that $3$ should be the smallest $k$ so that $r_k = k$. Notice that by $r_{a+1} = a+1$ for any $a$ will create at least two blocks of similar sizes: \[ \begin{tabular}{c|ccc|ccc|c} $k$&0&1&2&$\color{red}3$&4&5&$\color{blue}6$ \\ \hline $r_k$&0&&&$\color{red}3$&&&$\color{blue}6$ \\ \end{tabular}\]$\color{red} \rule{25cm}{0.4pt}$ From that, we can somewhat $\color{red} \textit{induct down}$ from the case $a = 5$ to the case $g = a' = 3$, being the common divisor of $6$ and $15$. This is because the placement of $1$ and $2$ in the second row (the $r_i$ row) $\color{red} \textit{must be in the first block}$, any later (for example $r_5 = 1$) would displace a $4$ in $r_8$, somewhere later than $a = 5$. When I say that this is an inductive argument, however, this doesn't mean that I can use the term $a = g-1$ (here denoted by $2$) to lower it further; I'd have to resort to the similarities in block size (in contrast with what $\#5$ does.) Now say that $1$ appears in $r_k$ where $k \ne 1$. Since it has to appear in the first block, we'd have $r_{lk} = l$ ($l \leq g-1$) as long as $lk$ does't exceed $c$. Then we can pick the first one to exceed $g$ and we are set. $\color{red} \rule{25cm}{0.4pt}$ A final remark: comparing with post $\#8$, after presumably establishing $a \geq b$ --- since we'd have to bound $b$ eventually if we were to use the (at least $c-b+1$) thing, the existence of similarly sized blocks may hint at the small size of $a$; as there are at least two of them. Then, noticing the similar behavior occuring between the left set $\{r_0,r_1,\ldots,r_a\}$ and $\{r_{a+1},\ldots,r_{c-1}\}$ (i.e. they are additive mod $c$, and each term are $b$ more than its previous one), one may as well WLOG the one from $a+1$ to $c-1$ to be the desired left set and operate on a $c-a-1$-sized block (which has size at most $\dfrac{c}{2}$). This will guarantee that the size of the left set ($a$) and the size of the summand $b$ which also is less than $a$ will not interfere with each other --- as $b \leq \dfrac{c}{2}$, a term as big as $c-b+1 \geq c-c\dfrac{c}{2}+1$ shouldn't appear in the small-sized left block $\{r_0,r_1,\ldots,r_a\}$. Note that without specifying $b$'s small size, the term $1-b$ or $c-b+1$ before $r^{b^{-1}}$ wouldn't be sus at all, due to the sheer large negative it has --- see my first configuration on the end of $\color{green} \text{Sec 2}$. There, it'd seem like $b+1 = r^{b^{-1}+1}$ would be a better candidate to derive a contradiction, but the realization that \[ b \leq a \Rightarrow b+1 \ \text{should be} \ \leq a \ \text{as well} \]should be enough to rethink that approach. $\color{red} \rule{25cm}{1pt}$ $\color{red} \text{Exact values: Series mod a polynomial.}$ Speaking of determining the set $\{r_0,\ldots,r_a\}$ precisely, I'd like to share a problem I've done (a short while after doing this problem and before writing my post) which illustrates the idea wonderfully: \[\begin{tabular}{p{12cm}} Let $p$ prime, $k$ natural number, and $\mathcal{A}$ be a finite set of integers with at least $p^k$ elements. Denote $N_{\text{even}}$ to be the set of $\mathcal{A}$'s subsets with even cardinality, and sum of elements divisible by $p^k$ --- and $N_{\text{odd}}$ the same way (with odd cardinality.) \\ \\ Prove that $|N_{\text{even}}| \equiv |N_{\text{odd}}| \pmod p$. \\ \\ This was initially done with $k = 1$, but the author generalized it to $p^k$ due to Kummer. \\ \end{tabular} \\ \]Since the best way to represent sum of elements is by manipulating power series, let $\mathcal{A} = \{a_1,\ldots,a_m\}$ with $m\geq p^k$ and construct the series \[ S_{\mathcal{A}} = (x^{a_1}-1)(x^{a_2}-1)\ldots(x^{a_m}-1)\]Why is this beautiful? Since the sum of ever subset of $\mathcal{A}$ is representable by multiplying $x^{\text{element included in that subset}}$ with each other (and discarding elements not included there by selecting the constant part), the problem is now equivalent to finding the total coefficient in powers divisible by $p^k$. And the constant should also be negative as the number of non-included terms in $N_{\text{even}}$ are equal in parity, and each addition of a subset (as a product of $x^{a_i}$) should have one magnitude (whether the sign is positive/negative depends on the number of included terms.) After absorbing those two observations by heart (sorry if that is too detailed, anyway) it should be easy by noting that \[ (x-1)^{p^k} \mid S_{\mathcal{A}} \]and that the coefficients of $(x-1)^{p^k}$ modulo $p$ (when grouped according to their powers mod $p^k$) is equal to $0 \pmod p$. Then the problem solves itself without needing to consider anything else (the construction of the series are the main difficulty.) I decided to include this in the Motivation to motivate the construction of $x^{ab}+\ldots+x^a+1$ dividing $x^c-1$ --- as the above problem basically says that \[ \text{some polynomial} \ ( \text{here it's} \ ) (x-1)^{p^k} \mid S_{\mathcal{A}}\]solves the whole thing, so does determining what adds with each other (here, $r_k$ adds with each other; hence the definition of $r_{i+1} \equiv r_i +b \pmod c$ in my Solution), what congruence/divisibility is there ($x^c-1$ is due to its powers removing itself modulo $c$ to be included among $r_k$) and how to manipulate them (easy, by noting that the end result of the powers simplifying itself should be equal to $\{0,1,\ldots,a\}$ in series form.) \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]