Consider the problem of representing the reciprocal of an integer as an Egyptian fraction where all the denominators are palindromes. i.e. write $$ \frac{1}{n} = \sum_{i} \frac{1}{a_i} $$ where $a_i$ is a palindrome (repeating $a_i$ is allowed). In base 10 it is possible to represent all reciprocal integers as a sum of reciprocal palindromes. For example $$ \frac{1}{87} = \frac{1}{88} + \frac{1}{8008} + \frac{1}{232232} + \frac{1}{696696} $$ This is not possible in bases that are primes or prime powers. I know how to prove this, so that's not the question. The question or challenge rather is whether anyone can represent the numbers from 10 to 100 with fewer reciprocal palindromes than I have managed. I will write $1/37=3/111$ as a short-hand for $1/37 = 1/111 + 1/111 + 1/111$, but I count this as a representation with 3 palindromes. Without further ado this is my list: \begin{eqnarray*} \frac{1}{10} & = & \frac{1}{22} + \frac{3}{55}\\\\ \frac{1}{11} & = & \frac{1}{11}\\\\ \frac{1}{12} & = & \frac{2}{33} + \frac{1}{44}\\\\ \frac{1}{13} & = & \frac{1}{22} + \frac{1}{33} + \frac{1}{858}\\\\ \frac{1}{14} & = & \frac{1}{22} + \frac{2}{77}\\\\ \frac{1}{15} & = & \frac{1}{33} + \frac{2}{55}\\\\ \frac{1}{16} & = & \frac{1}{22} + \frac{2}{121} + \frac{1}{2112} + \frac{1}{23232}\\\\ \frac{1}{17} & = & \frac{1}{55} + \frac{3}{77} + \frac{1}{595}\\\\ \frac{1}{18} & = & \frac{1}{22} + \frac{1}{99}\\\\ \frac{1}{19} & = & \frac{1}{33} + \frac{1}{99} + \frac{2}{171} + \frac{1}{1881}\\\\ \frac{1}{20} & = & \frac{2}{55} + \frac{1}{77} + \frac{1}{4004} + \frac{2}{5005}\\\\ \frac{1}{21} & = & \frac{1}{22} + \frac{1}{858} + \frac{1}{1001}\\\\ \frac{1}{22} & = & \frac{1}{22}\\\\ \frac{1}{23} & = & \frac{1}{33} + \frac{1}{99} + \frac{1}{414} + \frac{3}{4554}\\\\ \frac{1}{24} & = & \frac{1}{33} + \frac{1}{88}\\\\ \frac{1}{25} & = & \frac{1}{33} + \frac{1}{252} + \frac{1}{404} + \frac{1}{505} + \frac{1}{909} + \frac{1}{5775} (\bf {joriki})\\\\ \frac{1}{26} & = & \frac{1}{44} + \frac{1}{66} + \frac{1}{3003} + \frac{1}{4004}\\\\ \frac{1}{27} & = & \frac{1}{44} + \frac{1}{77} + \frac{1}{777} + \frac{1}{27972}\\\\ \frac{1}{28} & = & \frac{1}{44} + \frac{1}{77}\\\\ \frac{1}{29} & = & \frac{1}{44} + \frac{1}{88} + \frac{1}{2552}\\\\ \frac{1}{30} & = & \frac{1}{55} + \frac{1}{66}\\\\ \frac{1}{31} & = & \frac{1}{66} + \frac{1}{77} + \frac{1}{252} + \frac{1}{6776} + \frac{1}{270072}\\\\ \frac{1}{32} & = & \frac{1}{33} + \frac{2}{2112}\\\\ \frac{1}{33} & = & \frac{1}{33}\\\\ \frac{1}{34} & = & \frac{1}{44} + \frac{1}{242} + \frac{1}{484} + \frac{2}{4114}\\\\ \frac{1}{35} & = & \frac{1}{44} + \frac{1}{252} + \frac{1}{585} + \frac{1}{6006}\\\\ \frac{1}{36} & = & \frac{1}{44} + \frac{1}{202} + \frac{1}{9999}\\\\ \frac{1}{37} & = & \frac{3}{111}\\\\ \frac{1}{38} & = & \frac{1}{66} + \frac{1}{99} + \frac{2}{1881}\\\\ \frac{1}{39} & = & \frac{1}{44} + \frac{1}{616} + \frac{1}{858} + \frac{1}{8008}\\\\ \frac{1}{40} & = & \frac{1}{44} + \frac{1}{505} + \frac{1}{5555} + \frac{1}{8888}\\\\ \frac{1}{41} & = & \frac{1}{77} + \frac{1}{99} + \frac{1}{777} + \frac{1}{81918} + \frac{1}{333333} + \frac{1}{81999918}\\\\ \frac{1}{42} & = & \frac{1}{44} + \frac{1}{2002} + \frac{1}{3003} + \frac{1}{4004}\\\\ \frac{1}{43} & = & \frac{1}{44} + \frac{1}{2442} + \frac{1}{13431} + \frac{1}{26862} + \frac{1}{210012} + \frac{1}{420024} + \frac{1}{4620264}\\\\ \frac{1}{44} & = & \frac{1}{44}\\\\ \frac{1}{45} & = & \frac{1}{55} + \frac{1}{585} + \frac{2}{858}\\\\ \frac{1}{46} & = & \frac{1}{77} + \frac{1}{161} + \frac{1}{828} + \frac{1}{858} + \frac{1}{6006} + \frac{1}{828828}\\\\ \frac{1}{47} & = & \frac{3}{141}\\\\ \frac{1}{48} & = & \frac{2}{99} + \frac{1}{2112} + \frac{1}{6336}\\\\ \frac{1}{49} & = & \frac{1}{66} + \frac{1}{252} + \frac{1}{777} + \frac{1}{999999} + \frac{1}{2545452}\\\\ \frac{1}{50} & = & \frac{1}{66} + \frac{1}{252} + \frac{1}{2772} + \frac{3}{5775} (\bf joriki)\\\\ \frac{1}{51} & = & \frac{1}{99} + \frac{1}{171} + \frac{1}{323} + \frac{2}{3553}\\\\ \frac{1}{52} & = & \frac{1}{66} + \frac{1}{252} + \frac{1}{9009}\\\\ \frac{1}{53} & = & \frac{4}{212}\\\\ \frac{1}{54} & = & \frac{2}{111} + \frac{1}{2002} + \frac{1}{999999}\\\\ \frac{1}{55} & = & \frac{1}{55}\\\\ \frac{1}{56} & = & \frac{1}{88} + \frac{1}{252} + \frac{1}{444} + \frac{1}{3663}\\\\ \frac{1}{57} & = & \frac{3}{171}\\\\ \frac{1}{58} & = & \frac{4}{232}\\\\ \frac{1}{59} & = & \frac{2}{161} + \frac{1}{616} + \frac{2}{767} + \frac{1}{4004} + \frac{1}{21712} + \frac{1}{21733712}\\\\ \frac{1}{60} & = & \frac{1}{111} + \frac{1}{444} + \frac{3}{555}\\\\ \frac{1}{61} & = & \frac{1}{66} + \frac{1}{858} + \frac{1}{25452} + \frac{1}{28182} + \frac{1}{909909} + \frac{2}{4214124} + \frac{1}{17288271} + \frac{1}{46355364}\\\\ \frac{1}{62} & = & \frac{1}{66} + \frac{1}{2002} + \frac{1}{3003} + \frac{1}{7007} + \frac{1}{434434} (\bf {joriki})\\\\ \frac{1}{63} & = & \frac{1}{66} + \frac{2}{2772}\\\\ \frac{1}{64} & = & \frac{1}{66} + \frac{1}{2112}\\\\ \frac{1}{65} & = & \frac{1}{66} + \frac{1}{5005} + \frac{1}{55055} + \frac{1}{66066}\\\\ \frac{1}{66} & = & \frac{1}{66}\\\\ \frac{1}{67} & = & \frac{1}{77} + \frac{1}{1001} + \frac{1}{2002} + \frac{3}{7007} + \frac{1}{85358}\\\\ \frac{1}{68} & = & \frac{4}{272}\\\\ \frac{1}{69} & = & \frac{1}{77} + \frac{1}{1771} + \frac{2}{2772} + \frac{1}{4554}\\\\ \frac{1}{70} & = & \frac{1}{99} + \frac{1}{252} + \frac{1}{5445} + \frac{1}{44044} + \frac{1}{99099} ({\bf joriki})\\\\ \frac{1}{71} & = & \frac{1}{77} + \frac{1}{1221} + \frac{1}{7007} + \frac{1}{10101} + \frac{2}{75757} + \frac{1}{111111} + \frac{1}{777777} + \frac{5}{53888835}\\\\ \frac{1}{72} & = & \frac{1}{88} + \frac{1}{444} + \frac{1}{3663}\\\\ \frac{1}{73} & = & \frac{4}{292}\\\\ \frac{1}{74} & = & \frac{1}{111} + \frac{1}{222}\\\\ \frac{1}{75} & = & \frac{1}{77} + \frac{2}{5775}\\\\ \frac{1}{76} & = & \frac{1}{99} + \frac{1}{444} + \frac{1}{1881} + \frac{1}{3663}\\\\ \frac{1}{77} & = & \frac{1}{77}\\\\ \frac{1}{78} & = & \frac{1}{88} + \frac{1}{696} + \frac{1}{63336} + \frac{1}{232232}\\\\ \frac{1}{79} & = & \frac{1}{88} + \frac{1}{858} + \frac{1}{8008} + \frac{2}{474474}\\\\ \frac{1}{80} & = & \frac{2}{181} + \frac{1}{696} + \frac{1}{86768} + \frac{1}{507705} ({\bf joriki})\\\\ \frac{1}{81} & = & \frac{1}{88} + \frac{1}{1551} + \frac{1}{3663} + \frac{1}{15651} + \frac{1}{6006006} + \frac{1}{8008008}\\\\ & & + \frac{1}{11011011} + \frac{1}{99099099} + \frac{1}{141141141} + \frac{1}{333333333} + \frac{1}{999999999}\\\\ \frac{1}{82} & = & \frac{1}{111} + \frac{2}{656} + \frac{1}{11111} + \frac{1}{21312} + \frac{1}{2466642} + \frac{1}{236797632}\\\\ \frac{1}{83} & = & \frac{9}{747}\\\\ \frac{1}{84} & = & \frac{3}{252}\\\\ \frac{1}{85} & = & \frac{7}{595}\\\\ \frac{1}{86} & = & \frac{1}{88} + \frac{1}{4664} + \frac{1}{23532} + \frac{2}{401104} + \frac{1}{420024}\\\\ \frac{1}{87} & = & \frac{1}{88} + \frac{1}{8008} + \frac{1}{232232} + \frac{1}{696696}\\\\ \frac{1}{88} & = & \frac{1}{88}\\\\ \frac{1}{89} & = & \frac{11}{979}\\\\ \frac{1}{90} & = & \frac{1}{222} + \frac{1}{333} + \frac{2}{555}\\\\ \frac{1}{91} & = & \frac{1}{99} + \frac{2}{2772} + \frac{1}{6006}\\\\ \frac{1}{92} & = & \frac{4}{414} + \frac{1}{828}\\\\ \frac{1}{93} & = & \frac{1}{101} + \frac{1}{1331} + \frac{1}{9999} + \frac{1}{2970792} + \frac{1}{27277272} ({\bf joriki})\\\\ \frac{1}{94} & = & \frac{1}{141} + \frac{1}{282}\\\\ \frac{1}{95} & = & \frac{1}{222} + \frac{2}{494} + \frac{1}{525} + \frac{1}{20202} + \frac{1}{52725} ({\bf joriki})\\\\ \frac{1}{96} & = & \frac{1}{99} + \frac{2}{6336}\\\\ \frac{1}{97} & = & \frac{1}{111} + \frac{1}{777} + \frac{1}{111111} + \frac{1}{241142} + \frac{3}{26766762} + \frac{2}{221434122}\\\\ \frac{1}{98} & = & \frac{1}{99} + \frac{1}{9999} + \frac{1}{606606} + \frac{1}{707707}\\\\ \frac{1}{99} & = & \frac{1}{99}\\\\ \frac{1}{100} & = & \frac{1}{222} + \frac{1}{444} + \frac{1}{575} + \frac{1}{777} + \frac{2}{10101} + \frac{1}{52325}\\\\ \end{eqnarray*}
We can find all representation as a sum of two unit fractions $$ \frac{1}{n} = \frac{1}{a} + \frac{1}{b} $$ by use of the quadratic equation $(a-n)(b-n)=n^2$ obtained by cross-multiplying by $abn$ and adding $n^2$ to both sides. That means we can determine if $1/n$ can be represented as a sum of two reciprocal palindrome, so all entries with only 1, 2 or 3 palindromes can be assumed to be optimal. So, the interest is in finding shortest representation for the remaining integers.
To explain how I found a lot of the representation consider $n=19$. I find a palindrome that is a multiple of 19, in this case $1881 = 99 \cdot 19$ and find all palindromic divisors of 1881 larger than 19: 33, 99, 171, 1881. We then find find non-negative integers $a, b, c, d$ such that $$ \frac{1}{19} = \frac{a}{33} + \frac{b}{99}+\frac{1}{171}+\frac{d}{1881} $$ with the smallest sum $a+b+c+d$. Cross multiplying by 1881 gives the problem of minimizing $a+b+c+d$ subject to $$ 99 = 57a + 19b + 11c + d. $$ I found $a=b=d=1$ and $c=2$. In general this is can be solved as an integer linear program which is what I used for some of the larger problems. This method does however not work for multiples of 10 and is certainly not the only way to approach the problem. I would be interested if someone can do better for some of the entries or even come up with cooler representations with the same number of palindromes.
I wrote this Java code to do a systematic search for representations as follows. To find representations of length $l$, for the first $l-2$ denominators I tried all palindromic integers up to and including $11111$ (that’s the first $200$ palindromic integers), and then I found the last two denominators by a modified version of your quadratic equation: If $\frac pq$ was left to represent as $\frac1a+\frac1b$, I found all palindromic integer solutions of $(ap-q)(bp-q)=q^2$ by iterating over the divisors of $q^2$. Thus, the search up to a certain length would have found all representations of at most that length that include at most $2$ denominators greater than $11111$.
I found shorter representations for $\frac1{25}$, $\frac1{50}$ and $\frac1{95}$ (of length $6$) as well as $\frac1{62}$, $\frac1{70}$, $\frac1{80}$ and $\frac1{93}$ (of length $5$). $\frac1{25}$ actually has quite a number of representations with $6$ reciprocals.
\begin{eqnarray} \frac1{25} &=& \frac1{33} + \frac1{252} + \frac1{404} + \frac1{505} + \frac1{909} + \frac1{5775} \\ &=& \frac1{33} + \frac1{252} + \frac1{444} + \frac1{555} + \frac1{666} + \frac1{5775} \\ &=& \frac1{44} + \frac1{66} + \frac1{525} + \frac1{5445} + \frac1{44044} + \frac1{99099}\\ &=& \frac1{44} + \frac1{66} + \frac1{616} + \frac1{5005} + \frac1{5775} + \frac1{8008} \\ &=& \frac1{55} + \frac1{66} + \frac1{171} + \frac1{1881} + \frac1{5775} + \frac1{8778} \\ &=& \frac1{55} + \frac1{66} + \frac1{202} + \frac1{707} + \frac1{5775} + \frac1{7777} \\ &=& \frac1{55} + \frac1{66} + \frac1{252} + \frac1{444} + \frac1{3663} + \frac1{5775} \\ &=& \frac1{55} + \frac1{66} + \frac1{333} + \frac1{525} + \frac1{585} + \frac1{20202} \\ &=& \frac1{66} + \frac1{77} + \frac1{88} + \frac1{5005} + \frac1{5775} + \frac1{8008}\;, \\[8pt] \frac1{50}&=&\frac1{66} + \frac1{252} + \frac1{2772} + \frac3{5775}\;,\\[8pt] \frac1{62}&=&\frac1{66} + \frac1{2002} + \frac1{3003} + \frac1{7007} + \frac1{434434}\;,\\[8pt] \frac1{70} &=& \frac1{77} + \frac1{909} + \frac1{6006} + \frac1{33033} + \frac1{549945} \\ &=& \frac1{99} + \frac1{252} + \frac1{5445} + \frac1{44044} + \frac1{99099}\;,\\[8pt] \frac1{80} &=& \frac2{181} + \frac1{696} + \frac1{86768} + \frac1{507705} \\ &=& \frac1{202} + \frac1{232} + \frac1{505} + \frac1{808} + \frac1{46864} \;,\\[8pt] \frac1{93} &=& \frac1{101} + \frac1{1331} + \frac1{9999} + \frac1{2970792} + \frac1{27277272} \;,\\[8pt] \frac1{95} &=& \frac1{222} + \frac2{494} + \frac1{525} + \frac1{20202} + \frac1{52725} \;. \end{eqnarray}
For the following integers, I couldn’t find any representations up to length $5$: $41$, $43$, $59$, $61$, $67$, $71$, $81$, $82$, $83$, $85$, $86$, $89$, $97$, $100$. I tried again for length $6$ with a lower limit of $1111$ (the $100$th palindromic integer) on the $l-2$ denominators but again didn’t find any representations. Thus, for these numbers there are neither representations of length at most $5$ with at most $2$ denominators above $11111$, nor representations of length $6$ with at most $2$ denominators above $1111$.
For all remaining integers, your representations are as short as the shortest representations with at most $2$ denominators above $11111$. I found the following alternative representations with the same lengths as yours (and these are all there are with at most two denominators above $11111$):
\begin{eqnarray} \frac1{12} &=&\frac1{22} + \frac1{44} + \frac1{66}\;,\\[8pt] \frac1{17} &=&\frac1{22} + \frac1{88} + \frac1{595} + \frac1{5005} + \frac1{8008}\;,\\[8pt] \frac1{19} &=&\frac1{22} + \frac1{171} + \frac1{969} + \frac1{3663} + \frac1{41514}\;,\\[8pt] \frac1{20} &=&\frac1{22} + \frac1{252} + \frac1{2662} + \frac1{5445} + \frac1{66066} + \frac1{363363}\\ &=&\frac1{22} + \frac1{252} + \frac1{2772} + \frac1{5445} + \frac1{44044} + \frac1{99099}\\ &=&\frac1{22} + \frac1{252} + \frac1{3003} + \frac1{5445} + \frac1{22022} + \frac1{66066}\\ &=&\frac1{22} + \frac1{252} + \frac1{3003} + \frac1{5445} + \frac2{33033}\\ &=&\frac1{22} + \frac1{252} + \frac1{5005} + \frac2{5445} + \frac1{99099}\\ &=&\frac1{22} + \frac1{414} + \frac1{585} + \frac1{4554} + \frac1{5005} + \frac1{828828}\\ &=&\frac1{22} + \frac1{444} + \frac1{484} + \frac1{5445} + \frac1{23232} + \frac1{2578752}\\ &=&\frac1{22} + \frac1{444} + \frac1{585} + \frac1{3663} + \frac1{5005} + \frac1{9009}\\ &=&\frac1{22} + \frac1{444} + \frac2{999} + \frac1{3663} + \frac1{54945}\\ &=&\frac1{22} + \frac2{505} + \frac1{4444} + \frac2{5555}\\ &=&\frac1{22} + \frac1{555} + \frac1{585} + \frac1{1001} + \frac1{30303} + \frac1{444444}\\ &=&\frac1{33} + \frac2{121} + \frac1{484} + \frac1{909} + \frac1{549945}\\ &=&\frac1{44} + \frac1{55} + \frac1{202} + \frac2{505} + \frac1{5555}\\ &=&\frac1{44} + \frac1{66} + \frac1{99} + \frac1{585} + \frac1{5005} + \frac1{9009}\\ &=&\frac1{44} + \frac1{66} + \frac1{99} + \frac2{999} + \frac1{54945}\\ &=&\frac1{44} + \frac1{66} + \frac1{121} + \frac1{363} + \frac1{909} + \frac1{549945}\\ &=&\frac1{44} + \frac2{99} + \frac1{242} + \frac1{363} + \frac1{5445}\\ &=&\frac1{55} + \frac2{77} + \frac1{252} + \frac1{585} + \frac1{6006}\\ &=&\frac2{55} + \frac1{88} + \frac1{505} + \frac1{5555} + \frac1{8888}\;,\\[8pt] \frac1{23} &=&\frac1{33} + \frac1{88} + \frac1{808} + \frac1{2222} + \frac1{9999} + \frac1{41814}\\ &=&\frac1{33} + \frac1{99} + \frac1{606} + \frac1{909} + \frac1{3333} + \frac1{41814}\\ &=&\frac1{33} + \frac1{101} + \frac1{333} + \frac1{4444} + \frac1{41814} + \frac1{44844}\\ &=&\frac1{33} + \frac1{101} + \frac2{616} + \frac1{41814} + \frac1{279972}\\ &=&\frac1{33} + \frac1{161} + \frac1{252} + \frac1{414} + \frac1{2772} + \frac1{4554}\\ &=&\frac1{44} + \frac2{99} + \frac1{3333} + \frac1{4444} + \frac1{41814}\\ &=&\frac1{66} + \frac1{88} + \frac1{99} + \frac1{232} + \frac1{414} + \frac1{7337}\;,\\[8pt] \frac1{34} &=&\frac1{66} + \frac1{99} + \frac1{323} + \frac2{1881}\;,\\[8pt] \frac1{39} &=&\frac1{44} + \frac1{363} + \frac1{6776} + \frac1{88088}\\ &=&\frac1{77} + \frac1{88} + \frac1{858} + \frac1{8008}\;,\\[8pt] \frac1{42} &=&\frac1{44} + \frac3{2772}\\ &=&\frac1{77} + \frac1{99} + \frac2{2772}\\ &=&\frac2{101} + \frac1{252} + \frac1{25452}\\ &=&\frac2{111} + \frac1{222} + \frac1{777}\;,\\[8pt] \frac1{45} &=&\frac1{66} + \frac1{242} + \frac1{363} + \frac1{5445}\;,\\[8pt] \frac1{46} &=&\frac1{66} + \frac1{161} + \frac1{5005} + \frac1{5775} + \frac2{575575}\\ &=&\frac1{66} + \frac1{252} + \frac1{808} + \frac1{828} + \frac1{8888} + \frac1{16261}\\ &=&\frac1{66} + \frac1{252} + \frac1{828} + \frac1{1111} + \frac1{2222} + \frac1{16261}\\ &=&\frac1{66} + \frac1{414} + \frac2{616} + \frac1{1771} + \frac1{2772}\\ &=&\frac1{66} + \frac1{414} + \frac1{666} + \frac1{777} + \frac1{1221} + \frac1{1771}\\ &=&\frac1{77} + \frac1{161} + \frac1{828} + \frac1{1001} + \frac1{3003} + \frac1{828828}\\ &=&\frac1{77} + \frac2{252} + \frac1{1771} + \frac1{4004} + \frac1{828828}\\ &=&\frac1{99} + \frac1{161} + \frac1{252} + \frac1{828} + \frac1{4004} + \frac1{828828}\\ &=&\frac1{101} + \frac1{202} + \frac1{252} + \frac1{606} + \frac1{828} + \frac1{16261}\\ &=&\frac1{101} + \frac1{252} + \frac2{303} + \frac1{828} + \frac1{16261}\\ &=&\frac1{111} + \frac1{121} + \frac1{242} + \frac1{3003} + \frac1{2343432} + \frac1{6521771256}\\ &=&\frac1{111} + \frac1{161} + \frac1{252} + \frac1{777} + \frac1{828} + \frac1{17871}\\ &=&\frac1{111} + \frac1{161} + \frac1{333} + \frac1{444} + \frac1{828} + \frac1{17871}\\ &=&\frac1{161} + \frac3{252} + \frac1{414} + \frac1{828}\;,\\[8pt] \frac1{49} &=&\frac1{66} + \frac1{333} + \frac1{444} + \frac1{999999} + \frac1{2545452}\\ &=&\frac2{99} + \frac1{4884} + \frac1{999999} + \frac1{2545452}\;,\\[8pt] \frac1{51} &=&\frac3{171} + \frac2{969}\;,\\[8pt] \frac1{53} &=&\frac1{88} + \frac1{212} + \frac1{424} + \frac1{2332}\;,\\[8pt] \frac1{56} &=&\frac1{77} + \frac3{616}\\ &=&\frac1{88} + \frac1{171} + \frac1{1881} + \frac1{8778}\\ &=&\frac1{88} + \frac1{202} + \frac1{707} + \frac1{7777}\;,\\[8pt] \frac1{60} &=&\frac1{99} + \frac1{242} + \frac1{484} + \frac2{5445}\;,\\[8pt] \frac1{68} &=&\frac1{121} + \frac1{242} + \frac1{484} + \frac1{4114}\;,\\[8pt] \frac1{69} &=&\frac1{77} + \frac1{808} + \frac1{6666} + \frac1{8888} + \frac1{178871}\\ &=&\frac1{77} + \frac1{858} + \frac1{3003} + \frac1{178871} + \frac1{606606}\\ &=&\frac1{77} + \frac1{868} + \frac1{3003} + \frac1{48484} + \frac1{24266242}\\ &=&\frac1{77} + \frac1{909} + \frac1{2772} + \frac1{25452} + \frac1{178871}\\ &=&\frac1{77} + \frac1{909} + \frac1{3333} + \frac1{9999} + \frac1{178871}\\ &=&\frac1{77} + \frac1{1001} + \frac1{2002} + \frac1{178871} + \frac1{606606}\\ &=&\frac1{77} + \frac1{1111} + \frac1{2222} + \frac1{6666} + \frac1{178871}\\ &=&\frac1{77} + \frac1{1111} + \frac2{3333} + \frac1{178871}\\ &=&\frac1{88} + \frac1{333} + \frac1{8008} + \frac1{999999} + \frac1{4181814}\;,\\[8pt] \frac1{78} &=&\frac1{88} + \frac1{858} + \frac1{6006} + \frac1{8008}\\ &=&\frac1{88} + \frac1{1001} + \frac1{3003} + \frac1{8008}\\ &=&\frac1{121} + \frac1{242} + \frac1{3003} + \frac1{11011}\;,\\[8pt] \frac1{87} &=&\frac2{232} + \frac2{696}\;,\\[8pt] \frac1{91} &=&\frac2{252} + \frac1{333} + \frac1{20202}\;,\\[8pt] \frac1{92} &=&\frac1{99} + \frac1{3333} + \frac1{4444} + \frac1{4554} + \frac1{41814}\;,\\[8pt] \frac1{98} &=&\frac1{101} + \frac1{3333} + \frac1{606606} + \frac1{707707}\\ &=&\frac3{343} + \frac1{686}\;. \end{eqnarray}