# Finding the minimal number of members

I've been working on the following problem

For every issue in the Blue's association, a commission with 10 members (belonging the Blue's) is formed in order to solve the problem. The only condition is

There can't be two commissions having more than one member in common

The Blue's association has formed this year 40 commissions.

What's the minimal amount of members in the Blue's association?

I've only found out the following

For any commission you can form $$\binom{10}{2}=45$$ different pairs and none of them can appear in another commission.

Since 40 different commissions are formed, the minimal number of pairs is $$45\times 40=1800$$.

Denote by $$n$$ the number of members. Thus $$\binom{n}{2}≥1800\Rightarrow n>60$$



The minimal amount of members has to be 100 or less.

You can observe a distribution for 100 members here

My question:

Is 100 the answer or is there an ever smaller possible amount of members? If so, how can I prove it?

• Different but related question, perhaps some of the ideas there are helpful: math.stackexchange.com/questions/2879313/… – Servaes Dec 2 '18 at 18:57
• If there are two members that are in $6$ commisions, then they share at most one commission, and to fill these $11$ commissions then requires another $10\times9+1\times8=98$ distinct members, yielding a total of $100$ members if we count the two members we started with. So any example smaller than the one you found has at most one member in more than $5$ commissions. – Servaes Dec 2 '18 at 20:02
• By a similar (but simpler) argument, no member can be in more than $11$ commisions; to fill $11$ commissions sharing one member requires another $11\times9=99$ distinct members, yielding a total of $100$ members. – Servaes Dec 2 '18 at 20:06
• @Servaes If $a$ and $b$ are two members that are in $6$ commissions, how do you know that those $10\times9+1\times6=98$ members are all distinct? Why can't a commission containing $a$ overlap with a commission containing $b$? – bof Dec 5 '18 at 9:50
• @Servaes Note that, if the points of a projective plane of order $9$ are our "members" and the lines are our "commissions", then we have $91$ members and $91$ commissions, and each member belongs to $10$ commissions. – bof Dec 5 '18 at 11:32

This post exhibits a solution with $$82$$ members. Combined with the excellent answer by @Song, this means $$82$$ is indeed optimal.

Motivation: The excellent answer by @Song and the followup comments by @Servaes make me wonder... perhaps if we look for 41 commissions (not 40) then there is a solution with a great deal of symmetry:

• (a) 82 members (the optimal answer we seek)
• (b) 41 commissions (exceeds OP requirement)
• (c) each member associated with exactly 5 commissions (not part of OP)
• (d) each commission associated with exactly 10 members (equals OP requirement)
• (e) each 2 members have exactly 1 common commission (not part of OP)
• (f) each 2 commissions have exactly 1 common member (exceeds OP requirement)

This would be like a finite projective plane, but with 82 points and 41 lines. However, in a finite projective (respectively: affine) plane, the no. of points and no. of lines are equal (respectively: almost equal), and this is probably why a solution based on FPP only reaches 84. So I decided to look at related structures called Block Designs, Steiner Systems, etc. where there are typically many more "lines" than "points". After quite a bit of digging, I think I found the right structure:

The solution: It is a Steiner $$S(t=2,k=5,n=41)$$ system. A Steiner system is defined by the following properties:

• there are $$n=41$$ objects (these are the commissions)

• there are $$b$$ blocks (these are the members), each block (member) being a subset of objects (i.e. the commissions he/she is associated with)

• each block has $$k=5$$ objects (each member is associated with 5 commissions)

• every $$t=2$$ objects is contained in exactly 1 block (every 2 commissions have exactly 1 common member)

So this already satisfies (b), (c) and (f). Next, quoting from https://en.wikipedia.org/wiki/Steiner_system#Properties we have:

• $$b = {n \choose t} / {k \choose t} = (41 \times 40) / (5 \times 4) = 41 \times 2 = 82$$, satisfying (a)

• $$r = {n-1 \choose t-1} / {k-1 \choose t-1} = 40 / 4 = 10$$, where $$r$$ denotes "the number of blocks containing any given object", i.e. the number of members associated with any given commission, satisfying (d).

Thinking more, I don't think (e) can be satisfied. However, (e) is not needed for the OP, so it doesn't matter.

So finally we just need to prove that such a Steiner $$S(t=2,k=5,n=41)$$ system exists. This existence is non-trivial, but luckily more digging reveals:

• https://math.ccrwest.org/cover/steiner.html has a list of Steiner systems that are known to exist. $$S(2,5,41)$$ (the webpage sometimes lists the 3 parameters in different order) is not part of any infinite families listed, but if you go further down the page it is listed as a standalone example; clicking on that link goes to...

• https://math.ccrwest.org/cover/show_cover.php?v=41&k=5&t=2 which exhibits the system, created via "Cyclic construction" whatever that means.

I did not check the numbers thoroughly, but if I understand the webpage correctly, there should be 82 rows (members / blocks), each containing 5 numbers (commissions), all the numbers being 1 through 41 inclusive (the 41 commissions), each number (commission) should appear in 10 rows, and every pair-of-numbers should appear in 1 row.

I am not an expert in any of these, so if I had a mistake or misunderstanding above, my apologies. Perhaps someone more expert can check my work?

• +1 Good ideas and good find! I will have a look at the second link. As a side note, indeed (e) is impossible; there are $\binom{82}{2}=3321$ pairs of members. But there are $\binom{10}{2}=45$ pairs of members per commission, so $41\times45=1845$ pairs of members that share a commission. Alternatively, every member is in $5$ commissions, and hence in a commission with $5\times9=45$ members, so $\frac{82\times45}{2}=1845$ pairs of members that share a commission. Either way, there are too many pairs of members to all share a commission. – Servaes Dec 14 '18 at 0:55
• *Another, simpler, way to see that (e) is impossible; this would imply that every member shares a commission with $81$ members. But every member is in only $5$ commissions, and hence shares a commission with only $5\times9=45$ members. – Servaes Dec 14 '18 at 0:57
• @Servaes - yes of course. in fact, more abstractly / vaguely speaking, perhaps insisting on both (e) and (f) is WHY a finite projective plane has the same no. of lines and points. so if we introduce imbalance in (a) vs (b), and hence (c) vs (d), we must necessarily imbalance (e) vs (f) as well. this is all very vague of course. :) – antkam Dec 14 '18 at 1:07
• With duality between lines & points (or equiv: rows & cols of an incidence matrix), there is really no need to translate, but it's up to you. anyway, here's a link I found explaining (I think) the "cyclic construction" of some Steiner systems, although it doesn't mention $S(2,5,41)$ specifically: pitt.edu/~kaveh/Chap7-MATH1050.pdf – antkam Dec 14 '18 at 1:53
• Dan Gordon migrated the La Jolla covering design repository to his new site, so it might be a good idea to update your links. – hardmath Mar 16 at 0:46

Let $$i$$ denote each member of Blue's association and assume that there are $$N$$ members in total, that is, $$i=1,2,\cdots, N.$$ And let $$j,k=1,2,\ldots, 40$$ denote each of 40 commission. We will show that $$N$$ is at least $$82$$.

Consider the set $$S=\{(i,j,k)\;|\;1\leq i\leq N, 1\leq j Let $$d_i$$ denote the number of commissions that $$i$$ joined. We will calculate $$|S|$$ using double counting method. First, note that $$|S|=\sum_{(i,j,k)\in S}1 = \sum_{1\leq j since for each $$j, there is at most one $$i$$ in common. On the other hand, $$|S| = \sum_{1\leq i\leq N} \sum_{(j,k):(i,j,k)\in S}1 = \sum_{1\leq i\leq N} \binom{d_i}{2},$$ since for each $$i$$, the number of pairs $$(j,k)$$ that $$i$$ joined is $$\binom{d_i}{2}$$. We also have $$\sum_{1\leq i\leq N}d_i = 400,$$by the assumption.
Finally, note that the function $$f(x)= \binom{x}{2} = \frac{x^2-x}{2}$$ is convex. Thus by Jensen's inequality we have that $$\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq Nf\left(\frac{\sum_i d_i}{N}\right)=N\binom{\frac{400}{N}}{2}.$$ This gives us the bound $$40\cdot 39 \geq 400\cdot(\frac{400}{N}-1),$$and hence $$N \geq \frac{4000}{49} = 81.63\cdots$$ This establishes $$N\geq 82$$. However, I'm not sure if this bound is tight. I hope this will help.

$$\textbf{Note:}$$ If $$N=82$$ is tight, then above argument implies that $$d_i$$'s distribution is almost concentrated at $$\overline{d} = 400/82 \sim 5$$.
EDIT: @antkam's answer seemingly shows that $$N=82$$ is in fact optimal.

• +1 Fantastic argument! – Servaes Dec 10 '18 at 10:28
• A corollary of this argument is that in the best solution every commission must share a member with some other commission. – Will Fisher Dec 10 '18 at 20:02
• @WillFisher You mean, if there is a solution with $N=82$? In this case, it also follows that there are precisely $72$ members that are in $5$ commissions, and $10$ that are in $4$ commissions. Also, that every commission has precisely $9$ members in $5$ commissions and $1$ member in $4$ commissions. – Servaes Dec 10 '18 at 20:38
• I don't really understand this step: "Finally, note that the function $x\to \binom{x}{2} = \frac{x^2-x}{2}$ is convex. Thus, we have that $$\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq N\binom{\frac{\sum_i d_i}{N}}{2}=N\binom{\frac{400}{N}}{2}.$$" – Dr. Mathva Dec 13 '18 at 23:35
• I should have mentioned that it is by Jensen's inequality. The statement is in en.m.wikipedia.org/wiki/Jensen%27s_inequality. @Dr. Mathva – Song Dec 13 '18 at 23:44

This is just a partial answer. I will show that $$85$$ members suffice; I don't know if $$85$$ is the minimum.

Recall that a projective plane of order $$n$$ exists if $$n$$ is a prime power: it has $$n^2+n+1$$ points and $$n^2+n+1$$ lines; each line has $$n+1$$ points, and there are $$n+1$$ lines through each point; any pair of lines meets in a unique point, and any pair of points determines a unique line.

Consider a projective plane of order $$9$$; it has $$9^2+9+1=91$$ points and $$91$$ lines; there are $$10$$ points on each line and $$10$$ lines through each point. A set of points is in general position if no three of the points are collinear. Note that, if we have a set of $$t$$ points in general position, then the lines determined by those points (taken two at a time) cover a total of at most $$t+8\binom t2$$ points; as long as $$t\le5$$ then the number of covered points is at most $$5+8\binom52=85\lt91$$, so we can add another point to the set and still have them in general position. Thus we can find a set $$S$$ of $$6$$ points in general position.

Let the members of the Blue's association be the $$91-6=85$$ points which are not in $$S$$. The commissions are the lines which do not meet $$S$$; they have $$10$$ members each, and any two have exactly one member in common. Finally, by the in-and-out formula, the number of commissions is $$91-\binom61\cdot10+\binom62\cdot1=46.$$

P.S. Let $$m$$ be the minimum possible number of members. I showed above that $$m\le85$$. On the other hand, I have a small improvement on your lower bound $$m\ge61$$.

Suppose the $$i^\text{th}$$ member belongs to $$d_i$$ commissions; then $$\sum_{i=1}^md_i=400$$ since there are $$40$$ commissions with $$10$$ members each. Moreover $$d_i\le9$$ since $$m\le85\lt91$$. Let $$k=|\{i:d_i\ge5\}|$$. Then $$400=\sum_{i=1}^md_i\le4(m-k)+9k=4m+5k\le340+5k,$$ whence $$k\ge12$$; i.e., there are at least $$12$$ members who are on at least $$5$$ commissions. Choose two members $$i$$ and $$j$$ who are on at least $$5$$ commissions.

Case 1. There is a commission containing both $$i$$ and $$j$$.

First, there are $$10$$ members on the commission which $$i$$ and $$j$$ both belong to. Next $$i$$ belongs to $$4$$ more commissions, with $$36$$ additional members. Finally, $$j$$ belongs to $$4$$ more commissions, each of which contains at most one member of each of the $$5$$ commissions containg $$i$$, and at least $$5$$ members who haven't been counted yet, for a total of $$20$$ new members. This shows that $$m\ge10+36+20=66$$.

Case 2. There is no commission containing both $$i$$ and $$j$$.

In this case a similar argument shows that $$m\ge67$$.

This proves that $$m\ge66$$. Combining this with the upper bound shown earlier, we have $$66\le m\le85.$$

• +1 Great progress and nice ideas! Note that for every prime power $q$ there are arcs of $q+1$ points in $\Bbb{P}(\Bbb{F}_q)$. In particular you can take your set $S$ to contain $7$ points, yielding $42$ commissions and $m\leq84$. With $S$ containing $8$ points we get $39$ commissions and $m\leq83$, perhaps another commission can be fitted in somewhere? – Servaes Dec 5 '18 at 16:53
• Your bottom argument can be extended to show $m\ge 73$. – Will Fisher Dec 5 '18 at 17:04
• @Servaes I don't know much about projective planes. That fact about "arcs of $q+1$ points", is it easy or hard to prove? Is it true for all projective planes of prime power order, or just the field planes? – bof Dec 6 '18 at 10:20
• @WillFisher Nice. How do you prove it? – bof Dec 6 '18 at 10:21
• @bof In any projective plane over a finite field, any nondegenerate conic is an example of an arc of $q+1$ points. I've also checked that the same construction with a set $S$ of $8$ or more points cannot fit any other commission directly (i.e. without adjusting some of the existing $39$ or fewer commissions). – Servaes Dec 6 '18 at 12:46

Allow me to summarize and slightly refine the results in the current answers (if only to straighten out my own thoughts); they show that the minimum number of members $$m$$ satisfies $$82\leq m\leq84$$. They also imply strict conditions on any solution with $$m=82$$.

I also include my result that if $$m=83$$, then no member is in more than $$7$$ commissions. Much more can be said, but I do not have a definitive proof for the cases $$m=82$$ or $$m=83$$.

The upper bound $$m\leq84$$ comes from bof's construction in the projective plane of order $$9$$; the projective plane $$\Bbb{P}^2(\Bbb{F}_9)$$ consists of $$91$$ points on $$91$$ lines, with $$10$$ points on each line and $$10$$ lines through each point. Importantly, each pair of lines meets in precisely one point, and each pair of points is on precisely one line.

For $$7$$ distinct points in general position (no $$3$$ on a line, e.g. points on a smooth conic) there are precisely $$7\times10-\binom{7}{2}\times1=49$$ lines containing these points. Removing these $$7$$ points and the $$49$$ lines containing them leaves $$84$$ points and $$91-49=42$$ lines each containing $$10$$ points, and any pair of lines meets in at most one point. That is, we have $$84$$ members in $$42$$ commissions, with no $$2$$ commissions sharing more than one member, so $$m\leq84$$.

The lower bound $$m\geq82$$ comes from Song's answer; the number of pairs of commissions that share a member is at most $$\binom{40}{2}$$, as there are $$40$$ commissions. As every commission shares at most one member, this can also be counted as the number of pairs of commissions that each member is in. If the $$i$$-th member is in $$d_i$$ commissions, then it is in $$\binom{d_i}{2}$$ pairs of commissions and hence $$\sum_{i=1}^m\binom{d_i}{2}\leq\binom{40}{2}.\tag{1}$$ As there are $$40$$ commissions with $$10$$ members each, we also have $$\sum_{i=1}^md_i=400$$. In the inequality above we can bound the left hand side from below using the fact that for all positive integers $$x$$ we have $$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1.$$ This allows us to even out the $$d_i$$'s to find that $$\sum_{i=1}^m\binom{d_i}{2}\geq(m-n)\binom{x}{2}+n\binom{x+1}{2},\tag{2}$$ for some $$x$$ and $$n$$ with $$0\leq n, where $$(m-n)x+n(x+1)=\sum_{i=1}^md_i=400.$$ The latter simplifies to $$mx+n=400$$, which shows that $$x=\lfloor\frac{400}{m}\rfloor$$ and $$n=400-mx$$. Plugging this back in shows that $$\begin{eqnarray*} \binom{40}{2}&\geq&\sum_{i=1}^m\binom{d_i}{2} \geq(m-n)\binom{x}{2}+n\binom{x+1}{2}\\ &=&(m-(400-m\lfloor\tfrac{400}{m}\rfloor))\binom{\lfloor\frac{400}{m}\rfloor}{2}+(400-m\lfloor\tfrac{400}{m}\rfloor)\binom{\lfloor\frac{400}{m}\rfloor+1}{2}\\ &=&-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor^2-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor+400\lfloor\tfrac{400}{m}\rfloor. \end{eqnarray*}$$ The latter is strictly decreasing for $$m$$ in the interval $$[1,84]$$. The inequality is satisfied if and only if $$m\geq82$$, which proves the lower bound.

Let $$S$$ denote the number of times we need to apply the identity $$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1$$ to reduce the left hand side of $$(2)$$ to the right hand side. We can then write $$(2)$$ more precisely as $$\sum_{i=1}^m\binom{d_i}{2}=(m-n)\binom{x}{2}+n\binom{x+1}{2}+S.$$ Knowing that $$82\leq m\leq84$$ simplifies the above significantly, as then $$x=\lfloor\tfrac{400}{m}\rfloor=4$$ and $$n=400-4m$$. We find that $$780\geq\sum_{i=1}^m\binom{d_i}{2}=1600-10m+S.$$ In particular, for $$m=82$$ we find that $$S=0$$ and hence that there are precisely $$n=400-82\times4=72$$ members that are in $$4$$ commissions and $$10$$ members that are in $$5$$ commissions. We also see that we have equality in $$(1)$$, meaning that every pair of commissions shares a member. This implies $$\sum_{i\in C}(d_i-1)=39$$ for every commission $$C$$, from which it follows that every commission has precisely $$1$$ member that is in $$4$$ commissions, and $$9$$ members that are in $$5$$ commissions.

If $$m=83$$ then $$S\leq10$$, and there are at most $$10$$ pairs of commissions that do not share a member.

Here are a few unincorporated observations that may or may not be helpful. These concern restrictions on minimal examples with $$m<84$$, i.e. $$m=82$$ or $$m=83$$. They are all subsumed by the observations above for $$m=82$$, so I prove them only for $$m=83$$.

Observation 1: For all $$i$$ we have $$d_i\leq9$$.

To fill the commission of member $$i$$ requires $$9d_i+1$$ distinct members, including member $$i$$. We have $$9d_i+1\leq m=83$$ and hence $$d_i\leq9$$.

Observation 2: For all $$i$$ we have $$d_i\leq8$$.

To fill the commission of a member $$i$$ with $$d_i=9$$ requires $$9d_i+1=82$$ distinct members, leaving $$1$$ member remaining as $$m=83$$. Each of the remaining $$40-d_i=31$$ commissions has at most one member from teach of the $$d_i$$ commissions of $$i$$, and hence contains the remaining member. But this member is in at most $$9$$ commissions by observation $$1$$, a contradiction.

Observation 3: For any pair $$i$$, $$j$$ of members in a commission we have $$d_i+d_j\leq14$$.

If the inequality does not hold then without loss of generality $$d_i=8$$ and $$d_j\geq7$$. To fill the shared commission requires another $$8$$ members, and to fill the remaining $$7$$ commissions of member $$i$$ requires another $$9\times7=63$$ members. Each of the $$d_j-1$$ remaining commissions of $$j$$ contains at most $$7$$ members from the $$7$$ commissions of $$i$$, and hence at least $$2$$ new members. Hence we have a total of $$2+8+9\times(d_i-1)+2\times(d_j-1)\geq2+8+63+2\times6=85,$$ members, contradicting $$m=83$$.

Observation 4: For all $$i$$ we have $$d_i\leq7$$.

Suppose toward a contradiction that $$d_i=8$$ for some member $$i$$. To fill these $$d_i=8$$ commissions requires $$9d_i+1=73$$ distinct members, including member $$i$$, leaving $$10$$ members. Each of the remaining $$32$$ commissions has at most $$8$$ members from the $$d_i=8$$ commissions, hence at least $$2$$ members from the remaining $$10$$. Numbering these $$1$$ throught $$10$$ we find that $$\sum_{k=1}^{10}d_k\geq2\times32=64.$$ We distinguish two cases:

Case 1: If $$d_j=8$$ for some $$1\leq j\leq10$$ then $$j$$ shares a commission with at least $$8$$ other of these $$10$$ members, hence they all have $$d_k\leq6$$ by observation $$3$$. To satisfy the inequality there must be one more member $$j'$$ with $$d_{j'}=8$$, and the other $$8$$ have $$d_k=6$$.

We have $$11$$ members, including $$i$$, that together take up $$8+64=72$$ spots in the $$40$$ commissions. The remaining $$83-11=72$$ members then take up $$400-72=328$$ spots. As noted before, the sum $$\sum\binom{d_i}{2}$$ ranging over the remaining $$72$$ members is minimal when the values $$d_i$$ differ by at most $$1$$. This happens precisely when $$d_i=5$$ for $$40$$ members and $$d_i=4$$ for $$432$$ members. Then $$\sum_{k=1}^{83}\binom{d_i}{2}\geq3\binom{8}{2}+8\binom{6}{2}+40\binom{5}{2}+32\binom{4}{2}=796,$$ which exceeds the bound of $$\binom{40}{2}=780$$ we found before, a contradiction.

Case 2: If $$d_j\neq8$$ for all $$10$$ remaining members, then to satisfy $$\sum_{k=1}^{10}d_k\geq64$$ there must be a t least $$4$$ members with $$d_k=7$$. We also have $$\sum_{k=1}^{10}\leq70$$, and we proceed as before.

We have $$5$$ members, including $$i$$, that together take up $$8+28=36$$ spots in the $$40$$ commissions. Hence the remaining $$83-5=78$$ members take up $$400-36=364$$ spots. The sum $$\sum\binom{d_i}{2}$$ over the remaining $$78$$ members is minimized when the $$d_i$$ differ by at most $$1$$. This happens precisely if $$d_i=5$$ for $$52$$ members and $$d_i=4$$ for $$26$$ members, and we $$\sum_{k=1}^{83}\binom{d_k}{2}\geq\binom{8}{2}+4\binom{7}{2}+52\binom{5}{2}+26\binom{4}{2}=788,$$ again contradicting the upper bound of $$\binom{40}{2}=780$$.

Much more can be said, but my computer is already freezing up at this big an answer.

Here's a partial answer that increases the lower bound for any (not necessarily optimal) solution to $$m\ge 74$$.

Suppose there is a solution with $$m$$ members and we know that there are two members each in $$l+1$$ commissions, then $$m\ge 9(l+1)+(8-l)(l+1)+2.$$ This is because if member one is in $$\ge l+1$$ commissions, each commission needs to be filled with $$9$$ new members since these $$l+1$$ commissions already have maximum overlap. For the commissions that member two is in, each needs $$9$$ more members to be accounted for. We can't have any overlap between these commissions because they already have maximum overlap (that being member two). We can at best choose one member from each group with member one in it, giving us $$l+1$$ members, but the other $$9-(l+1)=8-l$$ are new. This gives $$9(l+1)+(8-l)(l+1)$$ members other than the two we started with. (Note that this is the best bound in $$l$$ possible).

Now, suppose $$m$$ members has a solution to the problem. Let $$d_i$$ be the how many commissions the $$i$$-th member is in. First note that $$m\ge 9d_i+1$$ for every $$i$$, so $$d_i\le \lfloor (m-1)/9\rfloor$$. Let $$k_l=|\{i\; :\; d_i>l\}|$$. Then $$400=\sum_{i=1}^md_i\le l(m-k_l)+\lfloor (m-1)/9\rfloor k_l.$$ Hence $$k_l\ge \frac{400-lm}{\lfloor (m-1)/9\rfloor -l}.$$ Since $$k_l$$ is an integer, if $$\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$$ then $$k_l\ge 2$$, meaning that there is at least $$2$$ members in at least $$l+1$$ commissions so by the above $$m\ge 9(l+1)+(8-l)(l+1)+2$$. Note that $$\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$$ exactly when $$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$. Hence we have that for all $$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$ we have $$m\ge 9(l+1)+(8-l)(l+1)+2.$$ For this to be satisfied gives $$m\ge 74.$$

• Very nice! I had some trouble understanding the part from "Hence we have that for all...". Perhaps you could make more explicit that if this inequality holds for $l$, then the next inequality must also hold for that $l$. – Servaes Dec 7 '18 at 14:29
• @Servaes I edited the answer. It is basically just solving for the $l$'s such that $k_l\ge 2$, letting the first paragraphs argument follow. – Will Fisher Dec 7 '18 at 16:06