{"year": "2017", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "RMM", "problem": "(a) Prove that every positive integer $n$ can be written uniquely in the form\n\n$$\nn=\\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}\n$$\n\nwhere $k \\geq 0$ and $0 \\leq m_{1}<m_{2}<\\cdots<m_{2 k+1}$ are integers.\nThis number $k$ is called the weight of $n$.\n(b) Find (in closed form) the difference between the number of positive integers at most $2^{2017}$ with even weight and the number of positive integers at most $2^{2017}$ with odd weight.\n\n## Vjekoslav Kovač, Croatia", "solution": "(a) We show by induction on the integer $M \\geq 0$ that every integer $n$ in the range $-2^{M}+1$ through $2^{M}$ can uniquely be written in the form $n=\\sum_{j=1}^{\\ell}(-1)^{j-1} 2^{m_{j}}$ for some integers $\\ell \\geq 0$ and $0 \\leq m_{1}<m_{2}<\\cdots<m_{\\ell} \\leq M$ (empty sums are 0 ); moreover, in this unique representation $\\ell$ is odd if $n>0$, and even if $n \\leq 0$. The integer $w(n)=\\lfloor\\ell / 2\\rfloor$ is called the weight of $n$.\n\nExistence once proved, uniqueness follows from the fact that there are as many such representations as integers in the range $-2^{M}+1$ through $2^{M}$, namely, $2^{M+1}$.\n\nTo prove existence, notice that the base case $M=0$ is clear, so let $M \\geq 1$ and let $n$ be an integer in the range $-2^{M}+1$ through $2^{M}$.\n\nIf $-2^{M}+1 \\leq n \\leq-2^{M-1}$, then $1 \\leq n+2^{M} \\leq 2^{M-1}$, so $n+2^{M}=\\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}$ for some integers $k \\geq 0$ and $0 \\leq m_{1}<\\cdots<m_{2 k+1} \\leq M-1$ by the induction hypothesis, and $n=\\sum_{j=1}^{2 k+2}(-1)^{j-1} 2^{m_{j}}$, where $m_{2 k+2}=M$.\n\nThe case $-2^{M-1}+1 \\leq n \\leq 2^{M-1}$ is covered by the induction hypothesis.\nFinally, if $2^{M-1}+1 \\leq n \\leq 2^{M}$, then $-2^{M-1}+1 \\leq n-2^{M} \\leq 0$, so $n-2^{M}=\\sum_{j=1}^{2 k}(-1)^{j-1} 2^{m_{j}}$ for some integers $k \\geq 0$ and $0 \\leq m_{1}<\\cdots<m_{2 k} \\leq M-1$ by the induction hypothesis, and $n=\\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}$, where $m_{2 k+1}=M$.\n(b) First Approach. Let $M \\geq 0$ be an integer. The solution for part (a) shows that the number of even (respectively, odd) weight integers in the range 1 through $2^{M}$ coincides with the number of subsets in $\\{0,1,2, \\ldots, M\\}$ whose cardinality has remainder 1 (respectively, 3) modulo 4. Therefore, the difference of these numbers is\n\n$$\n\\sum_{k=0}^{\\lfloor M / 2\\rfloor}(-1)^{k}\\binom{M+1}{2 k+1}=\\frac{(1+\\mathrm{i})^{M+1}-(1-\\mathrm{i})^{M+1}}{2 \\mathrm{i}}=2^{(M+1) / 2} \\sin \\frac{(M+1) \\pi}{4}\n$$\n\nwhere $\\mathrm{i}=\\sqrt{-1}$ is the imaginary unit. Thus, the required difference is $2^{1009}$.\nSecond Approach. For every integer $M \\geq 0$, let $A_{M}=\\sum_{n=-2^{M}+1}^{0}(-1)^{w(n)}$ and let $B_{M}=$ $\\sum_{n=1}^{2^{M}}(-1)^{w(n)}$; thus, $B_{M}$ evaluates the difference of the number of even weight integers in the range 1 through $2^{M}$ and the number of odd weight integers in that range.\n\nNotice that\n\n$$\nw(n)= \\begin{cases}w\\left(n+2^{M}\\right)+1 & \\text { if }-2^{M}+1 \\leq n \\leq-2^{M-1} \\\\ w\\left(n-2^{M}\\right) & \\text { if } 2^{M-1}+1 \\leq n \\leq 2^{M}\\end{cases}\n$$\n\nto get\n\n$$\n\\begin{aligned}\n& A_{M}=-\\sum_{n=-2^{M}+1}^{-2^{M-1}}(-1)^{w\\left(n+2^{M}\\right)}+\\sum_{n=-2^{M-1}+1}^{0}(-1)^{w(n)}=-B_{M-1}+A_{M-1}, \\\\\n& B_{M}=\\sum_{n=1}^{2^{M-1}}(-1)^{w(n)}+\\sum_{n=2^{M-1}+1}^{2^{M}}(-1)^{w\\left(n-2^{M}\\right)}=B_{M-1}+A_{M-1} .\n\\end{aligned}\n$$\n\nIteration yields\n\n$$\n\\begin{aligned}\nB_{M} & =A_{M-1}+B_{M-1}=\\left(A_{M-2}-B_{M-2}\\right)+\\left(A_{M-2}+B_{M-2}\\right)=2 A_{M-2} \\\\\n& =2 A_{M-3}-2 B_{M-3}=2\\left(A_{M-4}-B_{M-4}\\right)-2\\left(A_{M-4}+B_{M-4}\\right)=-4 B_{M-4}\n\\end{aligned}\n$$\n\nThus, $B_{2017}=(-4)^{504} B_{1}=2^{1008} B_{1}$; since $B_{1}=(-1)^{w(1)}+(-1)^{w(2)}=2$, it follows that $B_{2017}=$ $2^{1009}$ 。", "metadata": {"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}}
{"year": "2017", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "RMM", "problem": "Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k \\leq n$, and $k+1$ distinct integers $x_{1}, x_{2}, \\ldots, x_{k+1}$ such that\n\n$$\nP\\left(x_{1}\\right)+P\\left(x_{2}\\right)+\\cdots+P\\left(x_{k}\\right)=P\\left(x_{k+1}\\right)\n$$\n\nSemen Petrov, Russia\nNote. A polynomial is monic if the coefficient of the highest power is one.", "solution": "There is only one such integer, namely, $n=2$. In this case, if $P$ is a constant polynomial, the required condition is clearly satisfied; if $P=X+c$, then $P(c-1)+P(c+1)=$ $P(3 c)$; and if $P=X^{2}+q X+r$, then $P(X)=P(-X-q)$.\n\nTo rule out all other values of $n$, it is sufficient to exhibit a monic polynomial $P$ of degree at most $n$ with integer coefficients, whose restriction to the integers is injective, and $P(x) \\equiv 1$ $(\\bmod n)$ for all integers $x$. This is easily seen by reading the relation in the statement modulo $n$, to deduce that $k \\equiv 1(\\bmod n)$, so $k=1$, since $1 \\leq k \\leq n$; hence $P\\left(x_{1}\\right)=P\\left(x_{2}\\right)$ for some distinct integers $x_{1}$ and $x_{2}$, which contradicts injectivity.\n\nIf $n=1$, let $P=X$, and if $n=4$, let $P=X^{4}+7 X^{2}+4 X+1$. In the latter case, clearly, $P(x) \\equiv 1(\\bmod 4)$ for all integers $x$; and $P$ is injective on the integers, since $P(x)-P(y)=$ $(x-y)\\left((x+y)\\left(x^{2}+y^{2}+7\\right)+4\\right)$, and the absolute value of $(x+y)\\left(x^{2}+y^{2}+7\\right)$ is either 0 or at least 7 for integral $x$ and $y$.\n\nAssume henceforth $n \\geq 3, n \\neq 4$, and let $f_{n}=(X-1)(X-2) \\cdots(X-n)$. Clearly, $f_{n}(x) \\equiv$ $0(\\bmod n)$ for all integers $x$. If $n$ is odd, then $f_{n}$ is non-decreasing on the integers; and if, in addition, $n>3$, then $f_{n}(x) \\equiv 0(\\bmod n+1)$ for all integers $x$, since $f_{n}(0)=-n!=-1 \\cdot 2 \\cdot \\ldots$. $\\frac{n+1}{2} \\cdots \\cdot n \\equiv 0(\\bmod n+1)$.\n\nFinally, let $P=f_{n}+n X+1$ if $n$ is odd, and let $P=f_{n-1}+n X+1$ if $n$ is even. In either case, $P$ is strictly increasing, hence injective, on the integers, and $P(x) \\equiv 1(\\bmod n)$ for all integers $x$.\n\nRemark. The polynomial $P=f_{n}+n X+1$ works equally well for even $n>2$. To prove injectivity, notice that $P$ is strictly monotone, hence injective, on non-positive (respectively, positive) integers. Suppose, if possible, that $P(a)=P(b)$ for some integers $a \\leq 0$ and $b>0$. Notice that $P(a) \\geq P(0)=n!+1>n^{2}+1=P(n)$, since $n \\geq 4$, to infer that $b \\geq n+1$. It is therefore sufficient to show that $P(x)>P(n+1-x)>P(x-1)$ for all integers $x \\geq n+1$. The former inequality is trivial, since $f_{n}(x)=f_{n}(n+1-x)$ for even $n$. For the latter, write\n\n$$\n\\begin{aligned}\nP(n+1-x)-P(x-1) & =(x-1) \\cdots(x-n)-(x-2) \\cdots(x-n-1)+n(n+2-2 x) \\\\\n& =n((x-2) \\cdots(x-n)+(n-2)-2(x-2)) \\geq n(n-2)>0,\n\\end{aligned}\n$$\n\nsince $(x-3) \\cdots(x-n) \\geq 2$.", "metadata": {"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}}
{"year": "2017", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "RMM", "problem": "Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \\ldots, A_{k}$ of $X$ is tight if the union $A_{1} \\cup \\cdots \\cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection of proper non-empty subsets of $X$, no non-empty subcollection of which is tight.\n\nNote. A subset $A$ of $X$ is proper if $A \\neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.\n\n## Alexander Polyansky, Russia", "solution": ". (Ilya Bogdanov) The required maximum is $2 n-2$. To describe a ( $2 n-2$ )-element collection satisfying the required conditions, write $X=\\{1,2, \\ldots, n\\}$ and set $B_{k}=\\{1,2, \\ldots, k\\}$, $k=1,2, \\ldots, n-1$, and $B_{k}=\\{k-n+2, k-n+3, \\ldots, n\\}, k=n, n+1, \\ldots, 2 n-2$. To show that no subcollection of the $B_{k}$ is tight, consider a subcollection $\\mathcal{C}$ whose union $U$ is a proper subset of $X$, let $m$ be an element in $X \\backslash U$, and notice that $\\mathcal{C}$ is a subcollection of $\\left\\{B_{1}, \\ldots, B_{m-1}, B_{m+n-1}, \\ldots, B_{2 n-2}\\right\\}$, since the other $B$ 's are precisely those containing $m$. If $U$ contains elements less than $m$, let $k$ be the greatest such and notice that $B_{k}$ is the only member of $\\mathcal{C}$ containing $k$; and if $U$ contains elements greater than $m$, let $k$ be the least such and notice that $B_{k+n-2}$ is the only member of $\\mathcal{C}$ containing $k$. Consequently, $\\mathcal{C}$ is not tight.\n\nWe now proceed to show by induction on $n \\geq 2$ that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2 n-2$. The base case $n=2$ is clear, so let $n>2$ and suppose, if possible, that $\\mathcal{B}$ is a collection of $2 n-1$ proper non-empty subsets of $X$ containing no tight subcollection.\n\nTo begin, notice that $\\mathcal{B}$ has an empty intersection: if the members of $\\mathcal{B}$ shared an element $x$, then $\\mathcal{B}^{\\prime}=\\{B \\backslash\\{x\\}: B \\in \\mathcal{B}, B \\neq\\{x\\}\\}$ would be a collection of at least $2 n-2$ proper non-empty subsets of $X \\backslash\\{x\\}$ containing no tight subcollection, and the induction hypothesis would be contradicted.\n\nNow, for every $x$ in $X$, let $\\mathcal{B}_{x}$ be the (non-empty) collection of all members of $\\mathcal{B}$ not containing $x$. Since no subcollection of $\\mathcal{B}$ is tight, $\\mathcal{B}_{x}$ is not tight, and since the union of $\\mathcal{B}_{x}$ does not contain $x$, some $x^{\\prime}$ in $X$ is covered by a single member of $\\mathcal{B}_{x}$. In other words, there is a single set in $\\mathcal{B}$ covering $x^{\\prime}$ but not $x$. In this case, draw an arrow from $x$ to $x^{\\prime}$. Since there is at least one arrow from each $x$ in $X$, some of these arrows form a (minimal) cycle $x_{1} \\rightarrow x_{2} \\rightarrow \\cdots \\rightarrow x_{k} \\rightarrow x_{k+1}=x_{1}$ for some suitable integer $k \\geq 2$. Let $A_{i}$ be the unique member of $\\mathcal{B}$ containing $x_{i+1}$ but not $x_{i}$, and let $X^{\\prime}=\\left\\{x_{1}, x_{2}, \\ldots, x_{k}\\right\\}$.\n\nRemove $A_{1}, A_{2}, \\ldots, A_{k}$ from $\\mathcal{B}$ to obtain a collection $\\mathcal{B}^{\\prime}$ each member of which either contains or is disjoint from $X^{\\prime}$ : for if a member $B$ of $\\mathcal{B}^{\\prime}$ contained some but not all elements of $X^{\\prime}$, then $B$ should contain $x_{i+1}$ but not $x_{i}$ for some $i$, and $B=A_{i}$, a contradiction. This rules out the case $k=n$, for otherwise $\\mathcal{B}=\\left\\{A_{1}, A_{2}, \\ldots, A_{n}\\right\\}$, so $|\\mathcal{B}|<2 n-1$.\n\nTo rule out the case $k<n$, consider an extra element $x^{*}$ outside $X$ and let\n\n$$\n\\mathcal{B}^{*}=\\left\\{B: B \\in \\mathcal{B}^{\\prime}, B \\cap X^{\\prime}=\\varnothing\\right\\} \\cup\\left\\{\\left(B \\backslash X^{\\prime}\\right) \\cup\\left\\{x^{*}\\right\\}: B \\in \\mathcal{B}^{\\prime}, X^{\\prime} \\subseteq B\\right\\} ;\n$$\n\nthus, in each member of $\\mathcal{B}^{\\prime}$ containing $X^{\\prime}$, the latter is collapsed to singleton $x^{*}$. Notice that $\\mathcal{B}^{*}$ is a collection of proper non-empty subsets of $X^{*}=\\left(X \\backslash X^{\\prime}\\right) \\cup\\left\\{x^{*}\\right\\}$, no subcollection of which is tight. By the induction hypothesis, $\\left|\\mathcal{B}^{\\prime}\\right|=\\left|\\mathcal{B}^{*}\\right| \\leq 2\\left|X^{*}\\right|-2=2(n-k)$, so $|\\mathcal{B}| \\leq 2(n-k)+k=$ $2 n-k<2 n-1$, a final contradiction.", "metadata": {"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution 1"}}
{"year": "2017", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "RMM", "problem": "Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \\ldots, A_{k}$ of $X$ is tight if the union $A_{1} \\cup \\cdots \\cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection of proper non-empty subsets of $X$, no non-empty subcollection of which is tight.\n\nNote. A subset $A$ of $X$ is proper if $A \\neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.\n\n## Alexander Polyansky, Russia", "solution": ". Proceed again by induction on $n$ to show that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2 n-2$.\n\nConsider any collection $\\mathcal{B}$ of proper non-empty subsets of $X$ with no tight subcollection (we call such collection good). Assume that there exist $M, N \\in \\mathcal{B}$ such that $M \\cup N$ is distinct from $M, N$, and $X$. In this case, we will show how to modify $\\mathcal{B}$ so that it remains good, contains the same number of sets, but the total number of elements in the sets of $\\mathcal{B}$ increases.\n\nConsider a maximal (relative to set-theoretic inclusion) subcollection $\\mathcal{C} \\subseteq \\mathcal{B}$ such that the set $C=\\bigcup_{A \\in \\mathcal{C}} A$ is distinct from $X$ and from all members of $\\mathcal{C}$. Notice here that the union of any subcollection $\\mathcal{D} \\subset \\mathcal{B}$ cannot coincide with any $K \\in \\mathcal{B} \\backslash \\mathcal{D}$, otherwise $\\{K\\} \\cup \\mathcal{D}$ would be tight. Surely, $\\mathcal{C}$ exists (since $\\{M, N\\}$ is an example of a collection satisfying the requirements on $\\mathcal{C}$, except for maximality); moreover, $C \\notin \\mathcal{B}$ by the above remark.\n\nSince $C \\neq X$, there exists an $L \\in \\mathcal{C}$ and $x \\in L$ such that $L$ is the unique set in $\\mathcal{C}$ containing $x$. Now replace in $\\mathcal{B}$ the set $L$ by $C$ in order to obtain a new collection $\\mathcal{B}^{\\prime}$ (then $\\left.\\left|\\mathcal{B}^{\\prime}\\right|=|\\mathcal{B}|\\right)$. We claim that $\\mathcal{B}^{\\prime}$ is good.\n\nAssume, to the contrary, that $\\mathcal{B}^{\\prime}$ contained a tight subcollection $\\mathcal{T}$; clearly, $C \\in \\mathcal{T}$, otherwise $\\mathcal{B}$ is not good. If $\\mathcal{T} \\subseteq \\mathcal{C} \\cup\\{C\\}$, then $C$ is the unique set in $\\mathcal{T}$ containing $x$ which is impossible. Therefore, there exists $P \\in \\mathcal{T} \\backslash(\\mathcal{C} \\cup\\{C\\})$. By maximality of $\\mathcal{C}$, the collection $\\mathcal{C} \\cup\\{P\\}$ does not satisfy the requirements imposed on $\\mathcal{C}$; since $P \\cup C \\neq X$, this may happen only if $C \\cup P=P$, i.e., if $C \\subset P$. But then $\\mathcal{G}=(\\mathcal{T} \\backslash\\{C\\}) \\cup \\mathcal{C}$ is a tight subcollection in $\\mathcal{B}$ : all elements of $C$ are covered by $\\mathcal{G}$ at least twice (by $P$ and an element of $\\mathcal{C}$ ), and all the rest elements are covered by $\\mathcal{G}$ the same number of times as by $\\mathcal{T}$. A contradiction. Thus $\\mathcal{B}^{\\prime}$ is good.\n\nSuch modifications may be performed finitely many times, since the total number of elements of sets in $\\mathcal{B}$ increases. Thus, at some moment we arrive at a good collection $\\mathcal{B}$ for which the procedure no longer applies. This means that for every $M, N \\in \\mathcal{B}$, either $M \\cup N=X$ or one of them is contained in the other.\n\nNow let $M$ be a minimal (with respect to inclusion) set in $\\mathcal{B}$. Then each set in $\\mathcal{B}$ either contains $M$ or forms $X$ in union with $M$ (i.e., contains $X \\backslash M$ ). Now one may easily see that the two collections\n\n$$\n\\mathcal{B}_{+}=\\{A \\backslash M: A \\in \\mathcal{B}, M \\subset A, A \\neq M\\}, \\quad \\mathcal{B}_{-}=\\{A \\cap M: A \\in \\mathcal{B}, X \\backslash M \\subset A, A \\neq X \\backslash M\\}\n$$\n\nare good as collections of subsets of $X \\backslash M$ and $M$, respectively; thus, by the induction hypothesis, we have $\\left|\\mathcal{B}_{+}\\right|+\\left|\\mathcal{B}_{-}\\right| \\leq 2 n-4$.\n\nFinally, each set $A \\in \\mathcal{B}$ either produces a set in one of the two new collections, or coincides with $M$ or $X \\backslash M$. Thus $|\\mathcal{B}| \\leq\\left|\\mathcal{B}_{+}\\right|+\\left|\\mathcal{B}_{-}\\right|+2 \\leq 2 n-2$, as required.", "metadata": {"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution 2"}}
{"year": "2017", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "RMM", "problem": "Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \\ldots, A_{k}$ of $X$ is tight if the union $A_{1} \\cup \\cdots \\cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection of proper non-empty subsets of $X$, no non-empty subcollection of which is tight.\n\nNote. A subset $A$ of $X$ is proper if $A \\neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.\n\n## Alexander Polyansky, Russia", "solution": ". We provide yet another proof of the estimate $|\\mathcal{B}| \\leq 2 n-2$, using the notion of a good collection from Solution 2. Arguing indirectly, we assume that there exists a good collection $\\mathcal{B}$ with $|\\mathcal{B}| \\geq 2 n-1$, and choose one such for the minimal possible value of $n$. Clearly, $n>2$.\n\nFirstly, we perform a different modification of $\\mathcal{B}$. Choose any $x \\in X$, and consider the subcollection $\\mathcal{B}_{x}=\\{B: B \\in \\mathcal{B}, x \\notin B\\}$. By our assumption, $\\mathcal{B}_{x}$ is not tight. As the union of sets in $\\mathcal{B}_{x}$ is distinct from $X$, either this collection is empty, or there exists an element $y \\in X$ contained in a unique member $A_{x}$ of $\\mathcal{B}_{x}$. In the former case, we add the set $B_{x}=X \\backslash\\{x\\}$ to $\\mathcal{B}$, and in the latter we replace $A_{x}$ by $B_{x}$, to form a new collection $\\mathcal{B}^{\\prime}$. (Notice that if $B_{x} \\in \\mathcal{B}$, then $B_{x} \\in \\mathcal{B}_{x}$ and $y \\in B_{x}$, so $B_{x}=A_{x}$.)\n\nWe claim that the collection $\\mathcal{B}^{\\prime}$ is also good. Indeed, if $\\mathcal{B}^{\\prime}$ has a tight subcollection $\\mathcal{T}$, then $B_{x}$ should lie in $\\mathcal{T}$. Then, as the union of the sets in $\\mathcal{T}$ is distinct from $X$, we should have $\\mathcal{T} \\subseteq \\mathcal{B}_{x} \\cup\\left\\{B_{x}\\right\\}$. But in this case an element $y$ is contained in a unique member of $\\mathcal{T}$, namely $B_{x}$, so $\\mathcal{T}$ is not tight - a contradiction.\n\nPerform this procedure for every $x \\in X$, to get a good collection $\\mathcal{B}$ containing the sets $B_{x}=X \\backslash\\{x\\}$ for all $x \\in X$. Consider now an element $x \\in X$ such that $\\left|\\mathcal{B}_{x}\\right|$ is maximal. As we have mentioned before, there exists an element $y \\in X$ belonging to a unique member (namely, $B_{x}$ ) of $\\mathcal{B}_{x}$. Thus, $\\mathcal{B}_{x} \\backslash\\left\\{B_{x}\\right\\} \\subset \\mathcal{B}_{y}$; also, $B_{y} \\in \\mathcal{B}_{y} \\backslash \\mathcal{B}_{x}$. Thus we get $\\left|\\mathcal{B}_{y}\\right| \\geq\\left|\\mathcal{B}_{x}\\right|$, which by the maximality assumption yields the equality, which in turn means that $\\mathcal{B}_{y}=\\left(\\mathcal{B}_{x} \\backslash\\left\\{B_{x}\\right\\}\\right) \\cup\\left\\{B_{y}\\right\\}$.\n\nTherefore, each set in $\\mathcal{B} \\backslash\\left\\{B_{x}, B_{y}\\right\\}$ contains either both $x$ and $y$, or none of them. Collapsing $\\{x, y\\}$ to singleton $x^{*}$, we get a new collection of $|\\mathcal{B}|-2$ subsets of $(X \\backslash\\{x, y\\}) \\cup\\left\\{x^{*}\\right\\}$ containing no tight subcollection. This contradicts minimality of $n$.\n\nRemarks. 1. Removal of the condition that subsets be proper would only increase the maximum by 1. The 'non-emptiness' condition could also be omitted, since the empty set forms a tight collection by itself, but the argument is a bit too formal to be considered.\n2. There are many different examples of good collections of $2 n-2$ sets. E.g., applying the algorithm from the first part of Solution 2 to the example shown in Solution 1, one may get the following example: $B_{k}=\\{1,2, \\ldots, k\\}, k=1,2, \\ldots, n-1$, and $B_{k}=X \\backslash\\{k-n+1\\}$, $k=n, n+1, \\ldots, 2 n-2$.", "metadata": {"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution 3"}}