{
  "index": "1979-A-5",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-5\nDenote by \\( [x] \\) the greatest integer less than or equal to \\( x \\) and by \\( S(x) \\) the sequence \\( [x],[2 x],[3 x], \\ldots \\). Prove that there are distinct real solutions \\( \\alpha \\) and \\( \\beta \\) of the equation \\( x^{3}-10 x^{2}+29 x-25=0 \\) such that infinitely many positive integers appear both in \\( S(\\alpha) \\) and in \\( S(\\beta) \\).",
  "solution": "A-5.\nLet \\( f(x)=x^{3}-10 x^{2}+29 x-25 \\). Then the table\n\\[\n\\begin{array}{rrrrrr}\nx & 1 & 2 & 3 & 5 & 6 \\\\\nf(x) & -5 & 1 & -1 & -5 & 5\n\\end{array}\n\\]\nshows that \\( f(x)=0 \\) has three real solutions \\( a, b, c \\) with \\( 1<a<2,2<b<3,5<c<6 \\). The number of integers that the set \\( \\{1,2, \\ldots, n\\} \\) has in common with \\( S(a), S(b) \\), and \\( S(c) \\) is \\( [n / a],[n / b] \\), and \\( [n / c] \\), respectively. Since\n\\[\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}>\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=1\n\\]\none sees that\n\\[\n\\lim _{n \\rightarrow \\infty}\\left\\{\\left[\\frac{n}{a}\\right]+\\left[\\frac{n}{b}\\right]+\\left[\\frac{n}{c}\\right]-n\\right\\}=\\infty\n\\]\nand hence that an infinite number of positive integers appear in more than one of \\( S(a), S(b) \\), \\( S(c) \\). This implies that some pair of these sets must have an infinite intersection.",
  "vars": [
    "x",
    "n",
    "\\\\alpha",
    "\\\\beta"
  ],
  "params": [
    "S",
    "f",
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvar",
        "n": "intnum",
        "\\alpha": "alphaval",
        "\\beta": "betaval",
        "S": "seqfunc",
        "f": "polyfun",
        "a": "rootone",
        "b": "roottwo",
        "c": "rootthr"
      },
      "question": "Problem A-5\nDenote by \\( [realvar] \\) the greatest integer less than or equal to \\( realvar \\) and by \\( seqfunc(realvar) \\) the sequence \\( [realvar],[2 realvar],[3 realvar], \\ldots \\). Prove that there are distinct real solutions \\( alphaval \\) and \\( betaval \\) of the equation \\( realvar^{3}-10 realvar^{2}+29 realvar-25=0 \\) such that infinitely many positive integers appear both in \\( seqfunc(alphaval) \\) and in \\( seqfunc(betaval) \\).",
      "solution": "A-5.\nLet \\( polyfun(realvar)=realvar^{3}-10 realvar^{2}+29 realvar-25 \\). Then the table\n\\[\n\\begin{array}{rrrrrr}\nrealvar & 1 & 2 & 3 & 5 & 6 \\\\\npolyfun(realvar) & -5 & 1 & -1 & -5 & 5\n\\end{array}\n\\]\nshows that \\( polyfun(realvar)=0 \\) has three real solutions \\( rootone, roottwo, rootthr \\) with \\( 1<rootone<2,2<roottwo<3,5<rootthr<6 \\). The number of integers that the set \\( \\{1,2, \\ldots, intnum\\} \\) has in common with \\( seqfunc(rootone), seqfunc(roottwo) \\), and \\( seqfunc(rootthr) \\) is \\( [intnum / rootone],[intnum / roottwo] \\), and \\( [intnum / rootthr] \\), respectively. Since\n\\[\n\\frac{1}{rootone}+\\frac{1}{roottwo}+\\frac{1}{rootthr}>\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=1\n\\]\none sees that\n\\[\n\\lim _{intnum \\rightarrow \\infty}\\left\\{\\left[\\frac{intnum}{rootone}\\right]+\\left[\\frac{intnum}{roottwo}\\right]+\\left[\\frac{intnum}{rootthr}\\right]-intnum\\right\\}=\\infty\n\\]\nand hence that an infinite number of positive integers appear in more than one of \\( seqfunc(rootone), seqfunc(roottwo) \\), \\( seqfunc(rootthr) \\). This implies that some pair of these sets must have an infinite intersection."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "chocolate",
        "n": "pineapple",
        "\\alpha": "thunderbolt",
        "\\beta": "eaglewind",
        "S": "galaxyway",
        "f": "rainforest",
        "a": "moonlight",
        "b": "starlight",
        "c": "buttercup"
      },
      "question": "Problem A-5\nDenote by \\( [chocolate] \\) the greatest integer less than or equal to \\( chocolate \\) and by \\( galaxyway(chocolate) \\) the sequence \\( [chocolate],[2 chocolate],[3 chocolate], \\ldots \\). Prove that there are distinct real solutions \\( thunderbolt \\) and \\( eaglewind \\) of the equation \\( chocolate^{3}-10 chocolate^{2}+29 chocolate-25=0 \\) such that infinitely many positive integers appear both in \\( galaxyway(thunderbolt) \\) and in \\( galaxyway(eaglewind) \\).",
      "solution": "A-5.\nLet \\( rainforest(chocolate)=chocolate^{3}-10 chocolate^{2}+29 chocolate-25 \\). Then the table\n\\[\n\\begin{array}{rrrrrr}\nchocolate & 1 & 2 & 3 & 5 & 6 \\\\\nrainforest(chocolate) & -5 & 1 & -1 & -5 & 5\n\\end{array}\n\\]\nshows that \\( rainforest(chocolate)=0 \\) has three real solutions \\( moonlight, starlight, buttercup \\) with \\( 1<moonlight<2,2<starlight<3,5<buttercup<6 \\). The number of integers that the set \\( \\{1,2, \\ldots, pineapple\\} \\) has in common with \\( galaxyway(moonlight), galaxyway(starlight) \\), and \\( galaxyway(buttercup) \\) is \\( [pineapple / moonlight],[pineapple / starlight] \\), and \\( [pineapple / buttercup] \\), respectively. Since\n\\[\n\\frac{1}{moonlight}+\\frac{1}{starlight}+\\frac{1}{buttercup}>\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=1\n\\]\none sees that\n\\[\n\\lim _{pineapple \\rightarrow \\infty}\\left\\{\\left[\\frac{pineapple}{moonlight}\\right]+\\left[\\frac{pineapple}{starlight}\\right]+\\left[\\frac{pineapple}{buttercup}\\right]-pineapple\\right\\}=\\infty\n\\]\nand hence that an infinite number of positive integers appear in more than one of \\( galaxyway(moonlight), galaxyway(starlight) \\), \\( galaxyway(buttercup) \\). This implies that some pair of these sets must have an infinite intersection."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownvalue",
        "n": "fractional",
        "\\alpha": "omegaval",
        "\\beta": "psivalues",
        "S": "staticset",
        "f": "constant",
        "a": "rootless",
        "b": "stemless",
        "c": "leafless"
      },
      "question": "Problem A-5\nDenote by \\( [knownvalue] \\) the greatest integer less than or equal to \\( knownvalue \\) and by \\( staticset(knownvalue) \\) the sequence \\( [knownvalue],[2 knownvalue],[3 knownvalue], \\ldots \\). Prove that there are distinct real solutions \\( omegaval \\) and \\( psivalues \\) of the equation \\( knownvalue^{3}-10 knownvalue^{2}+29 knownvalue-25=0 \\) such that infinitely many positive integers appear both in \\( staticset(omegaval) \\) and in \\( staticset(psivalues) \\).",
      "solution": "A-5.\nLet \\( constant(knownvalue)=knownvalue^{3}-10 knownvalue^{2}+29 knownvalue-25 \\). Then the table\n\\[\n\\begin{array}{rrrrrr}\nknownvalue & 1 & 2 & 3 & 5 & 6 \\\\\nconstant(knownvalue) & -5 & 1 & -1 & -5 & 5\n\\end{array}\n\\]\nshows that \\( constant(knownvalue)=0 \\) has three real solutions \\( rootless, stemless, leafless \\) with \\( 1<rootless<2,2<stemless<3,5<leafless<6 \\). The number of integers that the set \\{1,2, \\ldots, fractional\\} has in common with \\( staticset(rootless), staticset(stemless) \\), and \\( staticset(leafless) \\) is \\( [fractional / rootless],[fractional / stemless] \\), and \\( [fractional / leafless] \\), respectively. Since\n\\[\n\\frac{1}{rootless}+\\frac{1}{stemless}+\\frac{1}{leafless}>\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=1\n\\]\none sees that\n\\[\n\\lim _{fractional \\rightarrow \\infty}\\left\\{\\left[\\frac{fractional}{rootless}\\right]+\\left[\\frac{fractional}{stemless}\\right]+\\left[\\frac{fractional}{leafless}\\right]-fractional\\right\\}=\\infty\n\\]\nand hence that an infinite number of positive integers appear in more than one of \\( staticset(rootless), staticset(stemless), staticset(leafless) \\). This implies that some pair of these sets must have an infinite intersection."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "n": "hjgrkslac",
        "\\alpha": "mlqswzjka",
        "\\beta": "xvklduqwe",
        "S": "npghwzrtc",
        "f": "lskdntmqp",
        "a": "jwtpsdfrm",
        "b": "sndmczloh",
        "c": "tqghvpkla"
      },
      "question": "Problem A-5\nDenote by \\( [qzxwvtnp] \\) the greatest integer less than or equal to \\( qzxwvtnp \\) and by \\( npghwzrtc(qzxwvtnp) \\) the sequence \\( [qzxwvtnp],[2 qzxwvtnp],[3 qzxwvtnp], \\ldots \\). Prove that there are distinct real solutions \\( mlqswzjka \\) and \\( xvklduqwe \\) of the equation \\( qzxwvtnp^{3}-10 qzxwvtnp^{2}+29 qzxwvtnp-25=0 \\) such that infinitely many positive integers appear both in \\( npghwzrtc(mlqswzjka) \\) and in \\( npghwzrtc(xvklduqwe) \\).",
      "solution": "A-5.\nLet \\( lskdntmqp(qzxwvtnp)=qzxwvtnp^{3}-10 qzxwvtnp^{2}+29 qzxwvtnp-25 \\). Then the table\n\\[\n\\begin{array}{rrrrrr}\nqzxwvtnp & 1 & 2 & 3 & 5 & 6 \\\\\nlskdntmqp(qzxwvtnp) & -5 & 1 & -1 & -5 & 5\n\\end{array}\n\\]\nshows that \\( lskdntmqp(qzxwvtnp)=0 \\) has three real solutions \\( jwtpsdfrm, sndmczloh, tqghvpkla \\) with \\( 1<jwtpsdfrm<2, 2<sndmczloh<3, 5<tqghvpkla<6 \\). The number of integers that the set \\( \\{1,2, \\ldots, hjgrkslac\\} \\) has in common with \\( npghwzrtc(jwtpsdfrm), npghwzrtc(sndmczloh) \\), and \\( npghwzrtc(tqghvpkla) \\) is \\( [hjgrkslac / jwtpsdfrm],[hjgrkslac / sndmczloh] \\), and \\( [hjgrkslac / tqghvpkla] \\), respectively. Since\n\\[\n\\frac{1}{jwtpsdfrm}+\\frac{1}{sndmczloh}+\\frac{1}{tqghvpkla}>\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=1\n\\]\none sees that\n\\[\n\\lim _{hjgrkslac \\rightarrow \\infty}\\left\\{\\left[\\frac{hjgrkslac}{jwtpsdfrm}\\right]+\\left[\\frac{hjgrkslac}{sndmczloh}\\right]+\\left[\\frac{hjgrkslac}{tqghvpkla}\\right]-hjgrkslac\\right\\}=\\infty\n\\]\nand hence that an infinite number of positive integers appear in more than one of \\( npghwzrtc(jwtpsdfrm), npghwzrtc(sndmczloh) \\), \\( npghwzrtc(tqghvpkla) \\). This implies that some pair of these sets must have an infinite intersection."
    },
    "kernel_variant": {
      "question": "Let $\\lfloor x\\rfloor$ denote the greatest integer not exceeding $x$.  \nFor every positive real number $x$ put  \n\\[\n  S(x)\\;=\\;\\bigl\\{\\lfloor kx\\rfloor : k\\in\\mathbb N\\bigr\\}.\n\\]\n\nConsider the quintic polynomial  \n\\[\n  F(x)\\;=\\;\\bigl(5x-6\\bigr)\\bigl(6x-7\\bigr)\\bigl(7x-8\\bigr)\\bigl(8x-9\\bigr)\\bigl(9x-10\\bigr).\n\\]\n\n1.\\;Show that the equation $F(x)=0$ has the five distinct real roots  \n\\[\n  \\beta_{1}= \\dfrac65,\\qquad\n  \\beta_{2}= \\dfrac76,\\qquad\n  \\beta_{3}= \\dfrac87,\\qquad\n  \\beta_{4}= \\dfrac98,\\qquad\n  \\beta_{5}= \\dfrac{10}{9},\n\\]\nand prove the strict chain of inequalities  \n\\[\n  1\\;<\\;\\beta_{5}\\;<\\;\\beta_{4}\\;<\\;\\beta_{3}\\;<\\;\\beta_{2}\\;<\\;\\beta_{1}\\;<\\;2.\n  \\tag{$\\dagger$}\n\\]\n\n2.\\;Prove that the simultaneous intersection  \n\\[\n  S(\\beta_{1})\\cap S(\\beta_{2})\\cap S(\\beta_{3})\\cap S(\\beta_{4})\\cap S(\\beta_{5})\n\\]\ncontains infinitely many integers.  Moreover, its natural lower density satisfies  \n\\[\n  \\liminf_{N\\to\\infty}\n  \\frac{\\bigl|\\bigl(S(\\beta_{1})\\cap\\cdots\\cap S(\\beta_{5})\\bigr)\\cap\\{1,\\dots,N\\}\\bigr|}{N}\n  \\;\\ge\\;\n  \\frac{893}{2520}.\n\\]\n\n(Any explicit positive lower bound earns full credit; the exact value $\\dfrac{893}{2520}$ is required only for maximum credit.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "\\textbf{Step 1.\\;The roots and their ordering}\n\nBecause $F(x)$ is given as a product of five linear factors, each factor yields one zero:\n\\[\n  5x-6=0,\\;6x-7=0,\\;7x-8=0,\\;8x-9=0,\\;9x-10=0\n  \\quad\\Longrightarrow\\quad\n  x\\;=\\;\\beta_{1},\\dots ,\\beta_{5}\\;\\text{as stated}.\n\\]\n\nWrite $\\beta_{6-k}=1+\\dfrac1k$ for $k=5,6,7,8,9$.  Since the map\n\\[\n  k\\longmapsto 1+\\frac1k\n\\]\nis strictly decreasing on positive $k$, we have\n\\[\n  1+\\frac19<1+\\frac18<1+\\frac17<1+\\frac16<1+\\frac15<2,\n\\]\nwhich is precisely the inequality chain $(\\dagger)$.\n\n\\medskip\n\\textbf{Step 2.\\;A global counting device}\n\nFor $n\\in\\mathbb N$ define  \n\\[\n  T(n):=\\sum_{i=1}^{5}\\Bigl\\lfloor\\frac{n}{\\beta_{i}}\\Bigr\\rfloor .\n\\]\nUsing $\\lfloor y\\rfloor=y-\\{y\\}$ with $0\\le\\{y\\}<1$ we obtain  \n\\[\n  \\Bigl\\lfloor\\frac{n}{\\beta_{i}}\\Bigr\\rfloor\n    \\;\\ge\\;\\frac{n}{\\beta_{i}}-1 .\n\\]\nHence  \n\\[\n  T(n)\\;\\ge\\;n\\sum_{i=1}^{5}\\frac1{\\beta_{i}}-5 .\n\\tag{1}\n\\]\n\nA direct calculation gives  \n\\[\n  \\sum_{i=1}^{5}\\frac1{\\beta_{i}}\n  \\;=\\;\\frac56+\\frac67+\\frac78+\\frac89+\\frac9{10}\n  \\;=\\;\\frac{10973}{2520}\n  \\;=\\;4+\\frac{893}{2520}.\n\\]\nSubstituting this into (1) we find\n\\[\n  T(n)\\;\\ge\\;4n+\\frac{893}{2520}\\,n-5.\n\\tag{2}\n\\]\n\n\\medskip\n\\textbf{Step 3.\\;Relating $T(n)$ to multiplicities of appearance}\n\nFor $m\\in\\mathbb N$ let  \n\\[\n  c(m):=\\#\\bigl\\{\\,i\\in\\{1,\\dots,5\\}:m\\in S(\\beta_{i})\\,\\bigr\\},\n\\]\nand define the cumulative count  \n\\[\n  C(n):=\\sum_{m=1}^{n}c(m)\n       \\;=\\;\\sum_{i=1}^{5}\\bigl|S(\\beta_{i})\\cap\\{1,\\dots,n\\}\\bigr|.\n\\tag{3}\n\\]\n\nBecause each $\\beta_{i}>1$, every sequence $S(\\beta_{i})$ is strictly increasing.  Consequently  \n\\[\n  \\Bigl\\lfloor\\frac{n}{\\beta_{i}}\\Bigr\\rfloor\n     \\le\\bigl|S(\\beta_{i})\\cap\\{1,\\dots,n\\}\\bigr|\n     \\le\\Bigl\\lfloor\\frac{n}{\\beta_{i}}\\Bigr\\rfloor+1 .\n\\]\nSumming over $i$ yields  \n\\[\n  T(n)\\;\\le\\;C(n)\\;\\le\\;T(n)+5 .\n\\tag{4}\n\\]\n\n\\medskip\n\\textbf{Step 4.\\;Infinitude and positive lower density of the common intersection}\n\nLet  \n\\[\n  D(N):=\\#\\bigl\\{\\,m\\le N:c(m)=5\\,\\bigr\\},\n\\tag{5}\n\\]\nthe number of integers $\\le N$ that occur in \\emph{all} five sequences.\n\nEvery integer $m\\le N$ contributes exactly $c(m)$ to $C(N)$.  Splitting into the cases $c(m)=5$ and $c(m)\\le4$ we obtain  \n\\[\n  C(N)\n  \\;=\\;\\sum_{m\\le N,c(m)=5}5+\\sum_{m\\le N,c(m)\\le4}c(m)\n  \\;\\le\\;5D(N)+4\\bigl(N-D(N)\\bigr)\n  \\;=\\;4N+D(N).\n\\tag{6}\n\\]\n\nFrom (4) and the lower estimate (2) we also have  \n\\[\n  C(N)\\;\\ge\\;T(N)\\;\\ge\\;4N+\\frac{893}{2520}\\,N-5.\n\\tag{7}\n\\]\n\nCombining (6) and (7) yields  \n\\[\n  4N+D(N)\\;\\ge\\;4N+\\frac{893}{2520}\\,N-5,\n  \\qquad\\Longrightarrow\\qquad\n  D(N)\\;\\ge\\;\\frac{893}{2520}\\,N-5.\n\\tag{8}\n\\]\n\nDividing (8) by $N$ and taking the $\\liminf$ we deduce  \n\\[\n  \\liminf_{N\\to\\infty}\\frac{D(N)}{N}\n  \\;\\ge\\;\\frac{893}{2520}\\;>\\;0.\n\\]\n\nIn particular $D(N)\\to\\infty$, proving that\n\\[\n  S(\\beta_{1})\\cap S(\\beta_{2})\\cap S(\\beta_{3})\\cap S(\\beta_{4})\\cap S(\\beta_{5})\n\\]\nis infinite and, because $D(N)$ grows linearly with $N$, has the stated positive lower density.  $\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.640466",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher degree, more variables: the quintic brings five sequences instead of three, and the target is a triple intersection, not merely a pairwise one.  \n2. Additional constraints: the problem demands locating three specific roots whose sequences intersect infinitely often, rather than merely proving some intersection exists among all roots.  \n3. Deeper counting: the argument must raise the threshold from average “>1’’ (original problem) to “>2’’ and translate that into multiplicity ≥ 3, introducing a sharper combinatorial step.  \n4. More sophisticated structure: contestants must manage five interacting Beatty-type sequences and use a refined pigeonhole argument on C(5,3) possible triples.  \n5. Consequently the solution chain (factorisation, reciprocal sum estimate, global-to-local counting, and a second pigeonhole layer) is substantially longer and conceptually denser than in the original kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\lfloor x\\rfloor$ denote the greatest integer not exceeding $x$.  \nFor every positive real number $x$ put  \n\\[\n  S(x)\\;=\\;\\bigl\\{\\lfloor kx\\rfloor : k\\in\\mathbb N\\bigr\\}.\n\\]\n\nConsider the quintic polynomial  \n\\[\n  F(x)\\;=\\;\\bigl(5x-6\\bigr)\\bigl(6x-7\\bigr)\\bigl(7x-8\\bigr)\\bigl(8x-9\\bigr)\\bigl(9x-10\\bigr).\n\\]\n\n1.  Show that the equation $F(x)=0$ has the five distinct real roots  \n\\[\n  \\beta_{1}= \\dfrac65,\\qquad\n  \\beta_{2}= \\dfrac76,\\qquad\n  \\beta_{3}= \\dfrac87,\\qquad\n  \\beta_{4}= \\dfrac98,\\qquad\n  \\beta_{5}= \\dfrac{10}{9},\n\\]\nand prove the strict chain of inequalities  \n\\[\n  1\\;<\\;\\beta_{5}\\;<\\;\\beta_{4}\\;<\\;\\beta_{3}\\;<\\;\\beta_{2}\\;<\\;\\beta_{1}\\;<\\;2.\n\\tag{$\\dagger$}\n\\]\n\n2.  Prove that the simultaneous intersection  \n\\[\n  S(\\beta_{1})\\cap S(\\beta_{2})\\cap S(\\beta_{3})\\cap S(\\beta_{4})\\cap S(\\beta_{5})\n\\]\ncontains infinitely many integers.  Moreover, its natural lower density satisfies  \n\\[\n  \\liminf_{N\\to\\infty}\n  \\frac{\\bigl|\\bigl(S(\\beta_{1})\\cap\\cdots\\cap S(\\beta_{5})\\bigr)\\cap\\{1,\\dots,N\\}\\bigr|}{N}\n  \\;\\ge\\;\n  \\frac{893}{2520}.\n\\]\n\n(Any explicit positive lower bound earns full credit; the exact value $\\dfrac{893}{2520}$ is required only for maximum credit.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "\\textbf{Step 1.  The roots and their ordering}\n\nBecause $F(x)$ is already written as a product of five linear factors, each factor yields one zero:\n\\[\n  5x-6=0,\\;6x-7=0,\\;7x-8=0,\\;8x-9=0,\\;9x-10=0\n  \\quad\\Longrightarrow\\quad\n  \\beta_{1},\\dots ,\\beta_{5}\\text{ as stated}.\n\\]\n\nPut $\\beta_{\\,10-k}=1+\\dfrac1k$ for $k=5,6,7,8,9$.  \nSince $k\\mapsto 1+\\dfrac1k$ is strictly decreasing,\n\\[\n  1+\\frac19<1+\\frac18<1+\\frac17<1+\\frac16<1+\\frac15<2,\n\\]\nwhich is exactly chain $(\\dagger)$.\n\n\\medskip\n\\textbf{Step 2.  A global counting device}\n\nFor $n\\in\\mathbb N$ define  \n\\[\n  T(n):=\\sum_{i=1}^{5}\\Bigl\\lfloor\\frac{n}{\\beta_{i}}\\Bigr\\rfloor .\n\\]\nUsing $\\lfloor y\\rfloor=y-\\{y\\}$ with $0\\le\\{y\\}<1$ we have  \n\\[\n  \\Bigl\\lfloor\\frac{n}{\\beta_{i}}\\Bigr\\rfloor\n    \\;\\ge\\;\\frac{n}{\\beta_{i}}-1 .\n\\]\nHence  \n\\[\n  T(n)\\;\\ge\\;n\\sum_{i=1}^{5}\\frac1{\\beta_{i}}-5 .\n\\tag{1}\n\\]\n\nA direct calculation gives  \n\\[\n  \\sum_{i=1}^{5}\\frac1{\\beta_{i}}\n  =\\frac56+\\frac67+\\frac78+\\frac89+\\frac9{10}\n  =\\frac{10973}{2520}\n  =4+\\frac{893}{2520}.\n\\]\nSubstituting into (1),\n\\[\n  T(n)\\;\\ge\\;4n+\\frac{893}{2520}\\,n-5.\n\\tag{2}\n\\]\n\n\\medskip\n\\textbf{Step 3.  Relating $T(n)$ to multiplicities of appearance}\n\nFor $m\\in\\mathbb N$ let  \n\\[\n  c(m):=\\#\\bigl\\{\\,i\\in\\{1,\\dots,5\\}:m\\in S(\\beta_{i})\\,\\bigr\\},\n\\]\nand define  \n\\[\n  C(n):=\\sum_{m=1}^{n}c(m)\n       \\;=\\;\\sum_{i=1}^{5}\\bigl|S(\\beta_{i})\\cap\\{1,\\dots,n\\}\\bigr|.\n\\tag{3}\n\\]\n\nBecause $\\beta_{i}>1$, each sequence $S(\\beta_{i})$ is strictly increasing.  Consequently  \n\\[\n  \\Bigl\\lfloor\\frac{n}{\\beta_{i}}\\Bigr\\rfloor\n     \\le\\bigl|S(\\beta_{i})\\cap\\{1,\\dots,n\\}\\bigr|\n     \\le\\Bigl\\lfloor\\frac{n}{\\beta_{i}}\\Bigr\\rfloor+1 .\n\\]\nSumming over $i$ yields  \n\\[\n  T(n)\\;\\le\\;C(n)\\;\\le\\;T(n)+5 .\n\\tag{4}\n\\]\n\n\\medskip\n\\textbf{Step 4.  Infinitude and positive lower density of the common intersection}\n\nLet  \n\\[\n  D(N):=\\#\\bigl\\{\\,m\\le N:c(m)=5\\,\\bigr\\},\n\\tag{5}\n\\]\nthe number of integers $\\le N$ that occur in \\emph{all} five sequences.\n\nEvery integer $m\\le N$ contributes exactly $c(m)$ to $C(N)$.  Splitting into the cases $c(m)=5$ and $c(m)\\le4$ we obtain  \n\\[\n  C(N)\n  \\;=\\;\\sum_{m\\le N,c(m)=5}5+\\sum_{m\\le N,c(m)\\le4}c(m)\n  \\;\\le\\;5D(N)+4\\bigl(N-D(N)\\bigr)\n  \\;=\\;4N+D(N).\n\\tag{6}\n\\]\n\nFrom (4) and the lower estimate (2) we also have  \n\\[\n  C(N)\\;\\ge\\;T(N)\\;\\ge\\;4N+\\frac{893}{2520}\\,N-5.\n\\tag{7}\n\\]\n\nCombining (6) and (7) gives  \n\\[\n  4N+D(N)\\;\\ge\\;4N+\\frac{893}{2520}\\,N-5,\n  \\qquad\\Longrightarrow\\qquad\n  D(N)\\;\\ge\\;\\frac{893}{2520}\\,N-5.\n\\tag{8}\n\\]\n\nDividing (8) by $N$ and taking $\\liminf$ we deduce  \n\\[\n  \\liminf_{N\\to\\infty}\\frac{D(N)}{N}\n  \\;\\ge\\;\\frac{893}{2520}\\;>\\;0.\n\\]\n\nIn particular $D(N)\\to\\infty$, proving that\n\\[\n  S(\\beta_{1})\\cap S(\\beta_{2})\\cap S(\\beta_{3})\\cap S(\\beta_{4})\\cap S(\\beta_{5})\n\\]\nis infinite and, because $D(N)\\to\\infty$, has the stated positive lower density.  This completes the solution. $\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.508708",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher degree, more variables: the quintic brings five sequences instead of three, and the target is a triple intersection, not merely a pairwise one.  \n2. Additional constraints: the problem demands locating three specific roots whose sequences intersect infinitely often, rather than merely proving some intersection exists among all roots.  \n3. Deeper counting: the argument must raise the threshold from average “>1’’ (original problem) to “>2’’ and translate that into multiplicity ≥ 3, introducing a sharper combinatorial step.  \n4. More sophisticated structure: contestants must manage five interacting Beatty-type sequences and use a refined pigeonhole argument on C(5,3) possible triples.  \n5. Consequently the solution chain (factorisation, reciprocal sum estimate, global-to-local counting, and a second pigeonhole layer) is substantially longer and conceptually denser than in the original kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}