{
  "index": "1982-A-5",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-5\nLet \\( a, b, c \\), and \\( d \\) be positive integers and\n\\[\nr=1-\\frac{a}{b}-\\frac{c}{d}\n\\]\n\nGiven that \\( a+c \\leqslant 1982 \\) and \\( r>0 \\), prove that\n\\[\nr>\\frac{1}{1983^{3}} .\n\\]",
  "solution": "A-5.\nWe are given that\n\\[\nr=1-\\frac{a}{b}-\\frac{c}{d}=\\frac{b d-a d-b c}{b d}>0 .\n\\]\n\nThus \\( b d-a d-b c \\) is a positive integer and so \\( r \\geqslant 1 / b d \\). We may assume without loss of generality that \\( b \\leqslant d \\). If \\( b \\leqslant d \\leqslant 1983, r \\geqslant 1983^{-2}>1983^{-3} \\). Since \\( a+c \\leqslant 1982 \\), if \\( 1983 \\leqslant b \\leqslant \\) \\( d \\), one has\n\\[\nr \\geqslant 1-\\frac{a}{1983}-\\frac{c}{1983} \\geqslant 1-\\frac{1982}{1983}=\\frac{1}{1983}>\\frac{1}{1983^{3}} .\n\\]\n\nThe remaining case is that with \\( b<1983<d \\). Then the \\( d \\) that minimizes \\( r \\) for fixed \\( a, b, c \\) is \\( 1+[b c /(b-a)] \\), where \\( [x] \\) is the greatest integer in \\( x \\). This \\( d \\) is at most \\( 1983 b \\) since \\( b-a \\geqslant 1 \\) and \\( c<1982 \\) and thus\n\\[\nr \\geqslant \\frac{1}{b d} \\geqslant \\frac{1}{1983 b^{2}}>\\frac{1}{1983^{3}} .\n\\]\n\nHence we have the desired inequality in all cases.",
  "vars": [
    "a",
    "b",
    "c",
    "d",
    "r"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "integera",
        "b": "integerb",
        "c": "integerc",
        "d": "integerd",
        "r": "residual"
      },
      "question": "Problem A-5\nLet \\( integera, integerb, integerc \\), and \\( integerd \\) be positive integers and\n\\[\nresidual=1-\\frac{integera}{integerb}-\\frac{integerc}{integerd}\n\\]\n\nGiven that \\( integera+integerc \\leqslant 1982 \\) and \\( residual>0 \\), prove that\n\\[\nresidual>\\frac{1}{1983^{3}} .\n\\]",
      "solution": "A-5.\nWe are given that\n\\[\nresidual=1-\\frac{integera}{integerb}-\\frac{integerc}{integerd}=\\frac{integerb integerd-integera integerd-integerb integerc}{integerb integerd}>0 .\n\\]\n\nThus \\( integerb integerd-integera integerd-integerb integerc \\) is a positive integer and so \\( residual \\geqslant 1 / integerb integerd \\). We may assume without loss of generality that \\( integerb \\leqslant integerd \\). If \\( integerb \\leqslant integerd \\leqslant 1983, residual \\geqslant 1983^{-2}>1983^{-3} \\). Since \\( integera+integerc \\leqslant 1982 \\), if \\( 1983 \\leqslant integerb \\leqslant \\) \\( integerd \\), one has\n\\[\nresidual \\geqslant 1-\\frac{integera}{1983}-\\frac{integerc}{1983} \\geqslant 1-\\frac{1982}{1983}=\\frac{1}{1983}>\\frac{1}{1983^{3}} .\n\\]\n\nThe remaining case is that with \\( integerb<1983<integerd \\). Then the \\( integerd \\) that minimizes \\( residual \\) for fixed \\( integera, integerb, integerc \\) is \\( 1+[integerb integerc /(integerb-integera)] \\), where \\( [x] \\) is the greatest integer in \\( x \\). This \\( integerd \\) is at most \\( 1983 integerb \\) since \\( integerb-integera \\geqslant 1 \\) and \\( integerc<1982 \\) and thus\n\\[\nresidual \\geqslant \\frac{1}{integerb integerd} \\geqslant \\frac{1}{1983 integerb^{2}}>\\frac{1}{1983^{3}} .\n\\]\n\nHence we have the desired inequality in all cases."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "pineapple",
        "b": "honeycomb",
        "c": "butterfly",
        "d": "framework",
        "r": "lighthouse"
      },
      "question": "Problem A-5\nLet \\( pineapple, honeycomb, butterfly \\), and \\( framework \\) be positive integers and\n\\[\nlighthouse = 1-\\frac{pineapple}{honeycomb}-\\frac{butterfly}{framework}\n\\]\n\nGiven that \\( pineapple + butterfly \\leqslant 1982 \\) and \\( lighthouse > 0 \\), prove that\n\\[\nlighthouse > \\frac{1}{1983^{3}} .\n\\]",
      "solution": "A-5.\nWe are given that\n\\[\nlighthouse = 1-\\frac{pineapple}{honeycomb}-\\frac{butterfly}{framework}=\\frac{honeycomb framework-pineapple framework-honeycomb butterfly}{honeycomb framework}>0 .\n\\]\n\nThus \\( honeycomb framework-pineapple framework-honeycomb butterfly \\) is a positive integer and so \\( lighthouse \\geqslant 1 / honeycomb framework \\). We may assume without loss of generality that \\( honeycomb \\leqslant framework \\). If \\( honeycomb \\leqslant framework \\leqslant 1983, lighthouse \\geqslant 1983^{-2}>1983^{-3} \\). Since \\( pineapple + butterfly \\leqslant 1982 \\), if \\( 1983 \\leqslant honeycomb \\leqslant \\) \\( framework \\), one has\n\\[\nlighthouse \\geqslant 1-\\frac{pineapple}{1983}-\\frac{butterfly}{1983} \\geqslant 1-\\frac{1982}{1983}=\\frac{1}{1983}>\\frac{1}{1983^{3}} .\n\\]\n\nThe remaining case is that with \\( honeycomb<1983<framework \\). Then the \\( framework \\) that minimizes \\( lighthouse \\) for fixed \\( pineapple, honeycomb, butterfly \\) is \\( 1+[ honeycomb butterfly /( honeycomb-pineapple ) ] \\), where \\( [x] \\) is the greatest integer in \\( x \\). This \\( framework \\) is at most \\( 1983 honeycomb \\) since \\( honeycomb-pineapple \\geqslant 1 \\) and \\( butterfly<1982 \\) and thus\n\\[\nlighthouse \\geqslant \\frac{1}{honeycomb framework} \\geqslant \\frac{1}{1983 honeycomb^{2}}>\\frac{1}{1983^{3}} .\n\\]\n\nHence we have the desired inequality in all cases."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "noninteger",
        "b": "irrational",
        "c": "negativeval",
        "d": "zeroelement",
        "r": "completeness"
      },
      "question": "Problem A-5\nLet \\( noninteger, irrational, negativeval \\), and \\( zeroelement \\) be positive integers and\n\\[\ncompleteness = 1-\\frac{noninteger}{irrational}-\\frac{negativeval}{zeroelement}\n\\]\n\nGiven that \\( noninteger + negativeval \\leqslant 1982 \\) and \\( completeness > 0 \\), prove that\n\\[\ncompleteness > \\frac{1}{1983^{3}} .\n\\]",
      "solution": "A-5.\nWe are given that\n\\[\ncompleteness = 1-\\frac{noninteger}{irrational}-\\frac{negativeval}{zeroelement}=\\frac{irrational zeroelement-noninteger zeroelement-irrational negativeval}{irrational zeroelement}>0 .\n\\]\n\nThus \\( irrational zeroelement-noninteger zeroelement-irrational negativeval \\) is a positive integer and so \\( completeness \\geqslant 1 / irrational zeroelement \\). We may assume without loss of generality that \\( irrational \\leqslant zeroelement \\). If \\( irrational \\leqslant zeroelement \\leqslant 1983, completeness \\geqslant 1983^{-2}>1983^{-3} \\). Since \\( noninteger + negativeval \\leqslant 1982 \\), if \\( 1983 \\leqslant irrational \\leqslant \\) \\( zeroelement \\), one has\n\\[\ncompleteness \\geqslant 1-\\frac{noninteger}{1983}-\\frac{negativeval}{1983} \\geqslant 1-\\frac{1982}{1983}=\\frac{1}{1983}>\\frac{1}{1983^{3}} .\n\\]\n\nThe remaining case is that with \\( irrational < 1983 < zeroelement \\). Then the \\( zeroelement \\) that minimizes \\( completeness \\) for fixed \\( noninteger, irrational, negativeval \\) is \\( 1+[irrational negativeval /(irrational - noninteger)] \\), where \\( [x] \\) is the greatest integer in \\( x \\). This \\( zeroelement \\) is at most \\( 1983 irrational \\) since \\( irrational - noninteger \\geqslant 1 \\) and \\( negativeval < 1982 \\) and thus\n\\[\ncompleteness \\geqslant \\frac{1}{irrational zeroelement} \\geqslant \\frac{1}{1983 irrational^{2}} > \\frac{1}{1983^{3}} .\n\\]\n\nHence we have the desired inequality in all cases."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "mbplxqre",
        "d": "vgdhcani",
        "r": "sxnfolew"
      },
      "question": "Problem A-5\nLet \\( qzxwvtnp, hjgrksla, mbplxqre \\), and \\( vgdhcani \\) be positive integers and\n\\[\nsxnfolew=1-\\frac{qzxwvtnp}{hjgrksla}-\\frac{mbplxqre}{vgdhcani}\n\\]\n\nGiven that \\( qzxwvtnp+mbplxqre \\leqslant 1982 \\) and \\( sxnfolew>0 \\), prove that\n\\[\nsxnfolew>\\frac{1}{1983^{3}} .\n\\]",
      "solution": "A-5.\nWe are given that\n\\[\nsxnfolew=1-\\frac{qzxwvtnp}{hjgrksla}-\\frac{mbplxqre}{vgdhcani}=\\frac{hjgrksla vgdhcani-qzxwvtnp vgdhcani-hjgrksla mbplxqre}{hjgrksla vgdhcani}>0 .\n\\]\n\nThus \\( hjgrksla vgdhcani-qzxwvtnp vgdhcani-hjgrksla mbplxqre \\) is a positive integer and so \\( sxnfolew \\geqslant 1 / hjgrksla vgdhcani \\). We may assume without loss of generality that \\( hjgrksla \\leqslant vgdhcani \\). If \\( hjgrksla \\leqslant vgdhcani \\leqslant 1983, sxnfolew \\geqslant 1983^{-2}>1983^{-3} \\). Since \\( qzxwvtnp+mbplxqre \\leqslant 1982 \\), if \\( 1983 \\leqslant hjgrksla \\leqslant vgdhcani \\), one has\n\\[\nsxnfolew \\geqslant 1-\\frac{qzxwvtnp}{1983}-\\frac{mbplxqre}{1983} \\geqslant 1-\\frac{1982}{1983}=\\frac{1}{1983}>\\frac{1}{1983^{3}} .\n\\]\n\nThe remaining case is that with \\( hjgrksla<1983<vgdhcani \\). Then the \\( vgdhcani \\) that minimizes \\( sxnfolew \\) for fixed \\( qzxwvtnp, hjgrksla, mbplxqre \\) is \\( 1+[hjgrksla mbplxqre /(hjgrksla-qzxwvtnp)] \\), where \\( [x] \\) is the greatest integer in \\( x \\). This \\( vgdhcani \\) is at most \\( 1983 hjgrksla \\) since \\( hjgrksla-qzxwvtnp \\geqslant 1 \\) and \\( mbplxqre<1982 \\) and thus\n\\[\nsxnfolew \\geqslant \\frac{1}{hjgrksla vgdhcani} \\geqslant \\frac{1}{1983 hjgrksla^{2}}>\\frac{1}{1983^{3}} .\n\\]\n\nHence we have the desired inequality in all cases."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\na_{1},a_{2},a_{3},a_{4},\\;b_{1},b_{2},b_{3},b_{4}\\in\\mathbb Z_{>0}\n\\]  \nsatisfy  \n\\[\n\\gcd(a_{i},b_{i})=1\\quad(i=1,2,3,4),\\qquad \n2\\le b_{1}\\le b_{2}\\le b_{3}\\le b_{4}\\le 2021 ,\n\\]  \ntogether with the divisibility chain  \n\\[\nb_{2}\\mid b_{3}\\mid b_{4}.\n\\]  \nPut  \n\\[\nS:=a_{1}+a_{2}+a_{3}+a_{4}\\le 2020,\\qquad  \nr:=1-\\frac{a_{1}}{b_{1}}-\\frac{a_{2}}{b_{2}}\n        -\\frac{a_{3}}{b_{3}}-\\frac{a_{4}}{b_{4}}>0 .\n\\]\n\n(a)\\;  Prove the universal estimate  \n\\[\n\\boxed{\\, r\\ge\\dfrac{1}{2019\\times 2021}\\,}\n\\]  \nand prove that the denominator $2019\\times 2021=4\\,080\\,399$ cannot be\nreplaced by any \\emph{smaller} positive integer that depends only on the\nupper bound $2021$ for the $b_{i}$.\n\n(b)\\;  Describe \\emph{all} $8$–tuples  \n\\[\n(a_{1},a_{2},a_{3},a_{4};\\,b_{1},b_{2},b_{3},b_{4})\n\\]  \nfor which  \n\\[\n\\boxed{\\, r=\\dfrac{1}{b_{1} b_{4}}\\,}\n\\]  \nholds.",
      "solution": "Throughout write  \n\\[\nL:=\\operatorname{lcm}(b_{1},b_{2},b_{3},b_{4}),\\qquad \nr=\\frac{k}{L}\\quad(k\\in\\mathbb Z_{>0}).\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\\textbf{1.\\;Part (a) – the best possible lower bound.}\n\nBecause  \n\\[\nk=L-\\sum_{i=1}^{4}a_{i}\\,\\frac{L}{b_{i}}\\ge 1 ,\n\\]\none has  \n\\[\nr=\\frac{k}{L}\\ge\\frac{1}{L}.                                     \\tag{1.1}\n\\]\nThe divisibility chain implies  \n\\[\nL=\\operatorname{lcm}(b_{1},b_{2},b_{3},b_{4})\n  =\\operatorname{lcm}(b_{1},b_{4})\\le b_{1}b_{4},                 \\tag{1.2}\n\\]\nhence  \n\\[\nr\\ge\\frac{1}{b_{1}b_{4}}.                                         \\tag{1.3}\n\\]\n\nIf $b_{1}\\le 2019$ then $r\\ge 1/(b_{1}b_{4})\\ge 1/(2019\\cdot 2021)$\nbecause $b_{4}\\le 2021$.  It remains to treat $b_{1}=2020$ and\n$b_{1}=2021$.\n\n\\emph{Case $b_{1}=2021$.}\\;  Then $b_{i}\\ge 2021$, so  \n\\[\nr=1-\\sum_{i=1}^{4}\\frac{a_{i}}{b_{i}}\n     \\ge 1-\\frac{S}{2021}\n     \\ge 1-\\frac{2020}{2021}= \\frac{1}{2021}\n     >\\frac{1}{2019\\cdot 2021}.\n\\]\n\n\\emph{Case $b_{1}=2020$.}  \nTwo sub–cases occur.\n\n(i)\\; $b_{2}=b_{3}=b_{4}=2020$.  Then  \n$r=1-S/2020\\ge 1/2020>1/(2019\\cdot 2021)$.\n\n(ii)\\;  At least one of $b_{2},b_{3},b_{4}$ equals $2021$.  The\ndivisibility relations force $b_{2}=b_{3}=b_{4}=2021$, and writing\n$T:=a_{2}+a_{3}+a_{4}$ one gets $T\\ge 3$ and $a_{1}=2020-T$.  Hence  \n\\[\nr\n=1-\\frac{a_{1}}{2020}-\\frac{T}{2021}\n=1-\\frac{2020-T}{2020}-\\frac{T}{2021}\n=\\frac{T}{2020}-\\frac{T}{2021}\n     =\\frac{T}{2020\\cdot 2021}\n     \\ge\\frac{3}{2020\\cdot 2021}\n     >\\frac{1}{2019\\cdot 2021 }.\n\\]\n\nCombining all cases proves  \n\\[\n\\boxed{\\,r\\ge\\frac{1}{2019\\cdot 2021}\\,}.\n\\]\n\n\\emph{Optimality.}\\;  The tuple  \n\\[\n(1009,1,1,1009;\\,2019,2021,2021,2021)\n\\]\nsatisfies all hypotheses and gives $r=1/(2019\\cdot 2021)$, so the\nconstant $2019\\cdot 2021$ is the smallest possible. \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\\textbf{2.\\;Preparations for part (b).}\n\nAssume from now on that  \n\\[\nr=\\frac{1}{b_{1} b_{4}}.                                          \\tag{2.1}\n\\]\nThen $k=1$ and hence  \n\\[\nL=b_{1}b_{4},\\qquad \\gcd(b_{1},b_{4})=1.                         \\tag{2.2}\n\\]\n\nBecause of $b_{2}\\mid b_{3}\\mid b_{4}$ define  \n\\[\nm:=\\frac{b_{4}}{b_{2}},\\qquad n:=\\frac{b_{4}}{b_{3}},\n\\qquad m,n\\in\\mathbb Z_{>0},\\; n\\mid m.                          \\tag{2.3}\n\\]\n\nEquality $k=1$ turns into  \n\\[\n1=b_{1}b_{4}-a_{1}b_{4}-b_{1}\\bigl(m a_{2}+n a_{3}+a_{4}\\bigr).   \\tag{2.4}\n\\]\n\nBecause $\\gcd(b_{1},b_{4})=1$ there exists a unique  \n\\[\nc\\in\\{1,2,\\dots ,b_{1}-1\\}\\quad\\text{with}\\quad \nc\\,b_{4}\\equiv 1\\pmod{b_{1}}.                                    \\tag{2.5}\n\\]\nReducing \\eqref{2.4} modulo $b_{1}$ gives $a_{1}=b_{1}-c$.  Substituting\nthis into \\eqref{2.4} yields the linear Diophantine equation  \n\\[\nm a_{2}+n a_{3}+a_{4}=T,\\qquad \nT:=\\frac{c\\,b_{4}-1}{b_{1}}.                                     \\tag{2.6}\n\\]\n\n\\emph{Basic properties of $T$.}\\;  Because $1\\le c\\le b_{1}-1$ one has  \n\\[\n0<T<b_{4},\\qquad \\gcd(T,b_{4})=1.                                \\tag{2.7}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\\textbf{3.\\;A useful inequality.}\n\nSuppose $m+n\\ge T$.  For any positive integers $a_{2},a_{3},a_{4}$ one\nwould then have $m a_{2}+n a_{3}+a_{4}\\ge m+n+1\\ge T+1$, contradicting\n\\eqref{2.6}.  Hence  \n\\[\n\\boxed{\\,m+n<T\\,}.                                               \\tag{2.8}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\\textbf{4.\\;Description of \\boldmath$a_{2},a_{3},a_{4}$.}\n\nEquation \\eqref{2.6} is linear.  Choose any pair  \n\\[\na_{2},a_{3}\\in\\mathbb Z_{>0}\n\\quad\\text{such that}\\quad\n\\begin{cases}\nm a_{2}+n a_{3}<T,\\\\[2mm]\n\\gcd(a_{2},b_{2})=\\gcd(a_{3},b_{3})=1,\n\\end{cases}                                                     \\tag{4.1}\n\\]\nand put  \n\\[\na_{4}:=T-m a_{2}-n a_{3}>0 .                                     \\tag{4.2}\n\\]\nBecause of \\eqref{4.1}–\\eqref{4.2} one gets automatically  \n\\[\nm a_{2}+n a_{3}+a_{4}=T,\\qquad a_{4}>0.                          \\tag{4.3}\n\\]\n\n\\emph{Coprimality with $b_{4}$.}\\;  Nothing in\n\\eqref{4.1}–\\eqref{4.2} forces $\\gcd(a_{4},b_{4})=1$.  This condition\nmust therefore be imposed explicitly:\n\\[\n\\gcd\\bigl(a_{4},\\,b_{4}\\bigr)=\n\\gcd\\bigl(T-m a_{2}-n a_{3},\\,b_{4}\\bigr)=1.                     \\tag{4.4}\n\\]\n\nConversely, any positive triple $(a_{2},a_{3},a_{4})$ that fulfills\n\\eqref{4.1}–\\eqref{4.4} satisfies the central equation\n\\eqref{2.6} and all individual coprimality requirements.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\\textbf{5.\\;Complete characterisation of the extremal tuples.}\n\nCollecting the preceding deductions one obtains a \\emph{necessary and\nsufficient} system.\n\n\\[\n\\boxed{\n\\begin{aligned}\nr=\\frac{1}{b_{1}b_{4}}\n&\\;\\Longleftrightarrow\\;\n\\text{there exist integers }c,m,n,T,a_{2},a_{3} \\text{ such that}\\\\[1mm]\n\\text{(i)}\\;& 2\\le b_{1}\\le b_{2}\\le b_{3}\\le b_{4}\\le 2021,\\\\\n           & b_{2}\\mid b_{3}\\mid b_{4},\\; \\gcd(b_{1},b_{4})=1;\\\\[1mm]\n\\text{(ii)}\\;& m:=b_{4}/b_{2},\\;n:=b_{4}/b_{3}\\quad(n\\mid m),\\;\n              m+n<T<b_{4};\\\\[1mm]\n\\text{(iii)}\\;& c\\in\\{1,\\dots ,b_{1}-1\\},\\quad c\\,b_{4}\\equiv 1\\pmod{b_{1}},\\\\\n              & T:=(c\\,b_{4}-1)/b_{1},\\; \\gcd(T,b_{4})=1;\\\\[1mm]\n\\text{(iv)}\\;& a_{2},a_{3}\\in\\mathbb Z_{>0},\\; \n              \\gcd(a_{2},b_{2})=\\gcd(a_{3},b_{3})=1,\\\\\n              & m a_{2}+n a_{3}<T;\\\\[1mm]\n\\text{(v)}\\;& a_{4}:=T-m a_{2}-n a_{3}>0,\\;\n              \\gcd(a_{4},b_{4})=1;\\\\[1mm]\n\\text{(vi)}\\;& a_{1}:=b_{1}-c>0,\\quad \n              a_{1}+a_{2}+a_{3}+a_{4}\\le 2020.\n\\end{aligned}\n}                                                             \\tag{5.1}\n\\]\n\nDirection “$\\Longrightarrow$’’ has already been proved in Sections\n2–4.  For the converse pick data obeying (i)–(vi); then the identities\n\\eqref{2.2}, \\eqref{2.6} and \\eqref{4.3} ensure $k=1$, whence\n$r=1/(b_{1}b_{4})$.  Thus the list \\eqref{5.1} indeed describes\n\\emph{every} extremal $8$–tuple. \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n\\textbf{6.\\;Example.}\n\nChoose  \n\\[\nb_{1}=2019,\\;b_{2}=b_{3}=b_{4}=2021,\\;\nc=1010,\\;m=n=1,\\;\na_{2}=2,\\;a_{3}=1 .\n\\]\nThen $T=(c b_{4}-1)/b_{1}=1011$ and $a_{4}=1011-2-1=1008$, giving the\ntuple  \n\\[\n(1009,2,1,1008;\\; 2019,2021,2021,2021),\n\\]\nwhich satisfies all conditions and attains\n$r=1/(2019\\cdot 2021)$; hence it appears in the classification.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.664716",
        "was_fixed": false,
        "difficulty_analysis": "1. More variables & higher‐dimensional optimisation  \n   • Moves from two to four rational summands.  \n   • Denominators are partially ordered and linked by divisibility, greatly enlarging the feasible lattice.\n\n2. Additional arithmetic structure  \n   • Reducedness (\\(\\gcd(a_{i},b_{i})=1\\)).  \n   • A chain of divisors \\(b_{2}\\mid b_{3}\\mid b_{4}\\).\n\n3. Deeper argument layers  \n   • Requires simultaneous control of an lcm and of the largest denominator.  \n   • Compulsory case distinction intertwining additive (sum of numerators) and multiplicative (divisibility, lcm) constraints.  \n   • Necessitates bounding \\(r\\) twice (via \\(1/D\\) and via writing all terms over \\(b_{4}\\)) and then dovetailing the bounds.\n\n4. Denominator explosion  \n   • Target bound involves the tenth power \\(2021^{10}\\) instead of the third or fourth power in earlier versions; proof must prevent denominator products of that magnitude.\n\n5. Extremal analysis  \n   • After proving the inequality one must examine the chain of equalities to see whether the bound is attainable—an extra step absent in the original problem.\n\nThese layers oblige the solver to juggle integrality, lcm estimates, divisibility chains, and tight additive–multiplicative interplay, well beyond the reach of the straightforward bounding arguments that solved the original and the first kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\na_{1},a_{2},a_{3},a_{4},\\;b_{1},b_{2},b_{3},b_{4}\\in\\mathbb Z_{>0}\n\\]  \nsatisfy  \n\\[\n\\gcd(a_{i},b_{i})=1\\quad(i=1,2,3,4),\\qquad \n2\\le b_{1}\\le b_{2}\\le b_{3}\\le b_{4}\\le 2021 ,\n\\]  \ntogether with the divisibility chain  \n\\[\nb_{2}\\mid b_{3}\\mid b_{4}.\n\\]  \nPut  \n\\[\nS:=a_{1}+a_{2}+a_{3}+a_{4}\\le 2020,\\qquad  \nr:=1-\\frac{a_{1}}{b_{1}}-\\frac{a_{2}}{b_{2}}\n        -\\frac{a_{3}}{b_{3}}-\\frac{a_{4}}{b_{4}}>0 .\n\\]\n\n(a)  Prove the universal estimate  \n\\[\n\\boxed{\\, r\\ge\\dfrac{1}{2019\\times 2021}\\,}\n\\]  \nand prove that the denominator $2019\\times 2021=4\\,080\\,399$ cannot be\nreplaced by any \\emph{smaller} positive integer that depends only on the\nupper bound $2021$ for the $b_{i}$.\n\n(b)  Determine every $8$–tuple  \n\\[\n(a_{1},a_{2},a_{3},a_{4};\\,b_{1},b_{2},b_{3},b_{4})\n\\]  \nfor which  \n\\[\n\\boxed{\\, r=\\dfrac{1}{b_{1}b_{4}}\\,}\n\\]  \nholds.",
      "solution": "Throughout write  \n\\[\nL:=\\operatorname{lcm}(b_{1},b_{2},b_{3},b_{4}),\\qquad \nr=\\frac{k}{L}\\quad(k\\in\\mathbb Z_{>0}).\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{1.\\;Part (a) – the sharp lower bound.}\n\nBecause $k=L-\\sum_{i=1}^{4}a_{i}\\,\\dfrac{L}{b_{i}}\\ge 1$,  \n\\[\nr=\\frac{k}{L}\\ge\\frac{1}{L}.                                     \\tag{1.1}\n\\]\nThe divisibility chain implies  \n\\[\nL=\\operatorname{lcm}(b_{1},b_{2},b_{3},b_{4})\n  =\\operatorname{lcm}(b_{1},b_{4})\\le b_{1}b_{4},                 \\tag{1.2}\n\\]\nso  \n\\[\nr\\ge\\frac{1}{b_{1}b_{4}}.                                         \\tag{1.3}\n\\]\n\nSince $b_{4}\\le 2021$, the bound $r\\ge\\dfrac{1}{b_{1}b_{4}}\\ge\n\\dfrac{1}{2019\\cdot 2021}$ is already in place when $b_{1}\\le 2019$.\nWe therefore analyse $b_{1}=2020$ and $b_{1}=2021$.\n\n\\emph{Case $b_{1}=2021$.}  \nThen $b_{i}\\ge 2021$ $(i=2,3,4)$, hence  \n\\[\nr=1-\\sum_{i=1}^{4}\\frac{a_{i}}{b_{i}}\n     \\ge 1-\\frac{S}{2021}\n     \\ge 1-\\frac{2020}{2021}= \\frac{1}{2021}\n     >\\frac{1}{2019\\cdot 2021}.\n\\]\n\n\\emph{Case $b_{1}=2020$.}  \n\n\\qquad\\emph{(i) If $b_{2}=b_{3}=b_{4}=2020$}, then $r=1-\\dfrac{S}{2020}\n\\ge\\dfrac{1}{2020}>\\dfrac{1}{2019\\cdot 2021}$.\n\n\\qquad\\emph{(ii) Otherwise at least one of $b_{2},b_{3},b_{4}$ equals\n$2021$.}  Divisibility enforces $b_{2}=b_{3}=b_{4}=2021$ (because\n$2020\\nmid 2021$).  Put $T:=a_{2}+a_{3}+a_{4}$.  Since $T\\ge 3$ and\n$a_{1}=2020-T$,  \n\\[\nr\n=1-\\frac{a_{1}}{2020}-\\frac{T}{2021}\n\\ge 1-\\frac{2017}{2020}-\\frac{3}{2021}\n     =\\frac{3}{2020\\cdot 2021}\n     >\\frac{1}{2019\\cdot 2021}.\n\\]\n\nCombining the three cases proves  \n\\[\nr\\ge\\dfrac{1}{2019\\cdot 2021}.\n\\]\n\n\\emph{Optimality.}  \nTake  \n\\[\n(1009,1,1,1009;\\,2019,2021,2021,2021),\n\\qquad r=\\frac{1}{2019\\cdot 2021}.\n\\]\nAny replacement of the denominator by a \\emph{smaller} integer would\ngive a strictly larger lower bound, contradicted by this example.\nTherefore the constant is best possible. \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{2.\\;Preparations for part (b).}\n\nAssume from now on that  \n\\[\nr=\\frac{1}{b_{1}b_{4}}.                                           \\tag{2.1}\n\\]\nThen $k=1$ and consequently  \n\\[\nL=b_{1}b_{4},\\qquad \\gcd(b_{1},b_{4})=1.                         \\tag{2.2}\n\\]\n\nBecause of the divisibility chain let  \n\\[\nm:=\\frac{b_{4}}{b_{2}},\\qquad n:=\\frac{b_{4}}{b_{3}}\n\\qquad(m,n\\in\\mathbb Z_{>0},\\; n\\mid m).                        \\tag{2.3}\n\\]\n\nEquality $k=1$ rewrites as  \n\\[\n1=b_{1}b_{4}-a_{1}b_{4}-b_{1}\\bigl(m a_{2}+n a_{3}+a_{4}\\bigr).   \\tag{2.4}\n\\]\n\nBecause $\\gcd(b_{1},b_{4})=1$, there exists a unique  \n\\[\nc\\in\\{1,2,\\dots ,b_{1}-1\\}\\quad\\text{with}\\quad \nc\\,b_{4}\\equiv 1\\pmod{b_{1}}.                                    \\tag{2.5}\n\\]\nReducing \\eqref{2.4} modulo $b_{1}$ gives $a_{1}=b_{1}-c$.  Insert this\ninto \\eqref{2.4} to obtain the Diophantine equation  \n\\[\nm a_{2}+n a_{3}+a_{4}=T,\\qquad \nT:=\\frac{c\\,b_{4}-1}{b_{1}}.                                     \\tag{2.6}\n\\]\n\n\\emph{Properties of $T$.}  \nBecause $1\\le c\\le b_{1}-1$ one has $0<T<b_{4}$.  A prime divisor of\n$T$ would divide the left side of \\eqref{2.4}, hence $\\gcd(T,b_{4})=1$:\n\\[\n0<T<b_{4},\\qquad \\gcd(T,b_{4})=1.                                \\tag{2.7}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{3.\\;Why $m+n<T$.}\n\nSuppose $m+n\\ge T$.  Then for any positive integers\n$a_{2},a_{3},a_{4}$ one has $m a_{2}+n a_{3}+a_{4}\\ge m+n+1\\ge T+1$,\ncontradicting \\eqref{2.6}.  Therefore  \n\\[\n\\boxed{\\,m+n<T\\,}.                                               \\tag{2.8}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{4.\\;Solving the central equation.}\n\nEquation \\eqref{2.6} is linear; all its positive solutions can be\ndescribed explicitly.  Choose any pair  \n\\[\na_{2},a_{3}\\in\\mathbb Z_{>0}\\quad\\text{with}\\quad\nm a_{2}+n a_{3}<T,                                              \\tag{4.1}\n\\]\nsubject to the coprimality conditions  \n\\[\n\\gcd(a_{2},b_{2})=\\gcd(a_{3},b_{3})=1.                           \\tag{4.2}\n\\]\nThen define  \n\\[\na_{4}:=T-m a_{2}-n a_{3}>0.                                     \\tag{4.3}\n\\]\n\nPositivity of $a_{4}$ is guaranteed by \\eqref{4.1}, while\n\\eqref{2.7} and the fact that $m,n\\mid b_{4}$ give  \n\\[\n\\gcd(a_{4},b_{4})=\\gcd(T-m a_{2}-n a_{3},b_{4})=\\gcd(T,b_{4})=1.\n\\]  \n\nConversely, any positive solution of \\eqref{2.6} clearly satisfies\n\\eqref{4.1}.  Hence \\eqref{4.1}–\\eqref{4.3} describe all admissible\ntriples $(a_{2},a_{3},a_{4})$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{5.\\;The bound $S\\le 2020$ is always attainable.}\n\nBecause $a_{1}=b_{1}-c$,  \n\\[\nS=b_{1}-c+a_{2}+a_{3}+a_{4}\n  =b_{1}-c+T-(m-1)a_{2}-(n-1)a_{3}.                             \\tag{5.1}\n\\]\n\nObserve that the coefficients of $a_{2}$ and $a_{3}$ in \\eqref{5.1}\nare \\emph{non–negative}: $m,n\\ge 1$ implies $m-1,n-1\\ge 0$.\nConsequently\n\n\\[\n\\text{Increasing }a_{2}\\text{ or }a_{3}\\text{ \\emph{decreases} the sum }S.   \\tag{5.2}\n\\]\n\nIf an admissible choice $(a_{2},a_{3})$ leads to $S>2020$, simply raise\neither $a_{2}$ or $a_{3}$ by $1$.  The new pair still satisfies\n\\eqref{4.1}, because  \n\\[\nm(a_{2}+1)+n a_{3}=m a_{2}+n a_{3}+m<T+m\\le T+(T-1)<2T,\n\\]\nand $m+n<T$ by \\eqref{2.8}; hence strict inequality in \\eqref{4.1}\npersists.  Repeating the procedure lowers $S$ step by step, and after\nat most $S-2020$ iterations one necessarily reaches $S\\le 2020$.\nDuring this process one can always pick the variable that remains\ncoprime to its corresponding denominator, so \\eqref{4.2} is preserved.\n\nThus\n\n\\[\n\\boxed{\\text{There is \\emph{always} an admissible choice with }S\\le 2020.}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{6.\\;Complete characterisation of the extremal tuples.}\n\nWe assemble the preceding pieces.\n\n\\[\n\\boxed{\n\\begin{aligned}\n&r=\\dfrac{1}{b_{1}b_{4}}\\ \\Longleftrightarrow\\ \n \\text{there exist integers }m,n,c,T,a_{2},a_{3}\\text{ such that}\\\\\n&\\text{(i) } 2\\le b_{1}\\le b_{2}\\le b_{3}\\le b_{4}\\le 2021,\\;\n              b_{2}\\mid b_{3}\\mid b_{4},\\;\n              \\gcd(b_{1},b_{4})=1;\\\\\n&\\text{(ii) } m:=b_{4}/b_{2},\\;n:=b_{4}/b_{3}\\ (n\\mid m),\\quad m+n<T<b_{4};\\\\\n&\\text{(iii) } c\\in\\{1,\\dots ,b_{1}-1\\},\\ c\\,b_{4}\\equiv 1\\pmod{b_{1}},\\\\\n&\\phantom{(iii)}\\ T:=(c\\,b_{4}-1)/b_{1},\\ \\gcd(T,b_{4})=1;\\\\\n&\\text{(iv) } a_{2},a_{3}\\in\\mathbb Z_{>0}\\ \\text{satisfy } \n              m a_{2}+n a_{3}<T,\\ \n              \\gcd(a_{2},b_{2})=\\gcd(a_{3},b_{3})=1;\\\\\n&\\text{(v) } a_{4}:=T-m a_{2}-n a_{3},\\qquad \n              a_{4}>0,\\ \\gcd(a_{4},b_{4})=1;\\\\\n&\\text{(vi) } a_{1}:=b_{1}-c,\\qquad \n              a_{1}+a_{2}+a_{3}+a_{4}\\le 2020.\n\\end{aligned}\n}                                                              \\tag{6.1}\n\\]\n\nEvery choice of data fulfilling (i)–(vi) produces an $8$–tuple that\nsatisfies all original conditions and attains the equality\n$r=1/(b_{1}b_{4})$.  Conversely, each extremal $8$–tuple induces\nparameters $(m,n,c,T,a_{2},a_{3})$ that obey (i)–(vi), so the list\n\\eqref{6.1} is exhaustive.  Different admissible pairs\n$(a_{2},a_{3})$ (or different $c$) may yield the same $8$–tuple;\nuniqueness of the parameterisation is not claimed. \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{7.\\;Example revisited.}\n\nThe tuple  \n\\[\n(1009,2,1,1008;\\,2019,2021,2021,2021)\n\\]\nis obtained by choosing $m=n=1$, $c=1010$, $T=1011$, $a_{2}=2$, \n$a_{3}=1$; it obeys (i)–(vi) and is therefore covered by the corrected\nclassification.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.521362",
        "was_fixed": false,
        "difficulty_analysis": "1. More variables & higher‐dimensional optimisation  \n   • Moves from two to four rational summands.  \n   • Denominators are partially ordered and linked by divisibility, greatly enlarging the feasible lattice.\n\n2. Additional arithmetic structure  \n   • Reducedness (\\(\\gcd(a_{i},b_{i})=1\\)).  \n   • A chain of divisors \\(b_{2}\\mid b_{3}\\mid b_{4}\\).\n\n3. Deeper argument layers  \n   • Requires simultaneous control of an lcm and of the largest denominator.  \n   • Compulsory case distinction intertwining additive (sum of numerators) and multiplicative (divisibility, lcm) constraints.  \n   • Necessitates bounding \\(r\\) twice (via \\(1/D\\) and via writing all terms over \\(b_{4}\\)) and then dovetailing the bounds.\n\n4. Denominator explosion  \n   • Target bound involves the tenth power \\(2021^{10}\\) instead of the third or fourth power in earlier versions; proof must prevent denominator products of that magnitude.\n\n5. Extremal analysis  \n   • After proving the inequality one must examine the chain of equalities to see whether the bound is attainable—an extra step absent in the original problem.\n\nThese layers oblige the solver to juggle integrality, lcm estimates, divisibility chains, and tight additive–multiplicative interplay, well beyond the reach of the straightforward bounding arguments that solved the original and the first kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}