{
  "index": "1958-2-A-4",
  "type": "ALG",
  "tag": [
    "ALG",
    "COMB"
  ],
  "difficulty": "",
  "question": "4. In assigning dormitory rooms, a college gives preference to pairs of students in this order:\n\\[\nA A, A B, A C, B B, B C, A D, C C, B D, C D, D D\n\\]\nin which \\( A A \\) means two seniors, \\( A B \\) means a senior and a junior, etc. Determine numerical values to assign to \\( A, B, C, D \\) so that the set of numbers \\( A \\) \\( +A, A+B, A+C, B+B \\), etc., corresponding to the order above will be in descending magnitude. Find the general solution and the solution in least positive integers.",
  "solution": "Solution. The inequalities to be solved are\n\\[\n\\begin{array}{c}\n2 A>A+B>A+C>2 B>B+C>A+D> \\\\\n2 C>B+D>C+D>2 D .\n\\end{array}\n\\]\n\nEvidently we must have \\( A>B>C>D \\), so put\n\\[\nA=\\alpha+B . B=\\beta+C . C=\\gamma+D .\n\\]\nwhere \\( \\alpha, \\beta \\) and \\( \\gamma \\) are positive. The first, second, fourth, eighth, and ninth inequalities in (1) are now satisfied, while the third. fifth, sixth, and seventh inequalities become, respectively\n\\[\n\\alpha>\\beta . \\quad \\gamma>\\alpha . \\quad \\alpha+\\beta>\\gamma . \\quad \\text { and } \\quad \\gamma>\\beta .\n\\]\n\nThe inequality \\( \\gamma>\\beta \\) is a consequence of \\( \\gamma>\\alpha \\) and \\( \\alpha>\\beta \\), so it can be dropped from this system. Now put \\( \\alpha=\\beta+\\delta, \\gamma=\\alpha+\\epsilon \\) where \\( \\delta \\) and \\( \\epsilon \\) are positive. Then \\( \\alpha+\\beta>\\gamma \\) becomes \\( \\beta>\\epsilon \\), so we can put \\( \\beta=\\epsilon+\\zeta \\) with !? positive. Then\n\\[\n\\begin{array}{c}\n\\alpha=\\delta+\\epsilon+\\zeta \\\\\n\\beta=\\epsilon+\\zeta \\\\\n\\gamma=\\delta+2 \\epsilon+\\zeta\n\\end{array}\n\\]\nand finally.\n\\[\n\\begin{array}{l}\nA=2 \\delta+4 \\epsilon+3 \\zeta+D \\\\\nB=\\delta+3 \\epsilon+2 \\zeta+D \\\\\nC=\\delta+2 \\epsilon+\\zeta+D .\n\\end{array}\n\\]\n\nHere \\( D \\) can be chosen arbitrarily, while \\( \\delta, \\epsilon \\), and \\( \\zeta \\) must be positive. It follows either from the derivation or from a direct check that if \\( A, B, C, D \\) satisfy (2). where \\( \\delta, \\epsilon \\) and \\( \\zeta \\) are positive. then they satisfy the inequalities (1).\n\nTo find integral solutions. we note that if we start with integral \\( A . B . C \\). \\( D \\) satisfying (1). then all the numbers \\( \\alpha, \\beta, \\gamma, \\delta, \\epsilon, \\zeta \\) will be positive integers. Hence the least solution in positive integers is obtained by choosing \\( \\delta=\\epsilon=\\zeta=D=1 \\). Hence \\( A=10 . B=7, C=5, D=1 \\) is the least solution in positive integers.",
  "vars": [
    "A",
    "B",
    "C",
    "D"
  ],
  "params": [
    "\\\\alpha",
    "\\\\beta",
    "\\\\gamma",
    "\\\\delta",
    "\\\\epsilon",
    "\\\\zeta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "seniorv",
        "B": "juniorv",
        "C": "sophomv",
        "D": "freshmv",
        "\\alpha": "greekal",
        "\\beta": "greekbe",
        "\\gamma": "greekga",
        "\\delta": "greekde",
        "\\epsilon": "greekep",
        "\\zeta": "greekze"
      },
      "question": "4. In assigning dormitory rooms, a college gives preference to pairs of students in this order:\n\\[\nseniorv seniorv, seniorv juniorv, seniorv sophomv, juniorv juniorv, juniorv sophomv, seniorv freshmv, sophomv sophomv, juniorv freshmv, sophomv freshmv, freshmv freshmv\n\\]\nin which \\( seniorv seniorv \\) means two seniors, \\( seniorv juniorv \\) means a senior and a junior, etc. Determine numerical values to assign to \\( seniorv, juniorv, sophomv, freshmv \\) so that the set of numbers \\( seniorv+seniorv, seniorv+juniorv, seniorv+sophomv, juniorv+juniorv \\), etc., corresponding to the order above will be in descending magnitude. Find the general solution and the solution in least positive integers.",
      "solution": "Solution. The inequalities to be solved are\n\\[\n\\begin{array}{c}\n2\\,seniorv > seniorv + juniorv > seniorv + sophomv > 2\\,juniorv > juniorv + sophomv > seniorv + freshmv > \\\\\n2\\,sophomv > juniorv + freshmv > sophomv + freshmv > 2\\,freshmv .\n\\end{array}\n\\]\n\nEvidently we must have \\( seniorv > juniorv > sophomv > freshmv \\), so put\n\\[\nseniorv = greekal + juniorv, \\quad\njuniorv = greekbe + sophomv, \\quad\nsophomv = greekga + freshmv .\n\\]\nwhere \\( greekal, greekbe \\) and \\( greekga \\) are positive. The first, second, fourth, eighth, and ninth inequalities in (1) are now satisfied, while the third, fifth, sixth, and seventh inequalities become, respectively\n\\[\ngreekal > greekbe, \\quad greekga > greekal, \\quad greekal + greekbe > greekga, \\quad \\text{and} \\quad greekga > greekbe .\n\\]\n\nThe inequality \\( greekga > greekbe \\) is a consequence of \\( greekga > greekal \\) and \\( greekal > greekbe \\), so it can be dropped from this system. Now put \\( greekal = greekbe + greekde, \\quad greekga = greekal + greekep \\) where \\( greekde \\) and \\( greekep \\) are positive. Then \\( greekal + greekbe > greekga \\) becomes \\( greekbe > greekep \\), so we can put \\( greekbe = greekep + greekze \\) with \\( greekze \\) positive. Then\n\\[\n\\begin{array}{c}\ngreekal = greekde + greekep + greekze ,\\\\\ngreekbe = greekep + greekze ,\\\\\ngreekga = greekde + 2\\,greekep + greekze .\n\\end{array}\n\\]\nand finally.\n\\[\n\\begin{array}{l}\nseniorv = 2\\,greekde + 4\\,greekep + 3\\,greekze + freshmv ,\\\\\njuniorv = greekde + 3\\,greekep + 2\\,greekze + freshmv ,\\\\\nsophomv = greekde + 2\\,greekep + greekze + freshmv .\n\\end{array}\n\\]\n\nHere \\( freshmv \\) can be chosen arbitrarily, while \\( greekde, greekep \\), and \\( greekze \\) must be positive. It follows either from the derivation or from a direct check that if \\( seniorv, juniorv, sophomv, freshmv \\) satisfy (2), where \\( greekde, greekep \\) and \\( greekze \\) are positive, then they satisfy the inequalities (1).\n\nTo find integral solutions, we note that if we start with integral \\( seniorv, juniorv, sophomv, freshmv \\) satisfying (1), then all the numbers \\( greekal, greekbe, greekga, greekde, greekep, greekze \\) will be positive integers. Hence the least solution in positive integers is obtained by choosing \\( greekde = greekep = greekze = freshmv = 1 \\). Hence \\( seniorv = 10, juniorv = 7, sophomv = 5, freshmv = 1 \\) is the least solution in positive integers."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "copperhead",
        "B": "laundromat",
        "C": "nightshade",
        "D": "flashpoint",
        "\\\\alpha": "playground",
        "\\\\beta": "seashells",
        "\\\\gamma": "sunflower",
        "\\\\delta": "rainstorm",
        "\\\\epsilon": "stargazer",
        "\\\\zeta": "moonlight"
      },
      "question": "4. In assigning dormitory rooms, a college gives preference to pairs of students in this order:\n\\[\ncopperhead\\ copperhead, \\, copperhead\\ laundromat, \\, copperhead\\ nightshade, \\, laundromat\\ laundromat, \\, laundromat\\ nightshade, \\, copperhead\\ flashpoint, \\, nightshade\\ nightshade, \\, laundromat\\ flashpoint, \\, nightshade\\ flashpoint, \\, flashpoint\\ flashpoint\n\\]\nin which \\( copperhead\\ copperhead \\) means two seniors, \\( copperhead\\ laundromat \\) means a senior and a junior, etc. Determine numerical values to assign to \\( copperhead,\\, laundromat,\\, nightshade,\\, flashpoint \\) so that the set of numbers \\( copperhead + copperhead,\\, copperhead+laundromat,\\, copperhead+nightshade,\\, laundromat+laundromat \\), etc., corresponding to the order above will be in descending magnitude. Find the general solution and the solution in least positive integers.",
      "solution": "Solution. The inequalities to be solved are\n\\[\n\\begin{array}{c}\n2\\,copperhead > copperhead+laundromat > copperhead+nightshade > 2\\,laundromat > laundromat+nightshade > copperhead+flashpoint > \\\\\n2\\,nightshade > laundromat+flashpoint > nightshade+flashpoint > 2\\,flashpoint .\n\\end{array}\n\\]\n\nEvidently we must have \\( copperhead>laundromat>nightshade>flashpoint \\), so put\n\\[\ncopperhead=playground+laundromat .\\; laundromat=seashells+nightshade .\\; nightshade=sunflower+flashpoint .\n\\]\nwhere \\( playground, seashells \\) and \\( sunflower \\) are positive. The first, second, fourth, eighth, and ninth inequalities in (1) are now satisfied, while the third, fifth, sixth, and seventh inequalities become, respectively\n\\[\nplayground>seashells . \\quad sunflower>playground . \\quad playground+seashells>sunflower . \\quad \\text { and } \\quad sunflower>seashells .\n\\]\n\nThe inequality \\( sunflower>seashells \\) is a consequence of \\( sunflower>playground \\) and \\( playground>seashells \\), so it can be dropped from this system. Now put \\( playground=seashells+rainstorm, \\; sunflower=playground+stargazer \\) where \\( rainstorm \\) and \\( stargazer \\) are positive. Then \\( playground+seashells>sunflower \\) becomes \\( seashells>stargazer \\), so we can put \\( seashells=stargazer+moonlight \\) with !? positive. Then\n\\[\n\\begin{array}{c}\nplayground=rainstorm+stargazer+moonlight \\\\\nseashells=stargazer+moonlight \\\\\nsunflower=rainstorm+2\\,stargazer+moonlight\n\\end{array}\n\\]\nand finally.\n\\[\n\\begin{array}{l}\ncopperhead=2\\,rainstorm+4\\,stargazer+3\\,moonlight+flashpoint \\\\\nlaundromat=rainstorm+3\\,stargazer+2\\,moonlight+flashpoint \\\\\nnightshade=rainstorm+2\\,stargazer+moonlight+flashpoint .\n\\end{array}\n\\]\n\nHere \\( flashpoint \\) can be chosen arbitrarily, while \\( rainstorm, stargazer \\), and \\( moonlight \\) must be positive. It follows either from the derivation or from a direct check that if \\( copperhead, laundromat, nightshade, flashpoint \\) satisfy (2), where \\( rainstorm, stargazer \\) and \\( moonlight \\) are positive, then they satisfy the inequalities (1).\n\nTo find integral solutions, we note that if we start with integral \\( copperhead, laundromat, nightshade, flashpoint \\) satisfying (1), then all the numbers \\( playground, seashells, sunflower, rainstorm, stargazer, moonlight \\) will be positive integers. Hence the least solution in positive integers is obtained by choosing \\( rainstorm=stargazer=moonlight=flashpoint=1 \\). Hence \\( copperhead=10,\\; laundromat=7,\\; nightshade=5,\\; flashpoint=1 \\) is the least solution in positive integers."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "freshman",
        "B": "seniorly",
        "C": "graduate",
        "D": "upperclass",
        "\\alpha": "omegaend",
        "\\beta": "alphabeg",
        "\\gamma": "deltamid",
        "\\delta": "staticval",
        "\\epsilon": "largesize",
        "\\zeta": "concrete"
      },
      "question": "4. In assigning dormitory rooms, a college gives preference to pairs of students in this order:\n\\[\nfreshman freshman, freshman seniorly, freshman graduate, seniorly seniorly, seniorly graduate, freshman upperclass, graduate graduate, seniorly upperclass, graduate upperclass, upperclass upperclass\n\\]\nin which \\( freshman freshman \\) means two seniors, \\( freshman seniorly \\) means a senior and a junior, etc. Determine numerical values to assign to \\( freshman, seniorly, graduate, upperclass \\) so that the set of numbers \\( freshman \\) \\( +freshman, freshman+seniorly, freshman+graduate, seniorly+seniorly \\), etc., corresponding to the order above will be in descending magnitude. Find the general solution and the solution in least positive integers.",
      "solution": "Solution. The inequalities to be solved are\n\\[\n\\begin{array}{c}\n2 freshman>freshman+seniorly>freshman+graduate>2 seniorly>seniorly+graduate>freshman+upperclass> \\\\\n2 graduate>seniorly+upperclass>graduate+upperclass>2 upperclass .\n\\end{array}\n\\]\n\nEvidently we must have \\( freshman>seniorly>graduate>upperclass \\), so put\n\\[\nfreshman=omegaend+seniorly .\\; seniorly=alphabeg+graduate .\\; graduate=deltamid+upperclass .\n\\]\nwhere \\( omegaend, alphabeg \\) and \\( deltamid \\) are positive. The first, second, fourth, eighth, and ninth inequalities in (1) are now satisfied, while the third. fifth, sixth, and seventh inequalities become, respectively\n\\[\nomegaend>alphabeg . \\quad deltamid>omegaend . \\quad omegaend+alphabeg>deltamid . \\quad \\text { and } \\quad deltamid>alphabeg .\n\\]\n\nThe inequality \\( deltamid>alphabeg \\) is a consequence of \\( deltamid>omegaend \\) and \\( omegaend>alphabeg \\), so it can be dropped from this system. Now put \\( omegaend=alphabeg+staticval, \\; deltamid=omegaend+largesize \\) where \\( staticval \\) and \\( largesize \\) are positive. Then \\( omegaend+alphabeg>deltamid \\) becomes \\( alphabeg>largesize \\), so we can put \\( alphabeg=largesize+concrete \\) with !? positive. Then\n\\[\n\\begin{array}{c}\nomegaend=staticval+largesize+concrete \\\\\nalphabeg=largesize+concrete \\\\\ndeltamid=staticval+2 largesize+concrete\n\\end{array}\n\\]\nand finally.\n\\[\n\\begin{array}{l}\nfreshman=2 staticval+4 largesize+3 concrete+upperclass \\\\\nseniorly=staticval+3 largesize+2 concrete+upperclass \\\\\ngraduate=staticval+2 largesize+concrete+upperclass .\n\\end{array}\n\\]\n\nHere \\( upperclass \\) can be chosen arbitrarily, while \\( staticval, largesize \\), and \\( concrete \\) must be positive. It follows either from the derivation or from a direct check that if \\( freshman, seniorly, graduate, upperclass \\) satisfy (2). where \\( staticval, largesize \\) and \\( concrete \\) are positive. then they satisfy the inequalities (1).\n\nTo find integral solutions. we note that if we start with integral \\( freshman . seniorly . graduate \\). \\( upperclass \\) satisfying (1). then all the numbers \\( omegaend, alphabeg, deltamid, staticval, largesize, concrete \\) will be positive integers. Hence the least solution in positive integers is obtained by choosing \\( staticval=largesize=concrete=upperclass=1 \\). Hence \\( freshman=10 . seniorly=7, graduate=5, upperclass=1 \\) is the least solution in positive integers."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "mfldepor",
        "D": "vknsqiut",
        "\\alpha": "pzocrine",
        "\\beta": "svakudom",
        "\\gamma": "welfsibn",
        "\\delta": "kluvemta",
        "\\epsilon": "draximop",
        "\\zeta": "nurbisyq"
      },
      "question": "4. In assigning dormitory rooms, a college gives preference to pairs of students in this order:\n\\[\nqzxwvtnp qzxwvtnp, qzxwvtnp hjgrksla, qzxwvtnp mfldepor, hjgrksla hjgrksla, hjgrksla mfldepor, qzxwvtnp vknsqiut, mfldepor mfldepor, hjgrksla vknsqiut, mfldepor vknsqiut, vknsqiut vknsqiut\n\\]\nin which \\( qzxwvtnp qzxwvtnp \\) means two seniors, \\( qzxwvtnp hjgrksla \\) means a senior and a junior, etc. Determine numerical values to assign to \\( qzxwvtnp, hjgrksla, mfldepor, vknsqiut \\) so that the set of numbers \\( qzxwvtnp \\) \\( +qzxwvtnp, qzxwvtnp+hjgrksla, qzxwvtnp+mfldepor, hjgrksla+hjgrksla \\), etc., corresponding to the order above will be in descending magnitude. Find the general solution and the solution in least positive integers.",
      "solution": "Solution. The inequalities to be solved are\n\\[\n\\begin{array}{c}\n2 qzxwvtnp>qzxwvtnp+hjgrksla>qzxwvtnp+mfldepor>2 hjgrksla>hjgrksla+mfldepor>qzxwvtnp+vknsqiut> \\\\\n2 mfldepor>hjgrksla+vknsqiut>mfldepor+vknsqiut>2 vknsqiut .\n\\end{array}\n\\]\n\nEvidently we must have \\( qzxwvtnp>hjgrksla>mfldepor>vknsqiut \\), so put\n\\[\nqzxwvtnp=pzocrine+hjgrksla . \\quad hjgrksla=svakudom+mfldepor . \\quad mfldepor=welfsibn+vknsqiut .\n\\]\nwhere \\( pzocrine, svakudom \\) and \\( welfsibn \\) are positive. The first, second, fourth, eighth, and ninth inequalities in (1) are now satisfied, while the third. fifth, sixth, and seventh inequalities become, respectively\n\\[\npzocrine>svakudom . \\quad welfsibn>pzocrine . \\quad pzocrine+svakudom>welfsibn . \\quad \\text { and } \\quad welfsibn>svakudom .\n\\]\n\nThe inequality \\( welfsibn>svakudom \\) is a consequence of \\( welfsibn>pzocrine \\) and \\( pzocrine>svakudom \\), so it can be dropped from this system. Now put \\( pzocrine=svakudom+kluvemta, welfsibn=pzocrine+draximop \\) where \\( kluvemta \\) and \\( draximop \\) are positive. Then \\( pzocrine+svakudom>welfsibn \\) becomes \\( svakudom>draximop \\), so we can put \\( svakudom=draximop+nurbisyq \\) with nurbisyq positive. Then\n\\[\n\\begin{array}{c}\npzocrine=kluvemta+draximop+nurbisyq \\\\\nsvakudom=draximop+nurbisyq \\\\\nwelfsibn=kluvemta+2 draximop+nurbisyq\n\\end{array}\n\\]\nand finally.\n\\[\n\\begin{array}{l}\nqzxwvtnp=2 kluvemta+4 draximop+3 nurbisyq+vknsqiut \\\\\nhjgrksla=kluvemta+3 draximop+2 nurbisyq+vknsqiut \\\\\nmfldepor=kluvemta+2 draximop+nurbisyq+vknsqiut .\n\\end{array}\n\\]\n\nHere \\( vknsqiut \\) can be chosen arbitrarily, while \\( kluvemta, draximop \\), and \\( nurbisyq \\) must be positive. It follows either from the derivation or from a direct check that if \\( qzxwvtnp, hjgrksla, mfldepor, vknsqiut \\) satisfy (2), where \\( kluvemta, draximop \\) and \\( nurbisyq \\) are positive, then they satisfy the inequalities (1).\n\nTo find integral solutions, we note that if we start with integral \\( qzxwvtnp, hjgrksla, mfldepor, vknsqiut \\) satisfying (1), then all the numbers \\( pzocrine, svakudom, welfsibn, kluvemta, draximop, nurbisyq \\) will be positive integers. Hence the least solution in positive integers is obtained by choosing \\( kluvemta=draximop=nurbisyq=vknsqiut=1 \\). Hence \\( qzxwvtnp=10 . hjgrksla=7, mfldepor=5, vknsqiut=1 \\) is the least solution in positive integers."
    },
    "kernel_variant": {
      "question": "In an intramural mathematics contest the five collegiate categories  \n\nSenior, Junior, Sophomore, Freshman, Pre-Freshman  \n\nare to be awarded, in this order, distinct positive even integers  \n\nP > Q > R > S > T.  \n\nA two–student team receives a rating equal to the sum of the points of its members.  \nThe organising committee demands that the fifteen possible (unordered) team–types have\nstrictly decreasing ratings in the sequence  \n PP, PQ, PR, QQ, PS, QR, RR, QS, PT, RS, SS, QT, RT, ST, TT. (★)\n\n(Thus, for instance, a team of one Junior and one Sophomore—type QR—must outrank any\nteam of two Sophomores—type RR.)\n\n1.  Determine all quintuples (P,Q,R,S,T) of distinct positive even integers that realise ordering (★).  \n2.  Find the unique quintuple for which the sum P+Q+R+S+T is minimal.",
      "solution": "Write  \n P > Q > R > S > T  (all positive even).  \n\nIntroduce the positive even gaps  \n\n α = P−Q, β = Q−R, γ = R−S, δ = S−T   (α,β,γ,δ ∈ 2ℕ).  (1)\n\nHence  \n\n P = T + δ + γ + β + α,  \n Q = T + δ + γ + β,  \n R = T + δ + γ,  \n S = T + δ.  (2)\n\n--------------------------------------------------------------------\nStep 1.  Translating the fifteen inequalities\n--------------------------------------------------------------------\nOrdering (★) is equivalent to the chain  \n\n2P > P+Q > P+R > 2Q > P+S > Q+R > 2R  \n   > Q+S > P+T > R+S > 2S > Q+T > R+T > S+T > 2T.  (3)\n\nSubstituting (2) and cancelling the common addend T gives the seven independent\nconditions  \n\n(a) α > β,   (b) α > γ,   (c) γ > β,  \n(d) α < β+γ, (e) δ > α,  (f) δ > β+γ,  (g) δ < α+β.  (4)\n\nBecause all variables are even, “>’’ means “at least 2 larger”.\nConversely, if (4) holds then (3) (hence (★)) holds, so (4) is\nnecessary and sufficient.\n\n--------------------------------------------------------------------\nStep 2.  General parametrisation\n--------------------------------------------------------------------\nChoose any positive even integers α,β,γ,δ,T satisfying (4) and define\nP,Q,R,S by (2).  The resulting quintuple (P,Q,R,S,T) consists of distinct\npositive even integers fulfilling (★).  Conversely, every quintuple that\nrealises (★) yields a unique 5-tuple (α,β,γ,δ,T) with (4).  Therefore\n\n All admissible quintuples arise from (4) and (2). □\n\n--------------------------------------------------------------------\nStep 3.  Minimising the total Σ = P+Q+R+S+T\n--------------------------------------------------------------------\nPut  \n\n Σ = P+Q+R+S+T = 5T + F, where F = 4δ + 3γ + 2β + α. (5)\n\nSince all numbers are even, the smallest possible base value is T = 2,\nso we must minimise F subject to (4).\n\nFor convenience write every even number as twice a positive integer:\n α = 2a, β = 2b, γ = 2c, δ = 2d with a,b,c,d ∈ ℕ and  \n a > b, a > c, c > b, a < b+c, d > a, d > b+c, d < a+b  (6)\n\nWe now determine the least quadruple (b,c,a,d).\n\n------------------------------------------------\n3.1  A lower bound on β  (=2b)\n------------------------------------------------\n•  b = 1 (β = 2)  \n Then c ≥ 2 (γ ≥ 4), a ≥ 3 (α ≥ 6) and d ≥ a+1 ≥ 4.  \n But d < a+b ≤ 4, contradiction.  \n•  b = 2 (β = 4)  \n We need c ≥ 3.  Put the smallest c = 3 (γ = 6).  \n Then a must satisfy a > c and a < b+c, hence a = 4 (α = 8) is forced.  \n The d-conditions give  \n  d > a = 4  and  d > b+c = 5 ⇒ d ≥ 6,  \n  d < a+b = 6         ⇒ d ≤ 5, contradiction.  \nHence b ≥ 3, i.e. β ≥ 6.\n\n------------------------------------------------\n3.2  Fixing β = 6  (b = 3) and minimising γ\n------------------------------------------------\nWith b = 3 we must have c ≥ 4 (γ ≥ 8).\n\n------------------------------------------------\n3.3  Try c = 4  (γ = 8) and minimise α\n------------------------------------------------\nFor c = 4 the inequalities demand\n a > c ⇒ a ≥ 5,  \n a < b+c = 7 ⇒ a ∈ {5,6}.  \n\n•  a = 5  (α = 10)  \n d > a ⇒ d ≥ 6,  \n d > b+c = 7 ⇒ d ≥ 8,  \n d < a+b = 8 ⇒ d ≤ 7, impossible.\n\n•  a = 6  (α = 12)  \n d > a = 6    ⇒ d ≥ 7,  \n d > b+c = 7  ⇒ d ≥ 8,  \n d < a+b = 9  ⇒ d ≤ 8, so d = 8 is forced.\n\nThe quadruple (b,c,a,d) = (3,4,6,8) — equivalently  \n β = 6, γ = 8, α = 12, δ = 16 — satisfies all of (6).\n\n------------------------------------------------\n3.4  No smaller F is possible\n------------------------------------------------\nAny increase in b, or in c, or in a, or in d raises at least one of the\ncoefficients 2β,3γ,α,4δ in F, so the above choice yields the absolute\nminimum of F.\n\nThus\n F_min = 4·16 + 3·8 + 2·6 + 12 = 112,  \n Σ_min = 5·2 + 112 = 122.\n\n--------------------------------------------------------------------\nStep 4.  The unique minimal quintuple\n--------------------------------------------------------------------\nWith T = 2 and (α,β,γ,δ) = (12,6,8,16) formula (2) gives  \n\n P = 44, Q = 32, R = 26, S = 18, T = 2.\n\nHence the minimal quintuple is (44,32,26,18,2) and its total is 122.\nUniqueness follows from the uniqueness of (b,c,a,d) obtained in 3.3.\n\n--------------------------------------------------------------------\nAnswer\n--------------------------------------------------------------------\n1.  All admissible quintuples are obtained by choosing positive even\n    α,β,γ,δ,T satisfying\n      α > β, α > γ, γ > β, α < β+γ,\n      δ > α, δ > β+γ, δ < α+β\n    and then setting  \n  P = T+δ+γ+β+α, Q = T+δ+γ+β, R = T+δ+γ, S = T+δ.\n\n2.  The unique quintuple with minimal sum is  \n  (P,Q,R,S,T) = (44, 32, 26, 18, 2), whose total is 122.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.494487",
        "was_fixed": false,
        "difficulty_analysis": "• More variables: the problem involves five unknowns instead of four, creating\n  fifteen rather than ten team-types and considerably enlarging the system of\n  inequalities.  \n\n• Additional constraints: the numbers must be even, and the imposed order is\n  not monotone in class seniority, so several inequalities are non-trivial.  \n\n• Structural depth: after a change of variables the problem reduces to seven\n  interdependent inequalities in four parameters, whose feasibility region is\n  pieced together by analysing sharp lower and upper bounds.  \n\n• Solution technique: obtaining the general solution requires recursive\n  parameterisation, inequality chaining, and optimisation within an integer\n  lattice – substantially more involved than the linear telescoping used in\n  the original.  \n\n• Minimality: proving uniqueness of the minimal even solution forces a\n  careful minimisation of a weighted linear form subject to intricate parity\n  and ordering constraints, again a step beyond the original exercise."
      }
    },
    "original_kernel_variant": {
      "question": "In an intramural mathematics contest the five collegiate categories  \n\nSenior, Junior, Sophomore, Freshman, Pre-Freshman  \n\nare to be awarded, in this order, distinct positive even integers  \n\nP > Q > R > S > T.  \n\nA two–student team receives a rating equal to the sum of the points of its members.  \nThe organising committee demands that the fifteen possible (unordered) team–types have\nstrictly decreasing ratings in the sequence  \n PP, PQ, PR, QQ, PS, QR, RR, QS, PT, RS, SS, QT, RT, ST, TT. (★)\n\n(Thus, for instance, a team of one Junior and one Sophomore—type QR—must outrank any\nteam of two Sophomores—type RR.)\n\n1.  Determine all quintuples (P,Q,R,S,T) of distinct positive even integers that realise ordering (★).  \n2.  Find the unique quintuple for which the sum P+Q+R+S+T is minimal.",
      "solution": "Write  \n P > Q > R > S > T  (all positive even).  \n\nIntroduce the positive even gaps  \n\n α = P−Q, β = Q−R, γ = R−S, δ = S−T   (α,β,γ,δ ∈ 2ℕ).  (1)\n\nHence  \n\n P = T + δ + γ + β + α,  \n Q = T + δ + γ + β,  \n R = T + δ + γ,  \n S = T + δ.  (2)\n\n--------------------------------------------------------------------\nStep 1.  Translating the fifteen inequalities\n--------------------------------------------------------------------\nOrdering (★) is equivalent to the chain  \n\n2P > P+Q > P+R > 2Q > P+S > Q+R > 2R  \n   > Q+S > P+T > R+S > 2S > Q+T > R+T > S+T > 2T.  (3)\n\nSubstituting (2) and cancelling the common addend T gives the seven independent\nconditions  \n\n(a) α > β,   (b) α > γ,   (c) γ > β,  \n(d) α < β+γ, (e) δ > α,  (f) δ > β+γ,  (g) δ < α+β.  (4)\n\nBecause all variables are even, “>’’ means “at least 2 larger”.\nConversely, if (4) holds then (3) (hence (★)) holds, so (4) is\nnecessary and sufficient.\n\n--------------------------------------------------------------------\nStep 2.  General parametrisation\n--------------------------------------------------------------------\nChoose any positive even integers α,β,γ,δ,T satisfying (4) and define\nP,Q,R,S by (2).  The resulting quintuple (P,Q,R,S,T) consists of distinct\npositive even integers fulfilling (★).  Conversely, every quintuple that\nrealises (★) yields a unique 5-tuple (α,β,γ,δ,T) with (4).  Therefore\n\n All admissible quintuples arise from (4) and (2). □\n\n--------------------------------------------------------------------\nStep 3.  Minimising the total Σ = P+Q+R+S+T\n--------------------------------------------------------------------\nPut  \n\n Σ = P+Q+R+S+T = 5T + F, where F = 4δ + 3γ + 2β + α. (5)\n\nSince all numbers are even, the smallest possible base value is T = 2,\nso we must minimise F subject to (4).\n\nFor convenience write every even number as twice a positive integer:\n α = 2a, β = 2b, γ = 2c, δ = 2d with a,b,c,d ∈ ℕ and  \n a > b, a > c, c > b, a < b+c, d > a, d > b+c, d < a+b  (6)\n\nWe now determine the least quadruple (b,c,a,d).\n\n------------------------------------------------\n3.1  A lower bound on β  (=2b)\n------------------------------------------------\n•  b = 1 (β = 2)  \n Then c ≥ 2 (γ ≥ 4), a ≥ 3 (α ≥ 6) and d ≥ a+1 ≥ 4.  \n But d < a+b ≤ 4, contradiction.  \n•  b = 2 (β = 4)  \n We need c ≥ 3.  Put the smallest c = 3 (γ = 6).  \n Then a must satisfy a > c and a < b+c, hence a = 4 (α = 8) is forced.  \n The d-conditions give  \n  d > a = 4  and  d > b+c = 5 ⇒ d ≥ 6,  \n  d < a+b = 6         ⇒ d ≤ 5, contradiction.  \nHence b ≥ 3, i.e. β ≥ 6.\n\n------------------------------------------------\n3.2  Fixing β = 6  (b = 3) and minimising γ\n------------------------------------------------\nWith b = 3 we must have c ≥ 4 (γ ≥ 8).\n\n------------------------------------------------\n3.3  Try c = 4  (γ = 8) and minimise α\n------------------------------------------------\nFor c = 4 the inequalities demand\n a > c ⇒ a ≥ 5,  \n a < b+c = 7 ⇒ a ∈ {5,6}.  \n\n•  a = 5  (α = 10)  \n d > a ⇒ d ≥ 6,  \n d > b+c = 7 ⇒ d ≥ 8,  \n d < a+b = 8 ⇒ d ≤ 7, impossible.\n\n•  a = 6  (α = 12)  \n d > a = 6    ⇒ d ≥ 7,  \n d > b+c = 7  ⇒ d ≥ 8,  \n d < a+b = 9  ⇒ d ≤ 8, so d = 8 is forced.\n\nThe quadruple (b,c,a,d) = (3,4,6,8) — equivalently  \n β = 6, γ = 8, α = 12, δ = 16 — satisfies all of (6).\n\n------------------------------------------------\n3.4  No smaller F is possible\n------------------------------------------------\nAny increase in b, or in c, or in a, or in d raises at least one of the\ncoefficients 2β,3γ,α,4δ in F, so the above choice yields the absolute\nminimum of F.\n\nThus\n F_min = 4·16 + 3·8 + 2·6 + 12 = 112,  \n Σ_min = 5·2 + 112 = 122.\n\n--------------------------------------------------------------------\nStep 4.  The unique minimal quintuple\n--------------------------------------------------------------------\nWith T = 2 and (α,β,γ,δ) = (12,6,8,16) formula (2) gives  \n\n P = 44, Q = 32, R = 26, S = 18, T = 2.\n\nHence the minimal quintuple is (44,32,26,18,2) and its total is 122.\nUniqueness follows from the uniqueness of (b,c,a,d) obtained in 3.3.\n\n--------------------------------------------------------------------\nAnswer\n--------------------------------------------------------------------\n1.  All admissible quintuples are obtained by choosing positive even\n    α,β,γ,δ,T satisfying\n      α > β, α > γ, γ > β, α < β+γ,\n      δ > α, δ > β+γ, δ < α+β\n    and then setting  \n  P = T+δ+γ+β+α, Q = T+δ+γ+β, R = T+δ+γ, S = T+δ.\n\n2.  The unique quintuple with minimal sum is  \n  (P,Q,R,S,T) = (44, 32, 26, 18, 2), whose total is 122.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.413788",
        "was_fixed": false,
        "difficulty_analysis": "• More variables: the problem involves five unknowns instead of four, creating\n  fifteen rather than ten team-types and considerably enlarging the system of\n  inequalities.  \n\n• Additional constraints: the numbers must be even, and the imposed order is\n  not monotone in class seniority, so several inequalities are non-trivial.  \n\n• Structural depth: after a change of variables the problem reduces to seven\n  interdependent inequalities in four parameters, whose feasibility region is\n  pieced together by analysing sharp lower and upper bounds.  \n\n• Solution technique: obtaining the general solution requires recursive\n  parameterisation, inequality chaining, and optimisation within an integer\n  lattice – substantially more involved than the linear telescoping used in\n  the original.  \n\n• Minimality: proving uniqueness of the minimal even solution forces a\n  careful minimisation of a weighted linear form subject to intricate parity\n  and ordering constraints, again a step beyond the original exercise."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}