{
  "index": "1971-A-5",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "A-5. A game of solitaire is played as follows. After each play, according to the outcome, the player receives either \\( a \\) or \\( b \\) points ( \\( a \\) and \\( b \\) are positive integers with \\( a \\) greater than \\( b) \\), and his score accumulates from play to play. It has been noticed that there are thirtyfive non-attainable scores and that one of these is 58 . Find \\( a \\) and \\( b \\).",
  "solution": "A-5 The attainable scores are those non-negative integers expressible in the form \\( x a+y b \\) with \\( x \\) and \\( y \\) non-negative integers. If \\( a \\) and \\( b \\) are not relatively prime there are infinitely many non-attainable scores. Hence \\( (a, b)=1 \\). It will be shown that the number of non-attainable scores is \\( \\frac{1}{2}(a-1)(b-1) \\).\n\nIf \\( m \\) is an attainable score, the line \\( a x+b y=m \\) passes through at least one lattice point in the closed first quadrant. Because \\( a \\) and \\( b \\) are relatively prime, the lattice points on a line \\( a x+b y=m \\) are at a horizontal distance of \\( b \\). The firstquadrant segment of \\( a x+b y=m \\) has a horizontal projection of \\( m / a \\) and thus every score \\( m \\geqq a b \\) is attainable. Every non-attainable score must satisfy \\( 0 \\leqq m<a b \\).\n\nIf \\( 0 \\leqq m<a b \\), the first-quadrant segment of the line \\( a x+b y=m \\) has a horizontal projection less than \\( b \\), and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points \\( (x, y) \\) with \\( 0 \\leqq a x+b y<a b \\) in the first quadrant and attainable scores with \\( 0 \\leqq m<a b \\). The closed rectangle \\( 0 \\leqq x \\leqq b \\), \\( 0 \\leqq y \\leqq a \\) contains \\( (a+1)(b+1) \\) lattice points, so the number of lattice points in the first quadrant with \\( 0 \\leqq a x+b y<a b \\) is \\( \\frac{1}{2}(a+1)(b+1)-1 \\). This is the number of attainable scores with \\( 0 \\leqq m<a b \\). Hence the number of non-attainable scores in this range (which is all of them) is \\( a b-\\frac{1}{2}(a+1)(b+1)+1=\\frac{1}{2}(a-1)(b-1) \\).\n\nIn our given example \\( 70=(a-1)(b-1)=1(70)=2(35)=5(14)=7(10) \\). The conditions \\( a>b,(a, b)=1 \\) yield two possibilities \\( a=71, b=2 \\) and \\( a=11, b=8 \\). Since \\( 58=71(0)+2(29) \\), the first of these alternatives is eliminated. The line \\( 11 x+8 y \\) \\( =58 \\) passes through \\( (6,-1) \\) and \\( (-2,10) \\) and thus does not pass through a lattice point in the first quadrant. The unique solution is \\( a=11, b=8 \\).",
  "vars": [
    "m",
    "x",
    "y"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "bigscore",
        "b": "smallscore",
        "m": "totalscore",
        "x": "playcount",
        "y": "otherplay"
      },
      "question": "A game of solitaire is played as follows. After each play, according to the outcome, the player receives either \\( bigscore \\) or \\( smallscore \\) points ( \\( bigscore \\) and \\( smallscore \\) are positive integers with \\( bigscore \\) greater than \\( smallscore) \\), and his score accumulates from play to play. It has been noticed that there are thirtyfive non-attainable scores and that one of these is 58 . Find \\( bigscore \\) and \\( smallscore \\).",
      "solution": "A-5 The attainable scores are those non-negative integers expressible in the form \\( playcount\\,bigscore+otherplay\\,smallscore \\) with \\( playcount \\) and \\( otherplay \\) non-negative integers. If \\( bigscore \\) and \\( smallscore \\) are not relatively prime there are infinitely many non-attainable scores. Hence \\( (bigscore,smallscore)=1 \\). It will be shown that the number of non-attainable scores is \\( \\frac{1}{2}(bigscore-1)(smallscore-1) \\).\n\nIf \\( totalscore \\) is an attainable score, the line \\( bigscore\\,playcount+smallscore\\,otherplay=totalscore \\) passes through at least one lattice point in the closed first quadrant. Because \\( bigscore \\) and \\( smallscore \\) are relatively prime, the lattice points on a line \\( bigscore\\,playcount+smallscore\\,otherplay=totalscore \\) are at a horizontal distance of \\( smallscore \\). The firstquadrant segment of \\( bigscore\\,playcount+smallscore\\,otherplay=totalscore \\) has a horizontal projection of \\( totalscore/bigscore \\) and thus every score \\( totalscore \\geqq bigscore\\,smallscore \\) is attainable. Every non-attainable score must satisfy \\( 0 \\leqq totalscore < bigscore\\,smallscore \\).\n\nIf \\( 0 \\leqq totalscore < bigscore\\,smallscore \\), the first-quadrant segment of the line \\( bigscore\\,playcount+smallscore\\,otherplay=totalscore \\) has a horizontal projection less than \\( smallscore \\), and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points \\( (playcount,otherplay) \\) with \\( 0 \\leqq bigscore\\,playcount+smallscore\\,otherplay < bigscore\\,smallscore \\) in the first quadrant and attainable scores with \\( 0 \\leqq totalscore < bigscore\\,smallscore \\). The closed rectangle \\( 0 \\leqq playcount \\leqq smallscore \\), \\( 0 \\leqq otherplay \\leqq bigscore \\) contains \\( (bigscore+1)(smallscore+1) \\) lattice points, so the number of lattice points in the first quadrant with \\( 0 \\leqq bigscore\\,playcount+smallscore\\,otherplay < bigscore\\,smallscore \\) is \\( \\frac{1}{2}(bigscore+1)(smallscore+1)-1 \\). This is the number of attainable scores with \\( 0 \\leqq totalscore < bigscore\\,smallscore \\). Hence the number of non-attainable scores in this range (which is all of them) is \\( bigscore\\,smallscore-\\frac{1}{2}(bigscore+1)(smallscore+1)+1=\\frac{1}{2}(bigscore-1)(smallscore-1) \\).\n\nIn our given example \\( 70=(bigscore-1)(smallscore-1)=1(70)=2(35)=5(14)=7(10) \\). The conditions \\( bigscore>smallscore,(bigscore,smallscore)=1 \\) yield two possibilities \\( bigscore=71,smallscore=2 \\) and \\( bigscore=11,smallscore=8 \\). Since \\( 58=71(0)+2(29) \\), the first of these alternatives is eliminated. The line \\( 11\\,playcount+8\\,otherplay=58 \\) passes through \\( (6,-1) \\) and \\( (-2,10) \\) and thus does not pass through a lattice point in the first quadrant. The unique solution is \\( bigscore=11,smallscore=8 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "pinecone",
        "b": "sailboat",
        "m": "blueberry",
        "x": "driftwood",
        "y": "sandstorm"
      },
      "question": "A-5. A game of solitaire is played as follows. After each play, according to the outcome, the player receives either \\( pinecone \\) or \\( sailboat \\) points ( \\( pinecone \\) and \\( sailboat \\) are positive integers with \\( pinecone \\) greater than \\( sailboat) \\), and his score accumulates from play to play. It has been noticed that there are thirtyfive non-attainable scores and that one of these is 58 . Find \\( pinecone \\) and \\( sailboat \\).",
      "solution": "A-5 The attainable scores are those non-negative integers expressible in the form \\( driftwood pinecone+sandstorm sailboat \\) with \\( driftwood \\) and \\( sandstorm \\) non-negative integers. If \\( pinecone \\) and \\( sailboat \\) are not relatively prime there are infinitely many non-attainable scores. Hence \\( (pinecone, sailboat)=1 \\). It will be shown that the number of non-attainable scores is \\( \\frac{1}{2}(pinecone-1)(sailboat-1) \\).\n\nIf \\( blueberry \\) is an attainable score, the line \\( pinecone driftwood+sailboat sandstorm=blueberry \\) passes through at least one lattice point in the closed first quadrant. Because \\( pinecone \\) and \\( sailboat \\) are relatively prime, the lattice points on a line \\( pinecone driftwood+sailboat sandstorm=blueberry \\) are at a horizontal distance of \\( sailboat \\). The firstquadrant segment of \\( pinecone driftwood+sailboat sandstorm=blueberry \\) has a horizontal projection of \\( blueberry / pinecone \\) and thus every score \\( blueberry \\geqq pinecone sailboat \\) is attainable. Every non-attainable score must satisfy \\( 0 \\leqq blueberry<pinecone sailboat \\).\n\nIf \\( 0 \\leqq blueberry<pinecone sailboat \\), the first-quadrant segment of the line \\( pinecone driftwood+sailboat sandstorm=blueberry \\) has a horizontal projection less than \\( sailboat \\), and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points \\( (driftwood, sandstorm) \\) with \\( 0 \\leqq pinecone driftwood+sailboat sandstorm<pinecone sailboat \\) in the first quadrant and attainable scores with \\( 0 \\leqq blueberry<pinecone sailboat \\). The closed rectangle \\( 0 \\leqq driftwood \\leqq sailboat \\), \\( 0 \\leqq sandstorm \\leqq pinecone \\) contains \\( (pinecone+1)(sailboat+1) \\) lattice points, so the number of lattice points in the first quadrant with \\( 0 \\leqq pinecone driftwood+sailboat sandstorm<pinecone sailboat \\) is \\( \\frac{1}{2}(pinecone+1)(sailboat+1)-1 \\). This is the number of attainable scores with \\( 0 \\leqq blueberry<pinecone sailboat \\). Hence the number of non-attainable scores in this range (which is all of them) is \\( pinecone sailboat-\\frac{1}{2}(pinecone+1)(sailboat+1)+1=\\frac{1}{2}(pinecone-1)(sailboat-1) \\).\n\nIn our given example \\( 70=(pinecone-1)(sailboat-1)=1(70)=2(35)=5(14)=7(10) \\). The conditions \\( pinecone>sailboat,(pinecone, sailboat)=1 \\) yield two possibilities \\( pinecone=71, sailboat=2 \\) and \\( pinecone=11, sailboat=8 \\). Since \\( 58=71(0)+2(29) \\), the first of these alternatives is eliminated. The line \\( 11 driftwood+8 sandstorm \\) \\( =58 \\) passes through \\( (6,-1) \\) and \\( (-2,10) \\) and thus does not pass through a lattice point in the first quadrant. The unique solution is \\( pinecone=11, sailboat=8 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "minorgain",
        "b": "majorgain",
        "m": "deficits",
        "x": "vertical",
        "y": "horizontal"
      },
      "question": "A game of solitaire is played as follows. After each play, according to the outcome, the player receives either \\( minorgain \\) or \\( majorgain \\) points ( \\( minorgain \\) and \\( majorgain \\) are positive integers with \\( minorgain \\) greater than \\( majorgain) \\), and his score accumulates from play to play. It has been noticed that there are thirtyfive non-attainable scores and that one of these is 58 . Find \\( minorgain \\) and \\( majorgain \\).",
      "solution": "A-5 The attainable scores are those non-negative integers expressible in the form \\( vertical minorgain+horizontal majorgain \\) with \\( vertical \\) and \\( horizontal \\) non-negative integers. If \\( minorgain \\) and \\( majorgain \\) are not relatively prime there are infinitely many non-attainable scores. Hence \\( (minorgain, majorgain)=1 \\). It will be shown that the number of non-attainable scores is \\( \\frac{1}{2}(minorgain-1)(majorgain-1) \\).\n\nIf \\( deficits \\) is an attainable score, the line \\( minorgain vertical+majorgain horizontal=deficits \\) passes through at least one lattice point in the closed first quadrant. Because \\( minorgain \\) and \\( majorgain \\) are relatively prime, the lattice points on a line \\( minorgain vertical+majorgain horizontal=deficits \\) are at a horizontal distance of \\( majorgain \\). The firstquadrant segment of \\( minorgain vertical+majorgain horizontal=deficits \\) has a horizontal projection of \\( deficits / minorgain \\) and thus every score \\( deficits \\geqq minorgain majorgain \\) is attainable. Every non-attainable score must satisfy \\( 0 \\leqq deficits<minorgain majorgain \\).\n\nIf \\( 0 \\leqq deficits<minorgain majorgain \\), the first-quadrant segment of the line \\( minorgain vertical+majorgain horizontal=deficits \\) has a horizontal projection less than \\( majorgain \\), and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points \\( (vertical, horizontal) \\) with \\( 0 \\leqq minorgain vertical+majorgain horizontal<minorgain majorgain \\) in the first quadrant and attainable scores with \\( 0 \\leqq deficits<minorgain majorgain \\). The closed rectangle \\( 0 \\leqq vertical \\leqq majorgain \\), \\( 0 \\leqq horizontal \\leqq minorgain \\) contains \\( (minorgain+1)(majorgain+1) \\) lattice points, so the number of lattice points in the first quadrant with \\( 0 \\leqq minorgain vertical+majorgain horizontal<minorgain majorgain \\) is \\( \\frac{1}{2}(minorgain+1)(majorgain+1)-1 \\). This is the number of attainable scores with \\( 0 \\leqq deficits<minorgain majorgain \\). Hence the number of non-attainable scores in this range (which is all of them) is \\( minorgain majorgain-\\frac{1}{2}(minorgain+1)(majorgain+1)+1=\\frac{1}{2}(minorgain-1)(majorgain-1) \\).\n\nIn our given example \\( 70=(minorgain-1)(majorgain-1)=1(70)=2(35)=5(14)=7(10) \\). The conditions \\( minorgain>majorgain,(minorgain, majorgain)=1 \\) yield two possibilities \\( minorgain=71, majorgain=2 \\) and \\( minorgain=11, majorgain=8 \\). Since \\( 58=71(0)+2(29) \\), the first of these alternatives is eliminated. The line \\( 11 vertical+8 horizontal \\) \\( =58 \\) passes through \\( (6,-1) \\) and \\( (-2,10) \\) and thus does not pass through a lattice point in the first quadrant. The unique solution is \\( minorgain=11, majorgain=8 \\)."
    },
    "garbled_string": {
      "map": {
        "m": "lfdasjqe",
        "x": "pvmrqkto",
        "y": "sntuwgie",
        "a": "qzxwvtnp",
        "b": "hjgrksla"
      },
      "question": "A-5. A game of solitaire is played as follows. After each play, according to the outcome, the player receives either \\( qzxwvtnp \\) or \\( hjgrksla \\) points ( \\( qzxwvtnp \\) and \\( hjgrksla \\) are positive integers with \\( qzxwvtnp \\) greater than \\( hjgrksla) \\), and his score accumulates from play to play. It has been noticed that there are thirtyfive non-attainable scores and that one of these is 58 . Find \\( qzxwvtnp \\) and \\( hjgrksla \\).",
      "solution": "A-5 The attainable scores are those non-negative integers expressible in the form \\( pvmrqkto qzxwvtnp+sntuwgie hjgrksla \\) with \\( pvmrqkto \\) and \\( sntuwgie \\) non-negative integers. If \\( qzxwvtnp \\) and \\( hjgrksla \\) are not relatively prime there are infinitely many non-attainable scores. Hence \\( (qzxwvtnp, hjgrksla)=1 \\). It will be shown that the number of non-attainable scores is \\( \\frac{1}{2}(qzxwvtnp-1)(hjgrksla-1) \\).\n\nIf \\( lfdasjqe \\) is an attainable score, the line \\( qzxwvtnp pvmrqkto+hjgrksla sntuwgie=lfdasjqe \\) passes through at least one lattice point in the closed first quadrant. Because \\( qzxwvtnp \\) and \\( hjgrksla \\) are relatively prime, the lattice points on a line \\( qzxwvtnp pvmrqkto+hjgrksla sntuwgie=lfdasjqe \\) are at a horizontal distance of \\( hjgrksla \\). The firstquadrant segment of \\( qzxwvtnp pvmrqkto+hjgrksla sntuwgie=lfdasjqe \\) has a horizontal projection of \\( lfdasjqe / qzxwvtnp \\) and thus every score \\( lfdasjqe \\geqq qzxwvtnp hjgrksla \\) is attainable. Every non-attainable score must satisfy \\( 0 \\leqq lfdasjqe<qzxwvtnp hjgrksla \\).\n\nIf \\( 0 \\leqq lfdasjqe<qzxwvtnp hjgrksla \\), the first-quadrant segment of the line \\( qzxwvtnp pvmrqkto+hjgrksla sntuwgie=lfdasjqe \\) has a horizontal projection less than \\( hjgrksla \\), and so contains at most one lattice point. Thus there is a one-to-one correspondence between lattice points \\( (pvmrqkto, sntuwgie) \\) with \\( 0 \\leqq qzxwvtnp pvmrqkto+hjgrksla sntuwgie<qzxwvtnp hjgrksla \\) in the first quadrant and attainable scores with \\( 0 \\leqq lfdasjqe<qzxwvtnp hjgrksla \\). The closed rectangle \\( 0 \\leqq pvmrqkto \\leqq hjgrksla \\), \\( 0 \\leqq sntuwgie \\leqq qzxwvtnp \\) contains \\( (qzxwvtnp+1)(hjgrksla+1) \\) lattice points, so the number of lattice points in the first quadrant with \\( 0 \\leqq qzxwvtnp pvmrqkto+hjgrksla sntuwgie<qzxwvtnp hjgrksla \\) is \\( \\frac{1}{2}(qzxwvtnp+1)(hjgrksla+1)-1 \\). This is the number of attainable scores with \\( 0 \\leqq lfdasjqe<qzxwvtnp hjgrksla \\). Hence the number of non-attainable scores in this range (which is all of them) is \\( qzxwvtnp hjgrksla-\\frac{1}{2}(qzxwvtnp+1)(hjgrksla+1)+1=\\frac{1}{2}(qzxwvtnp-1)(hjgrksla-1) \\).\n\nIn our given example \\( 70=(qzxwvtnp-1)(hjgrksla-1)=1(70)=2(35)=5(14)=7(10) \\). The conditions \\( qzxwvtnp>hjgrksla,(qzxwvtnp, hjgrksla)=1 \\) yield two possibilities \\( qzxwvtnp=71, hjgrksla=2 \\) and \\( qzxwvtnp=11, hjgrksla=8 \\). Since \\( 58=71(0)+2(29) \\), the first of these alternatives is eliminated. The line \\( 11 pvmrqkto+8 sntuwgie \\) \\( =58 \\) passes through \\( (6,-1) \\) and \\( (-2,10) \\) and thus does not pass through a lattice point in the first quadrant. The unique solution is \\( qzxwvtnp=11, hjgrksla=8 \\)."
    },
    "kernel_variant": {
      "question": "In a solitaire game the player’s cumulative score increases, move by move, by either a or b points, where a and b are coprime integers satisfying a > b > 1.  \nLet U denote the set of positive integers that can never occur as a total score (the “unattainable scores”).  It is known that  \n\n1. |U| = 65,  \n2. the arithmetic mean of the elements of U is 47,  \n3. the number 23 itself is unattainable.  \n\nDetermine the ordered pair (a, b).\n\n--------------------------------------------------------------------",
      "solution": "Notation.  An integer n is attainable iff n = a x + b y for some non–negative integers x, y.  We assume throughout that gcd(a, b) = 1 and a > b > 1.\n\nStep 1.  Classical Frobenius facts.  \nFor coprime positive integers a, b one has  \n\n(i)  the largest unattainable integer:  g = ab − a − b;  \n(ii) the number N of unattainable integers:  N = ½(a − 1)(b − 1);  \n(iii) the sum Σ of all unattainable integers:  \n         Σ = [(a − 1)(b − 1)(2ab − a − b − 1)] / 12.  \n\n(All three are derived by counting lattice points in the right triangle ax + by < ab.)\n\nStep 2.  Using the cardinality |U| = 65.  \nWith N = 65 we get\n\n½(a − 1)(b − 1) = 65  ⇒ (a − 1)(b − 1) = 130.  (1)\n\nFactor 130 and list all divisor pairs d₁ > d₂:\n\n(130, 1), (65, 2), (26, 5), (13, 10).\n\nAdding 1 to each component yields the candidate pairs (a, b):\n\n(131, 2), (66, 3), (27, 6), (14, 11).  (2)\n\nEnforce gcd(a, b) = 1:  \ngcd(66, 3) = 3 and gcd(27, 6) = 3, so the middle two pairs are invalid.  \nTwo viable pairs remain:\n\n(a, b) = (131, 2) or (a, b) = (14, 11).  (3)\n\nStep 3.  Employing the mean μ = 47.  \nLet μ be the mean of the unattainable numbers.  From (iii),\n\nΣ = μ |U| = 47 · 65 = 3055.  (4)\n\nInsert Σ and (1) into formula (iii):\n\n130 · (2ab − a − b − 1) / 12 = 3055  \n ⇒ 2ab − a − b − 1 = 3055·12 / 130 = 282  \n ⇒ 2ab − a − b = 283.  (5)\n\nTest the two candidates from (3):\n\n• (a, b) = (131, 2): 2·131·2 − 131 − 2 = 524 − 133 = 391 ≠ 283;  \n• (a, b) = (14, 11): 2·14·11 − 14 − 11 = 308 − 25 = 283 ✓.\n\nThus (a, b) = (14, 11) is the only pair fulfilling the first two conditions.\n\nStep 4.  Checking that 23 is unattainable.  \nSuppose 14x + 11y = 23 with x, y ≥ 0.  \nReduce modulo 11:\n\n14x ≡ 3x ≡ 23 ≡ 1 (mod 11)  \n ⇒ 3x ≡ 1 (mod 11).\n\nBecause 3⁻¹ ≡ 4 (mod 11), we have x ≡ 4 (mod 11) ⇒ x = 4 + 11k (k ≥ 0).  \nFor k = 0, x = 4 gives 14x = 56 > 23; for k ≥ 1, 14x grows further.  No non-negative solution (x, y) exists, so 23 is indeed unattainable.\n\nAll three given conditions are therefore satisfied uniquely by\n\n(a, b) = (14, 11).\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.598431",
        "was_fixed": false,
        "difficulty_analysis": "• The original problem required only one datum (the total count of unattainable scores) and a single exhibited unattainable number.  \n• The enhanced variant adds a second global statistic—the arithmetic mean—which forces the use of the advanced closed-form formula for the total sum of unattainable numbers.  This introduces an extra unknown-elimination equation, making mere inspection impossible.  \n• Solving demands simultaneous handling of three classical Frobenius invariants (count, sum and specific membership), not just the count.  Consequently the solver must marshal number-theoretic factorisation, lattice-point enumeration results, and modular arithmetic in concert—considerably more sophisticated than the single-formula argument that sufficed for the original task."
      }
    },
    "original_kernel_variant": {
      "question": "In a solitaire game the player’s cumulative score increases, move by move, by either a or b points, where a and b are coprime integers satisfying a > b > 1.  \nLet U denote the set of positive integers that can never occur as a total score (the “unattainable scores”).  It is known that  \n\n1. |U| = 65,  \n2. the arithmetic mean of the elements of U is 47,  \n3. the number 23 itself is unattainable.  \n\nDetermine the ordered pair (a, b).\n\n--------------------------------------------------------------------",
      "solution": "Notation.  An integer n is attainable iff n = a x + b y for some non–negative integers x, y.  We assume throughout that gcd(a, b) = 1 and a > b > 1.\n\nStep 1.  Classical Frobenius facts.  \nFor coprime positive integers a, b one has  \n\n(i)  the largest unattainable integer:  g = ab − a − b;  \n(ii) the number N of unattainable integers:  N = ½(a − 1)(b − 1);  \n(iii) the sum Σ of all unattainable integers:  \n         Σ = [(a − 1)(b − 1)(2ab − a − b − 1)] / 12.  \n\n(All three are derived by counting lattice points in the right triangle ax + by < ab.)\n\nStep 2.  Using the cardinality |U| = 65.  \nWith N = 65 we get\n\n½(a − 1)(b − 1) = 65  ⇒ (a − 1)(b − 1) = 130.  (1)\n\nFactor 130 and list all divisor pairs d₁ > d₂:\n\n(130, 1), (65, 2), (26, 5), (13, 10).\n\nAdding 1 to each component yields the candidate pairs (a, b):\n\n(131, 2), (66, 3), (27, 6), (14, 11).  (2)\n\nEnforce gcd(a, b) = 1:  \ngcd(66, 3) = 3 and gcd(27, 6) = 3, so the middle two pairs are invalid.  \nTwo viable pairs remain:\n\n(a, b) = (131, 2) or (a, b) = (14, 11).  (3)\n\nStep 3.  Employing the mean μ = 47.  \nLet μ be the mean of the unattainable numbers.  From (iii),\n\nΣ = μ |U| = 47 · 65 = 3055.  (4)\n\nInsert Σ and (1) into formula (iii):\n\n130 · (2ab − a − b − 1) / 12 = 3055  \n ⇒ 2ab − a − b − 1 = 3055·12 / 130 = 282  \n ⇒ 2ab − a − b = 283.  (5)\n\nTest the two candidates from (3):\n\n• (a, b) = (131, 2): 2·131·2 − 131 − 2 = 524 − 133 = 391 ≠ 283;  \n• (a, b) = (14, 11): 2·14·11 − 14 − 11 = 308 − 25 = 283 ✓.\n\nThus (a, b) = (14, 11) is the only pair fulfilling the first two conditions.\n\nStep 4.  Checking that 23 is unattainable.  \nSuppose 14x + 11y = 23 with x, y ≥ 0.  \nReduce modulo 11:\n\n14x ≡ 3x ≡ 23 ≡ 1 (mod 11)  \n ⇒ 3x ≡ 1 (mod 11).\n\nBecause 3⁻¹ ≡ 4 (mod 11), we have x ≡ 4 (mod 11) ⇒ x = 4 + 11k (k ≥ 0).  \nFor k = 0, x = 4 gives 14x = 56 > 23; for k ≥ 1, 14x grows further.  No non-negative solution (x, y) exists, so 23 is indeed unattainable.\n\nAll three given conditions are therefore satisfied uniquely by\n\n(a, b) = (14, 11).\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.479322",
        "was_fixed": false,
        "difficulty_analysis": "• The original problem required only one datum (the total count of unattainable scores) and a single exhibited unattainable number.  \n• The enhanced variant adds a second global statistic—the arithmetic mean—which forces the use of the advanced closed-form formula for the total sum of unattainable numbers.  This introduces an extra unknown-elimination equation, making mere inspection impossible.  \n• Solving demands simultaneous handling of three classical Frobenius invariants (count, sum and specific membership), not just the count.  Consequently the solver must marshal number-theoretic factorisation, lattice-point enumeration results, and modular arithmetic in concert—considerably more sophisticated than the single-formula argument that sufficed for the original task."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}