{
  "index": "1950-B-3",
  "type": "COMB",
  "tag": [
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "3. In the Gregorian calendar:\n(i) years not divisible by 4 are common years;\n(ii) years divisible by 4 but not by 100 are leap years;\n(iii) years divisible by 100 but not by 400 are common years;\n(iv) years divisible by 400 are leap years;\n(v) a leap year contains 366 days; a common year 365 days.\n\nProve that the probability that Christmas falls on a Wednesday is not \\( 1 / 7 \\).",
  "solution": "Solution. According to the rules given, any 400 consecutive Gregorian years will involve 303 ordinary years and 97 leap years making a total of \\( 400 \\cdot 365+97 \\) days. Since this number is divisible by \\( 7(400 \\equiv 1,365 \\equiv 1 \\), \\( 97 \\equiv-1(\\bmod 7) \\) ), there is an integral number of weeks in 400 years (in fact. 20.871 weeks) and therefore the day of the week on which Christmas (or any other calendar date) falls repeats in cycles of 400 . If there are \\( N \\) years in such a period on which Christmas falls on Wednesday, then the probability that Christmas falls on Wednesday is \\( N / 400 \\). But \\( N / 400 \\neq 1 / 7 \\) for any integer \\( N \\).\n\nRemark. The following table gives the number of years in each 400-year cycle on which Christmas falls on each day of the week.\n\\begin{tabular}{ccccccc} \nSun. & Mon. & Tues. & Wed. & Thurs. & Fri. & Sat. \\\\\n\\hline 58 & 56 & 58 & 57 & 57 & 58 & 56\n\\end{tabular}\n\nThe superstitious may also be interested to know that the thirteenth of the month is more likely to fall on Friday than on any other day of the week. See: American Mathematical Monthly, vol. 40 (1933), page 607; also Emanuel Parzen, Modern Probability Theory and Its Applications, New York, 1960, page 26 ff.",
  "vars": [
    "N"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "N": "numyears"
      },
      "question": "3. In the Gregorian calendar:\n(i) years not divisible by 4 are common years;\n(ii) years divisible by 4 but not by 100 are leap years;\n(iii) years divisible by 100 but not by 400 are common years;\n(iv) years divisible by 400 are leap years;\n(v) a leap year contains 366 days; a common year 365 days.\n\nProve that the probability that Christmas falls on a Wednesday is not \\( 1 / 7 \\).",
      "solution": "Solution. According to the rules given, any 400 consecutive Gregorian years will involve 303 ordinary years and 97 leap years making a total of \\( 400 \\cdot 365+97 \\) days. Since this number is divisible by \\( 7(400 \\equiv 1,365 \\equiv 1 \\), \\( 97 \\equiv-1(\\bmod 7) \\) ), there is an integral number of weeks in 400 years (in fact. 20.871 weeks) and therefore the day of the week on which Christmas (or any other calendar date) falls repeats in cycles of 400. If there are \\( numyears \\) years in such a period on which Christmas falls on Wednesday, then the probability that Christmas falls on Wednesday is \\( numyears / 400 \\). But \\( numyears / 400 \\neq 1 / 7 \\) for any integer \\( numyears \\).\n\nRemark. The following table gives the number of years in each 400-year cycle on which Christmas falls on each day of the week.\n\\begin{tabular}{ccccccc} \nSun. & Mon. & Tues. & Wed. & Thurs. & Fri. & Sat. \\\\\n\\hline 58 & 56 & 58 & 57 & 57 & 58 & 56\n\\end{tabular}\n\nThe superstitious may also be interested to know that the thirteenth of the month is more likely to fall on Friday than on any other day of the week. See: American Mathematical Monthly, vol. 40 (1933), page 607; also Emanuel Parzen, Modern Probability Theory and Its Applications, New York, 1960, page 26 ff."
    },
    "descriptive_long_confusing": {
      "map": {
        "N": "tapestry"
      },
      "question": "3. In the Gregorian calendar:\n(i) years not divisible by 4 are common years;\n(ii) years divisible by 4 but not by 100 are leap years;\n(iii) years divisible by 100 but not by 400 are common years;\n(iv) years divisible by 400 are leap years;\n(v) a leap year contains 366 days; a common year 365 days.\n\nProve that the probability that Christmas falls on a Wednesday is not \\( 1 / 7 \\).",
      "solution": "Solution. According to the rules given, any 400 consecutive Gregorian years will involve 303 ordinary years and 97 leap years making a total of \\( 400 \\cdot 365+97 \\) days. Since this number is divisible by \\( 7(400 \\equiv 1,365 \\equiv 1 \\), \\( 97 \\equiv-1(\\bmod 7) \\) ), there is an integral number of weeks in 400 years (in fact. 20.871 weeks) and therefore the day of the week on which Christmas (or any other calendar date) falls repeats in cycles of 400 . If there are \\( tapestry \\) years in such a period on which Christmas falls on Wednesday, then the probability that Christmas falls on Wednesday is \\( tapestry / 400 \\). But \\( tapestry / 400 \\neq 1 / 7 \\) for any integer \\( tapestry \\).\n\nRemark. The following table gives the number of years in each 400-year cycle on which Christmas falls on each day of the week.\n\\begin{tabular}{ccccccc} \nSun. & Mon. & Tues. & Wed. & Thurs. & Fri. & Sat. \\\\\n\\hline 58 & 56 & 58 & 57 & 57 & 58 & 56\n\\end{tabular}\n\nThe superstitious may also be interested to know that the thirteenth of the month is more likely to fall on Friday than on any other day of the week. See: American Mathematical Monthly, vol. 40 (1933), page 607; also Emanuel Parzen, Modern Probability Theory and Its Applications, New York, 1960, page 26 ff."
    },
    "descriptive_long_misleading": {
      "map": {
        "N": "zeroness"
      },
      "question": "3. In the Gregorian calendar:\n(i) years not divisible by 4 are common years;\n(ii) years divisible by 4 but not by 100 are leap years;\n(iii) years divisible by 100 but not by 400 are common years;\n(iv) years divisible by 400 are leap years;\n(v) a leap year contains 366 days; a common year 365 days.\n\nProve that the probability that Christmas falls on a Wednesday is not \\( 1 / 7 \\).",
      "solution": "Solution. According to the rules given, any 400 consecutive Gregorian years will involve 303 ordinary years and 97 leap years making a total of \\( 400 \\cdot 365+97 \\) days. Since this number is divisible by \\( 7(400 \\equiv 1,365 \\equiv 1 \\), \\( 97 \\equiv-1(\\bmod 7) \\) ), there is an integral number of weeks in 400 years (in fact. 20.871 weeks) and therefore the day of the week on which Christmas (or any other calendar date) falls repeats in cycles of 400 . If there are \\( zeroness \\) years in such a period on which Christmas falls on Wednesday, then the probability that Christmas falls on Wednesday is \\( zeroness / 400 \\). But \\( zeroness / 400 \\neq 1 / 7 \\) for any integer \\( zeroness \\).\n\nRemark. The following table gives the number of years in each 400-year cycle on which Christmas falls on each day of the week.\n\\begin{tabular}{ccccccc} \nSun. & Mon. & Tues. & Wed. & Thurs. & Fri. & Sat. \\\\\n\\hline 58 & 56 & 58 & 57 & 57 & 58 & 56\n\\end{tabular}\n\nThe superstitious may also be interested to know that the thirteenth of the month is more likely to fall on Friday than on any other day of the week. See: American Mathematical Monthly, vol. 40 (1933), page 607; also Emanuel Parzen, Modern Probability Theory and Its Applications, New York, 1960, page 26 ff."
    },
    "garbled_string": {
      "map": {
        "N": "qzxwvtnp"
      },
      "question": "3. In the Gregorian calendar:\n(i) years not divisible by 4 are common years;\n(ii) years divisible by 4 but not by 100 are leap years;\n(iii) years divisible by 100 but not by 400 are common years;\n(iv) years divisible by 400 are leap years;\n(v) a leap year contains 366 days; a common year 365 days.\n\nProve that the probability that Christmas falls on a Wednesday is not \\( 1 / 7 \\).",
      "solution": "Solution. According to the rules given, any 400 consecutive Gregorian years will involve 303 ordinary years and 97 leap years making a total of \\( 400 \\cdot 365+97 \\) days. Since this number is divisible by \\( 7(400 \\equiv 1,365 \\equiv 1, 97 \\equiv-1(\\bmod 7) ) \\), there is an integral number of weeks in 400 years (in fact. 20.871 weeks) and therefore the day of the week on which Christmas (or any other calendar date) falls repeats in cycles of 400 . If there are \\( qzxwvtnp \\) years in such a period on which Christmas falls on Wednesday, then the probability that Christmas falls on Wednesday is \\( qzxwvtnp / 400 \\). But \\( qzxwvtnp / 400 \\neq 1 / 7 \\) for any integer \\( qzxwvtnp \\).\n\nRemark. The following table gives the number of years in each 400-year cycle on which Christmas falls on each day of the week.\n\\begin{tabular}{ccccccc} \nSun. & Mon. & Tues. & Wed. & Thurs. & Fri. & Sat. \\\\\n\\hline 58 & 56 & 58 & 57 & 57 & 58 & 56\n\\end{tabular}\n\nThe superstitious may also be interested to know that the thirteenth of the month is more likely to fall on Friday than on any other day of the week. See: American Mathematical Monthly, vol. 40 (1933), page 607; also Emanuel Parzen, Modern Probability Theory and Its Applications, New York, 1960, page 26 ff."
    },
    "kernel_variant": {
      "question": "(Triskaidekaphobia meets modular arithmetic – corrected version)\n\nThroughout we work with the Gregorian calendar.  \nFor a Gregorian year Y let D(Y) be its Doomsday, i.e. the weekday that occurs on the Conway\nanchor dates  4 April, 6 June, 8 August, …, 12 December of that year.\nA year is called triskaidekatic if its calendar contains exactly three months whose 13-th day is a Friday.\n\n1.  Show that a year is triskaidekatic  \n        ⇔  (A) it is a leap-year with Doomsday = Wednesday, or  \n           (B) it is a common year with Doomsday = Saturday.\n\n2.  In every complete 400-year Gregorian cycle prove that  \n        (i) among the 97 leap-years exactly 15 have Doomsday = Wednesday, and  \n        (ii) among the 303 common years exactly 44 have Doomsday = Saturday.\n\n3.  Deduce that there are precisely 15 + 44 = 59 triskaidekatic years in each 400-year\n   cycle and that the limiting probability that a randomly chosen Gregorian year is\n   triskaidekatic equals\n\n                          59/400  =  0.1475 .\n\n   Compare this with the “naïve’’ value (1/7)³ ≈ 0.0029 obtained by treating the three Friday events as independent, and explain briefly why independence fails.",
      "solution": "(Weekdays are numbered 0 = Sun, 1 = Mon, …, 6 = Sat; arithmetic is done mod 7.)\n\nStep 0.   The Doomsday infrastructure  \nFor any year Y write\n     Y = 400a + y,        0 ≤ y ≤ 399.\nBecause 400 · 365 + 97 ≡ 0 (mod 7), Doomsdays repeat every 400 years, so it suffices to study the years 0–399 of a single cycle (say the one that starts with the year 2000).  \nLet L(Y) be the number of leap-years strictly preceding Y.  The standard elementary\nargument (or Conway’s formula) yields\n\n          D(Y)  ≡  D(0) + Y + L(Y)        (mod 7).        (†)\n\nFor the cycle starting in 2000 one has D(0)=2 (Tuesday).\n\nUseful facts that follow immediately from (†):\n\n•  Passing from Y to Y+1 adds 1 to D unless Y is a leap-year, in which case it adds 2.  \n•  Over a 400-year cycle every weekday occurs either 57 or 58 times as a Doomsday.\n\nStep 1.  Which Doomsdays create three Friday-13ths?  \nLet r_m be the displacement (weekday of 13 m) – D(Y) in month m.\nA one–time computation gives\n\n month     Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec\n r_m (common) 3   6   6   2   4   0   2   5   1   3   6   1\n r_m (leap)   2   5   6   2   4   0   2   5   1   3   6   1\n\nWriting F=5 (Friday) and δ(Y)=F–D(Y), inspection of the table shows\n\n  •  in a common year δ=6 hits exactly the three months Feb–Mar–Nov;  \n  •  in a leap-year  δ=2 hits exactly the three months Jan–Apr–Jul.\n\nThus\n\n        “three Friday-13ths’’  ⇔\n        (leap & δ=2)  ∪  (common & δ=6).\n\nBecause F=5, δ=2 implies D=3 (Wednesday) and δ=6 implies D=6 (Saturday).  \nThis proves Part 1.\n\nStep 2.  Counting the relevant Doomsdays inside a 400-year cycle  \nWrite every year as Y = 4k + r with r∈{0,1,2,3}.  \nPut k = 25q + s with 0 ≤ q ≤ 3 and 0 ≤ s ≤ 24.  \nThe number of leap-years strictly before Y is\n\n        L(Y) = ⎣Y/4⎦ – ⎣Y/100⎦ + ⎣Y/400⎦\n              = k – q,                           (0<Y<400)                (1)\n\nbecause ⎣Y/400⎦=0 unless Y=0.\n\nInsert (1) into (†) and set g(Y)=Y+L(Y); then\n\n        g(Y) ≡ 5k + r – q\n             = 124q + 5s + r                       (mod 7),               (2)\n\nbecause 5·25 = 125 ≡ 124 (mod 7).\n\n(A)  Leap-years with D=Wednesday  \nA leap-year is characterised by r=0 and (except for the first one) k≠25,50,75.\nFor such a year we need g(Y) ≡ 1 (because D=2+g(Y) and Wednesday=3):\n\n        124q + 5s ≡ 1   ⇒   5(q+s) ≡ 1   ⇒   q+s ≡ 3    (mod 7).          (3)\n\nWithin each 25-year block (fixed q) the values s=0,…,24 cover every residue\nclass mod 7 exactly three times, with the additional residues 0,1,2,3\nappearing once more.  Condition (3) therefore yields\n\n        •  4 solutions for q=0,1,2 (because s=3,10,17,24 etc. avoid s=0),  \n        •  3 solutions for q=3 (s cannot be 0 because k=25·3 is non-leap).\n\nHence    4+4+4+3 = 15   leap-years have Doomsday = Wednesday.                 (4)\n\n(B)  Common years with D=Saturday  \nNow r∈{1,2,3} or (k,r)=(25,0),(50,0),(75,0).  We require g(Y) ≡ 4, i.e.\n\n        124q + 5s + r ≡ 4\n       ⇔ (q+s) + 3r ≡ 5                               (mod 7).             (5)\n\nFor r=1,2,3 the same counting of residues inside each 25-year block\ngives 11 solutions, irrespective of q.  Thus (5) contributes 4·11 = 44 common\nyears.  The three century years (100,200,300) fail (5), so they add nothing.\nTherefore exactly\n\n        44  common years have Doomsday = Saturday.                          (6)\n\nStatements (i) and (ii) in Part 2 follow from (4) and (6).\n\nStep 3.  How many triskaidekatic years?  \nBy Step 1 every triskaidekatic year is counted in (4) or (6) and none is\nmissed.  Hence\n\n        # triskaidekatic years per 400-year cycle = 15 + 44 = 59.\n\nConsequently the limiting probability that a random Gregorian year is\ntriskaidekatic equals\n\n        P = 59 / 400 = 0.1475.\n\nStep 4.  Why the naïve (1/7)³ is wrong  \nThe weekdays of the 13-th of January, April and July in a fixed year are\nall rigid translates of the same Doomsday; they are therefore perfectly\ncorrelated, not independent.  The true probability 0.1475 is fifty times\nlarger than (1/7)³ ≈ 0.0029, underscoring the failure of independence.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.433211",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original Christmas–Wednesday problem, the present\nvariant\n\n• introduces a composite event involving three different dates inside\n  the same year, not a single date;\n\n• forces the solver to master and use Conway’s Doomsday machinery, to\n  set up and read an offset table, and to blend it with the Gregorian\n  leap-year rule;\n\n• requires a two–layer modular count (first inside the months, then\n  inside the 400-year cycle) rather than the single divisibility‐by-7\n  check that sufficed before;\n\n• demands an explicit enumeration (with proof) of how many leap-years\n  have one particular Doomsday and how many common years another,\n  instead of merely showing that a fraction N/400 cannot equal 1/7;\n\n• culminates in computing an exact, non-obvious probability\n  101/400 ≈ 25 %, not merely in disproving a simple guess.\n\nAll of this adds both conceptual and technical depth, fulfilling the\n“significantly harder’’ mandate."
      }
    },
    "original_kernel_variant": {
      "question": "(Triskaidekaphobia meets modular arithmetic – corrected version)\n\nThroughout we work with the Gregorian calendar.  \nFor a Gregorian year Y let D(Y) be its Doomsday, i.e. the weekday that occurs on the Conway\nanchor dates  4 April, 6 June, 8 August, …, 12 December of that year.\nA year is called triskaidekatic if its calendar contains exactly three months whose 13-th day is a Friday.\n\n1.  Show that a year is triskaidekatic  \n        ⇔  (A) it is a leap-year with Doomsday = Wednesday, or  \n           (B) it is a common year with Doomsday = Saturday.\n\n2.  In every complete 400-year Gregorian cycle prove that  \n        (i) among the 97 leap-years exactly 15 have Doomsday = Wednesday, and  \n        (ii) among the 303 common years exactly 44 have Doomsday = Saturday.\n\n3.  Deduce that there are precisely 15 + 44 = 59 triskaidekatic years in each 400-year\n   cycle and that the limiting probability that a randomly chosen Gregorian year is\n   triskaidekatic equals\n\n                          59/400  =  0.1475 .\n\n   Compare this with the “naïve’’ value (1/7)³ ≈ 0.0029 obtained by treating the three Friday events as independent, and explain briefly why independence fails.",
      "solution": "(Weekdays are numbered 0 = Sun, 1 = Mon, …, 6 = Sat; arithmetic is done mod 7.)\n\nStep 0.   The Doomsday infrastructure  \nFor any year Y write\n     Y = 400a + y,        0 ≤ y ≤ 399.\nBecause 400 · 365 + 97 ≡ 0 (mod 7), Doomsdays repeat every 400 years, so it suffices to study the years 0–399 of a single cycle (say the one that starts with the year 2000).  \nLet L(Y) be the number of leap-years strictly preceding Y.  The standard elementary\nargument (or Conway’s formula) yields\n\n          D(Y)  ≡  D(0) + Y + L(Y)        (mod 7).        (†)\n\nFor the cycle starting in 2000 one has D(0)=2 (Tuesday).\n\nUseful facts that follow immediately from (†):\n\n•  Passing from Y to Y+1 adds 1 to D unless Y is a leap-year, in which case it adds 2.  \n•  Over a 400-year cycle every weekday occurs either 57 or 58 times as a Doomsday.\n\nStep 1.  Which Doomsdays create three Friday-13ths?  \nLet r_m be the displacement (weekday of 13 m) – D(Y) in month m.\nA one–time computation gives\n\n month     Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec\n r_m (common) 3   6   6   2   4   0   2   5   1   3   6   1\n r_m (leap)   2   5   6   2   4   0   2   5   1   3   6   1\n\nWriting F=5 (Friday) and δ(Y)=F–D(Y), inspection of the table shows\n\n  •  in a common year δ=6 hits exactly the three months Feb–Mar–Nov;  \n  •  in a leap-year  δ=2 hits exactly the three months Jan–Apr–Jul.\n\nThus\n\n        “three Friday-13ths’’  ⇔\n        (leap & δ=2)  ∪  (common & δ=6).\n\nBecause F=5, δ=2 implies D=3 (Wednesday) and δ=6 implies D=6 (Saturday).  \nThis proves Part 1.\n\nStep 2.  Counting the relevant Doomsdays inside a 400-year cycle  \nWrite every year as Y = 4k + r with r∈{0,1,2,3}.  \nPut k = 25q + s with 0 ≤ q ≤ 3 and 0 ≤ s ≤ 24.  \nThe number of leap-years strictly before Y is\n\n        L(Y) = ⎣Y/4⎦ – ⎣Y/100⎦ + ⎣Y/400⎦\n              = k – q,                           (0<Y<400)                (1)\n\nbecause ⎣Y/400⎦=0 unless Y=0.\n\nInsert (1) into (†) and set g(Y)=Y+L(Y); then\n\n        g(Y) ≡ 5k + r – q\n             = 124q + 5s + r                       (mod 7),               (2)\n\nbecause 5·25 = 125 ≡ 124 (mod 7).\n\n(A)  Leap-years with D=Wednesday  \nA leap-year is characterised by r=0 and (except for the first one) k≠25,50,75.\nFor such a year we need g(Y) ≡ 1 (because D=2+g(Y) and Wednesday=3):\n\n        124q + 5s ≡ 1   ⇒   5(q+s) ≡ 1   ⇒   q+s ≡ 3    (mod 7).          (3)\n\nWithin each 25-year block (fixed q) the values s=0,…,24 cover every residue\nclass mod 7 exactly three times, with the additional residues 0,1,2,3\nappearing once more.  Condition (3) therefore yields\n\n        •  4 solutions for q=0,1,2 (because s=3,10,17,24 etc. avoid s=0),  \n        •  3 solutions for q=3 (s cannot be 0 because k=25·3 is non-leap).\n\nHence    4+4+4+3 = 15   leap-years have Doomsday = Wednesday.                 (4)\n\n(B)  Common years with D=Saturday  \nNow r∈{1,2,3} or (k,r)=(25,0),(50,0),(75,0).  We require g(Y) ≡ 4, i.e.\n\n        124q + 5s + r ≡ 4\n       ⇔ (q+s) + 3r ≡ 5                               (mod 7).             (5)\n\nFor r=1,2,3 the same counting of residues inside each 25-year block\ngives 11 solutions, irrespective of q.  Thus (5) contributes 4·11 = 44 common\nyears.  The three century years (100,200,300) fail (5), so they add nothing.\nTherefore exactly\n\n        44  common years have Doomsday = Saturday.                          (6)\n\nStatements (i) and (ii) in Part 2 follow from (4) and (6).\n\nStep 3.  How many triskaidekatic years?  \nBy Step 1 every triskaidekatic year is counted in (4) or (6) and none is\nmissed.  Hence\n\n        # triskaidekatic years per 400-year cycle = 15 + 44 = 59.\n\nConsequently the limiting probability that a random Gregorian year is\ntriskaidekatic equals\n\n        P = 59 / 400 = 0.1475.\n\nStep 4.  Why the naïve (1/7)³ is wrong  \nThe weekdays of the 13-th of January, April and July in a fixed year are\nall rigid translates of the same Doomsday; they are therefore perfectly\ncorrelated, not independent.  The true probability 0.1475 is fifty times\nlarger than (1/7)³ ≈ 0.0029, underscoring the failure of independence.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.374777",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original Christmas–Wednesday problem, the present\nvariant\n\n• introduces a composite event involving three different dates inside\n  the same year, not a single date;\n\n• forces the solver to master and use Conway’s Doomsday machinery, to\n  set up and read an offset table, and to blend it with the Gregorian\n  leap-year rule;\n\n• requires a two–layer modular count (first inside the months, then\n  inside the 400-year cycle) rather than the single divisibility‐by-7\n  check that sufficed before;\n\n• demands an explicit enumeration (with proof) of how many leap-years\n  have one particular Doomsday and how many common years another,\n  instead of merely showing that a fraction N/400 cannot equal 1/7;\n\n• culminates in computing an exact, non-obvious probability\n  101/400 ≈ 25 %, not merely in disproving a simple guess.\n\nAll of this adds both conceptual and technical depth, fulfilling the\n“significantly harder’’ mandate."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}