{
  "index": "1984-A-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-6. Let \\( n \\) be a positive integer, and let \\( f(n) \\) denote the last nonzero digit in the decimal expansion of \\( n! \\). For instance, \\( f(5)=2 \\).\n(a) Show that if \\( a_{1}, a_{2}, \\ldots, a_{k} \\) are distinct nonnegative integers, then \\( f\\left(5^{a_{1}}+5^{a_{2}}+\\cdots+5^{a_{k}}\\right) \\) depends only on the sum \\( a_{1}+a_{2}+\\cdots+a_{k} \\).\n(b) Assuming part (a), we can define\n\\[\ng(s)=f\\left(5^{a_{1}}+5^{a_{2}}+\\cdots+5^{a_{k}}\\right)\n\\]\nwhere \\( s=a_{1}+a_{2}+\\cdots+a_{k} \\). Find the least positive integer \\( p \\) for which\n\\[\ng(s)=g(s+p), \\quad \\text { for all } s \\geqslant 1\n\\]\nor else show that no such \\( p \\) exists.",
  "solution": "A-6.\n(a) All congruences are modulo 10 .\n\nLemma. \\( f(5 n) \\equiv 2^{n} f(n) \\).\nProof. We have\n(*)\n\\[\n(5 n)!=10^{n} n!\\prod_{i=0}^{n-1} \\frac{(5 i+1)(5 i+2)(5 i+3)(5 i+4)}{2} .\n\\]\n\nIf \\( i \\) is even, then\n\\[\n\\frac{1}{2}(5 i+1)(5 i+2)(5 i+3)(5 i+4) \\equiv \\frac{1}{2}(1 \\cdot 2 \\cdot 3 \\cdot 4) \\equiv 2,\n\\]\nand if \\( i \\) is odd, then\n\\[\n\\frac{1}{2}(5 i+1)(5 i+2)(5 i+3)(5 i+4) \\equiv \\frac{1}{2}(6 \\cdot 7 \\cdot 8 \\cdot 9) \\equiv 2\n\\]\n\nThus the entire product above is congruent to \\( 2^{n} \\). From (*) it is clear that the largest power of 10 dividing ( \\( 5 n \\) )! is the same as the largest power of 10 dividing \\( 10^{n} n \\) !, and the proof follows.\n\nWe now show by induction on \\( 5^{a_{1}}+\\cdots+5^{a_{k}} \\) that\n\\[\nf\\left(5^{a_{1}}+\\cdots+5^{a_{k}}\\right) \\equiv 2^{a_{1}+\\cdots+a_{k}}\n\\]\n(which depends only on \\( a_{1}+\\cdots+a_{k} \\) as desired).\nThis is true for \\( 5^{a_{1}}+\\cdots+5^{a_{k}}=1 \\), since \\( f\\left(5^{0}\\right) \\equiv 2^{0} \\equiv 1 \\).\nCase 1. All \\( a_{i}>0 \\). By the lemma and induction,\n\\[\n\\begin{aligned}\nf\\left(5^{a_{1}}+\\cdots+5^{a_{k}}\\right) & \\equiv 2^{5^{a_{1}-1}+\\cdots+5^{a_{k}-1}} f\\left(5^{a_{1}-1}+\\cdots+5^{a_{k}-1}\\right) \\\\\n& \\equiv 2^{k} \\cdot 2^{\\left(a_{1}-1\\right)+\\cdots+\\left(a_{k}-1\\right)} \\quad\\left(\\text { since } 2^{5^{\\prime}} \\equiv 2 \\text { for } i \\geqslant 0\\right) \\\\\n& \\equiv 2^{a_{1}+\\cdots+a_{k}} .\n\\end{aligned}\n\\]\n\nCase 2. Some \\( a_{1}=0 \\), say \\( a_{1}=0 \\). Now\n\\[\n(1+5 m)!=(1+5 m)(5 m)!,\n\\]\nso \\( f(1+5 m) \\equiv(1+5 m) f(5 m) \\). But \\( f(5 m) \\) is even for \\( m \\geqslant 1 \\) since \\( (5 m)! \\) is divisible by a higher power of 2 than of 5 . But\n\\[\n(1+5 m) \\cdot(2 j) \\equiv 2 j,\n\\]\nso \\( f(1+5 m) \\equiv f(5 m) \\). Letting \\( m=5^{a_{2}-1}+\\cdots+5^{a_{k}-1} \\), the proof follows by induction.\n(b) The least \\( p \\geqslant 1 \\) for which \\( 2^{s+p} \\equiv 2^{2} \\) for all \\( s \\geqslant 1 \\) is \\( p=4 \\).",
  "vars": [
    "n",
    "f",
    "a_1",
    "a_2",
    "a_k",
    "a_i",
    "k",
    "g",
    "s",
    "p",
    "i",
    "j",
    "m"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "posintn",
        "f": "lastdigit",
        "a_1": "firstexp",
        "a_2": "secondex",
        "a_k": "kthexpon",
        "a_i": "indexexpo",
        "k": "counttot",
        "g": "sumfunct",
        "s": "sumsindex",
        "p": "periodic",
        "i": "loopindx",
        "j": "loopindy",
        "m": "auxiliar"
      },
      "question": "Problem A-6. Let \\( \\text{posintn} \\) be a positive integer, and let \\( \\text{lastdigit}(\\text{posintn}) \\) denote the last nonzero digit in the decimal expansion of \\( \\text{posintn}! \\). For instance, \\( \\text{lastdigit}(5)=2 \\).\n(a) Show that if \\( \\text{firstexp}, \\text{secondex}, \\ldots, \\text{kthexpon} \\) are distinct nonnegative integers, then\n\\[\n\\text{lastdigit}\\left(5^{\\text{firstexp}}+5^{\\text{secondex}}+\\cdots+5^{\\text{kthexpon}}\\right)\n\\]\ndepends only on the sum \\( \\text{firstexp}+\\text{secondex}+\\cdots+\\text{kthexpon} \\).\n(b) Assuming part (a), we can define\n\\[\n\\text{sumfunct}(\\text{sumsindex})=\\text{lastdigit}\\left(5^{\\text{firstexp}}+5^{\\text{secondex}}+\\cdots+5^{\\text{kthexpon}}\\right)\n\\]\nwhere \\( \\text{sumsindex}=\\text{firstexp}+\\text{secondex}+\\cdots+\\text{kthexpon} \\). Find the least positive integer \\( \\text{periodic} \\) for which\n\\[\n\\text{sumfunct}(\\text{sumsindex})=\\text{sumfunct}(\\text{sumsindex}+\\text{periodic}), \\quad \\text { for all } \\text{sumsindex} \\geqslant 1\n\\]\nor else show that no such \\( \\text{periodic} \\) exists.",
      "solution": "A-6.\n(a) All congruences are modulo 10.\n\nLemma. \\( \\text{lastdigit}(5\\,\\text{posintn}) \\equiv 2^{\\text{posintn}} \\text{lastdigit}(\\text{posintn}) \\).\n\nProof. We have\n(*)\n\\[\n(5 \\text{posintn})!=10^{\\text{posintn}} \\; \\text{posintn}!\\prod_{\\text{loopindx}=0}^{\\text{posintn}-1} \\frac{(5 \\text{loopindx}+1)(5 \\text{loopindx}+2)(5 \\text{loopindx}+3)(5 \\text{loopindx}+4)}{2} .\n\\]\n\nIf \\( \\text{loopindx} \\) is even, then\n\\[\n\\frac{1}{2}(5 \\text{loopindx}+1)(5 \\text{loopindx}+2)(5 \\text{loopindx}+3)(5 \\text{loopindx}+4) \\equiv \\frac{1}{2}(1 \\cdot 2 \\cdot 3 \\cdot 4) \\equiv 2,\n\\]\nand if \\( \\text{loopindx} \\) is odd, then\n\\[\n\\frac{1}{2}(5 \\text{loopindx}+1)(5 \\text{loopindx}+2)(5 \\text{loopindx}+3)(5 \\text{loopindx}+4) \\equiv \\frac{1}{2}(6 \\cdot 7 \\cdot 8 \\cdot 9) \\equiv 2.\n\\]\n\nThus the entire product above is congruent to \\( 2^{\\text{posintn}} \\). From (*) it is clear that the largest power of 10 dividing \\( (5 \\text{posintn})! \\) is the same as the largest power of 10 dividing \\( 10^{\\text{posintn}}\\, \\text{posintn}! \\), and the proof follows.\n\nWe now show by induction on \\( 5^{\\text{firstexp}}+\\cdots+5^{\\text{kthexpon}} \\) that\n\\[\n\\text{lastdigit}\\left(5^{\\text{firstexp}}+\\cdots+5^{\\text{kthexpon}}\\right) \\equiv 2^{\\text{firstexp}+\\cdots+\\text{kthexpon}}\n\\]\n(which depends only on \\( \\text{firstexp}+\\cdots+\\text{kthexpon} \\) as desired).\nThis is true for \\( 5^{\\text{firstexp}}+\\cdots+5^{\\text{kthexpon}}=1 \\), since \\( \\text{lastdigit}\\left(5^{0}\\right) \\equiv 2^{0} \\equiv 1 \\).\n\nCase 1. All \\( \\text{firstexp},\\ldots,\\text{kthexpon} >0 \\). By the lemma and induction,\n\\[\n\\begin{aligned}\n\\text{lastdigit}\\left(5^{\\text{firstexp}}+\\cdots+5^{\\text{kthexpon}}\\right) & \\equiv 2^{5^{\\text{firstexp}-1}+\\cdots+5^{\\text{kthexpon}-1}} \\text{lastdigit}\\left(5^{\\text{firstexp}-1}+\\cdots+5^{\\text{kthexpon}-1}\\right) \\\\\n& \\equiv 2^{\\text{counttot}} \\cdot 2^{\\left(\\text{firstexp}-1\\right)+\\cdots+\\left(\\text{kthexpon}-1\\right)} \\quad\\left(\\text { since } 2^{5^{\\text{loopindx}}} \\equiv 2 \\text { for } \\text{loopindx} \\geqslant 0\\right) \\\\\n& \\equiv 2^{\\text{firstexp}+\\cdots+\\text{kthexpon}} .\n\\end{aligned}\n\\]\n\nCase 2. Some \\( \\text{firstexp}=0 \\), say \\( \\text{firstexp}=0 \\). Now\n\\[\n(1+5 \\text{auxiliar})!=(1+5 \\text{auxiliar})(5 \\text{auxiliar})!,\n\\]\nso \\( \\text{lastdigit}(1+5 \\text{auxiliar}) \\equiv(1+5 \\text{auxiliar}) \\text{lastdigit}(5 \\text{auxiliar}) \\). But \\( \\text{lastdigit}(5 \\text{auxiliar}) \\) is even for \\( \\text{auxiliar} \\geqslant 1 \\) since \\( (5 \\text{auxiliar})! \\) is divisible by a higher power of 2 than of 5. But\n\\[\n(1+5 \\text{auxiliar}) \\cdot(2 \\text{loopindy}) \\equiv 2 \\text{loopindy},\n\\]\nso \\( \\text{lastdigit}(1+5 \\text{auxiliar}) \\equiv \\text{lastdigit}(5 \\text{auxiliar}) \\). Letting \\( \\text{auxiliar}=5^{\\text{secondex}-1}+\\cdots+5^{\\text{kthexpon}-1} \\), the proof follows by induction.\n\n(b) The least \\( \\text{periodic} \\geqslant 1 \\) for which \\( 2^{\\text{sumsindex}+\\text{periodic}} \\equiv 2^{2} \\) for all \\( \\text{sumsindex} \\geqslant 1 \\) is \\( \\text{periodic}=4 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pineapple",
        "f": "tortoise",
        "a_1": "cinnamon",
        "a_2": "blueberry",
        "a_k": "macadamia",
        "a_i": "butternut",
        "k": "lemonade",
        "g": "armadillo",
        "s": "kangaroo",
        "p": "marshland",
        "i": "snowflake",
        "j": "rainstorm",
        "m": "thunderer"
      },
      "question": "Problem A-6. Let \\( pineapple \\) be a positive integer, and let \\( tortoise(pineapple) \\) denote the last nonzero digit in the decimal expansion of \\( pineapple! \\). For instance, \\( tortoise(5)=2 \\).\n(a) Show that if \\( cinnamon, blueberry, \\ldots, macadamia \\) are distinct nonnegative integers, then \\( tortoise\\left(5^{cinnamon}+5^{blueberry}+\\cdots+5^{macadamia}\\right) \\) depends only on the sum \\( cinnamon+blueberry+\\cdots+macadamia \\).\n(b) Assuming part (a), we can define\n\\[\narmadillo(kangaroo)=tortoise\\left(5^{cinnamon}+5^{blueberry}+\\cdots+5^{macadamia}\\right)\n\\]\nwhere \\( kangaroo=cinnamon+blueberry+\\cdots+macadamia \\). Find the least positive integer \\( marshland \\) for which\n\\[\narmadillo(kangaroo)=armadillo(kangaroo+marshland), \\quad \\text { for all } kangaroo \\geqslant 1\n\\]\nor else show that no such \\( marshland \\) exists.",
      "solution": "A-6.\n(a) All congruences are modulo 10.\n\nLemma. \\( tortoise(5 pineapple) \\equiv 2^{pineapple} \\, tortoise(pineapple) \\).\nProof. We have\n(* )\n\\[\n(5 pineapple)!=10^{pineapple} \\, pineapple!\\prod_{snowflake=0}^{pineapple-1} \\frac{(5 \\, snowflake+1)(5 \\, snowflake+2)(5 \\, snowflake+3)(5 \\, snowflake+4)}{2} .\n\\]\nIf \\( snowflake \\) is even, then\n\\[\n\\frac{1}{2}(5 \\, snowflake+1)(5 \\, snowflake+2)(5 \\, snowflake+3)(5 \\, snowflake+4) \\equiv \\frac{1}{2}(1 \\cdot 2 \\cdot 3 \\cdot 4) \\equiv 2,\n\\]\nand if \\( snowflake \\) is odd, then\n\\[\n\\frac{1}{2}(5 \\, snowflake+1)(5 \\, snowflake+2)(5 \\, snowflake+3)(5 \\, snowflake+4) \\equiv \\frac{1}{2}(6 \\cdot 7 \\cdot 8 \\cdot 9) \\equiv 2.\n\\]\nThus the entire product above is congruent to \\( 2^{pineapple} \\). From (*) it is clear that the largest power of 10 dividing (\\( 5 pineapple \\) )! is the same as the largest power of 10 dividing \\( 10^{pineapple} \\, pineapple! \\), and the proof follows.\n\nWe now show by induction on \\( 5^{cinnamon}+\\cdots+5^{macadamia} \\) that\n\\[\ntortoise\\left(5^{cinnamon}+\\cdots+5^{macadamia}\\right) \\equiv 2^{cinnamon+\\cdots+macadamia},\\]\nwhich depends only on \\( cinnamon+\\cdots+macadamia \\) as desired.\nThis is true for \\( 5^{cinnamon}+\\cdots+5^{macadamia}=1 \\), since \\( tortoise\\left(5^{0}\\right) \\equiv 2^{0} \\equiv 1 \\).\n\nCase 1. All \\( cinnamon, blueberry, \\ldots, macadamia>0 \\). By the lemma and induction,\n\\[\n\\begin{aligned}\ntortoise\\left(5^{cinnamon}+\\cdots+5^{macadamia}\\right) & \\equiv 2^{5^{cinnamon-1}+\\cdots+5^{macadamia-1}} \\, tortoise\\left(5^{cinnamon-1}+\\cdots+5^{macadamia-1}\\right) \\\\ & \\equiv 2^{lemonade} \\cdot 2^{(cinnamon-1)+\\cdots+(macadamia-1)} \\quad(\\text{since } 2^{5^{\\prime}} \\equiv 2 \\text{ for } snowflake \\geqslant 0) \\\\ & \\equiv 2^{cinnamon+\\cdots+macadamia} .\n\\end{aligned}\n\\]\n\nCase 2. Some \\( cinnamon=0 \\), say \\( cinnamon=0 \\). Now\n\\[\n(1+5 \\, thunderer)!=(1+5 \\, thunderer)(5 \\, thunderer)!,\n\\]\nso \\( tortoise(1+5 \\, thunderer) \\equiv (1+5 \\, thunderer) \\, tortoise(5 \\, thunderer) \\). But \\( tortoise(5 \\, thunderer) \\) is even for \\( thunderer \\geqslant 1 \\) since \\( (5 \\, thunderer)! \\) is divisible by a higher power of 2 than of 5. But\n\\[\n(1+5 \\, thunderer) \\cdot (2 \\, rainstorm) \\equiv 2 \\, rainstorm,\n\\]\nso \\( tortoise(1+5 \\, thunderer) \\equiv tortoise(5 \\, thunderer) \\). Letting \\( thunderer = 5^{blueberry-1}+\\cdots+5^{macadamia-1} \\), the proof follows by induction.\n\n(b) The least \\( marshland \\geqslant 1 \\) for which \\( 2^{kangaroo+marshland} \\equiv 2^{2} \\) for all \\( kangaroo \\geqslant 1 \\) is \\( marshland = 4 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "fractionalval",
        "f": "wholesumdigit",
        "a_1": "apexindexone",
        "a_2": "apexindextwo",
        "a_k": "apexindexkay",
        "a_i": "apexindexvar",
        "k": "singularvalue",
        "g": "inversefunc",
        "s": "difference",
        "p": "aperiodic",
        "i": "maximalindex",
        "j": "residualvar",
        "m": "microamount"
      },
      "question": "Problem A-6. Let \\( fractionalval \\) be a positive integer, and let \\( wholesumdigit(fractionalval) \\) denote the last nonzero digit in the decimal expansion of \\( fractionalval! \\). For instance, \\( wholesumdigit(5)=2 \\).\n(a) Show that if \\( apexindexone, apexindextwo, \\ldots, apexindexkay \\) are distinct nonnegative integers, then \\( wholesumdigit\\left(5^{apexindexone}+5^{apexindextwo}+\\cdots+5^{apexindexkay}\\right) \\) depends only on the sum \\( apexindexone+apexindextwo+\\cdots+apexindexkay \\).\n(b) Assuming part (a), we can define\n\\[\ninversefunc(difference)=wholesumdigit\\left(5^{apexindexone}+5^{apexindextwo}+\\cdots+5^{apexindexkay}\\right)\n\\]\nwhere \\( difference=apexindexone+apexindextwo+\\cdots+apexindexkay \\). Find the least positive integer \\( aperiodic \\) for which\n\\[\ninversefunc(difference)=inversefunc(difference+aperiodic), \\quad \\text { for all } difference \\geqslant 1\n\\]\nor else show that no such \\( aperiodic \\) exists.",
      "solution": "A-6.\n(a) All congruences are modulo 10 .\n\nLemma. \\( wholesumdigit(5 fractionalval) \\equiv 2^{fractionalval} wholesumdigit(fractionalval) \\).\n\nProof. We have\n(*)\n\\[\n(5 fractionalval)!=10^{fractionalval} fractionalval!\\prod_{maximalindex=0}^{fractionalval-1} \\frac{(5 maximalindex+1)(5 maximalindex+2)(5 maximalindex+3)(5 maximalindex+4)}{2} .\n\\]\n\nIf \\( maximalindex \\) is even, then\n\\[\n\\frac{1}{2}(5 maximalindex+1)(5 maximalindex+2)(5 maximalindex+3)(5 maximalindex+4) \\equiv \\frac{1}{2}(1 \\cdot 2 \\cdot 3 \\cdot 4) \\equiv 2,\n\\]\nand if \\( maximalindex \\) is odd, then\n\\[\n\\frac{1}{2}(5 maximalindex+1)(5 maximalindex+2)(5 maximalindex+3)(5 maximalindex+4) \\equiv \\frac{1}{2}(6 \\cdot 7 \\cdot 8 \\cdot 9) \\equiv 2\n\\]\n\nThus the entire product above is congruent to \\( 2^{fractionalval} \\). From (*) it is clear that the largest power of 10 dividing \\( (5 fractionalval)! \\) is the same as the largest power of 10 dividing \\( 10^{fractionalval} fractionalval! \\), and the proof follows.\n\nWe now show by induction on \\( 5^{apexindexone}+\\cdots+5^{apexindexkay} \\) that\n\\[\nwholesumdigit\\left(5^{apexindexone}+\\cdots+5^{apexindexkay}\\right) \\equiv 2^{apexindexone+\\cdots+apexindexkay}\n\\]\n(which depends only on \\( apexindexone+\\cdots+apexindexkay \\) as desired).\nThis is true for \\( 5^{apexindexone}+\\cdots+5^{apexindexkay}=1 \\), since \\( wholesumdigit\\left(5^{0}\\right) \\equiv 2^{0} \\equiv 1 \\).\n\nCase 1. All \\( apexindexvar>0 \\). By the lemma and induction,\n\\[\n\\begin{aligned}\nwholesumdigit\\left(5^{apexindexone}+\\cdots+5^{apexindexkay}\\right) & \\equiv 2^{5^{apexindexone-1}+\\cdots+5^{apexindexkay-1}} wholesumdigit\\left(5^{apexindexone-1}+\\cdots+5^{apexindexkay-1}\\right) \\\\\n& \\equiv 2^{singularvalue} \\cdot 2^{\\left(apexindexone-1\\right)+\\cdots+\\left(apexindexkay-1\\right)} \\quad\\left(\\text { since } 2^{maximalindex} \\equiv 2 \\text { for } maximalindex \\geqslant 0\\right) \\\\\n& \\equiv 2^{apexindexone+\\cdots+apexindexkay} .\n\\end{aligned}\n\\]\n\nCase 2. Some \\( apexindexone=0 \\), say \\( apexindexone=0 \\). Now\n\\[\n(1+5 microamount)!=(1+5 microamount)(5 microamount)!,\n\\]\nso \\( wholesumdigit(1+5 microamount) \\equiv(1+5 microamount) wholesumdigit(5 microamount) \\). But \\( wholesumdigit(5 microamount) \\) is even for \\( microamount \\geqslant 1 \\) since \\( (5 microamount)! \\) is divisible by a higher power of 2 than of 5 . But\n\\[\n(1+5 microamount) \\cdot(2 residualvar) \\equiv 2 residualvar,\n\\]\nso \\( wholesumdigit(1+5 microamount) \\equiv wholesumdigit(5 microamount) \\). Letting \\( microamount=5^{apexindextwo-1}+\\cdots+5^{apexindexkay-1} \\), the proof follows by induction.\n\n(b) The least \\( aperiodic \\geqslant 1 \\) for which \\( 2^{difference+aperiodic} \\equiv 2^{2} \\) for all \\( difference \\geqslant 1 \\) is \\( aperiodic=4 \\)."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "f": "hjgrksla",
        "a_1": "nlskdver",
        "a_2": "gjtkwpza",
        "a_k": "wqmxzptb",
        "a_i": "pkstvuqo",
        "k": "vblmsgqf",
        "g": "zfyodmpr",
        "s": "rjsnwevo",
        "p": "dktvhwlo",
        "i": "xspqgzne",
        "j": "cfrdkenu",
        "m": "uytahvrl"
      },
      "question": "Problem A-6. Let \\( qzxwvtnp \\) be a positive integer, and let \\( hjgrksla(qzxwvtnp) \\) denote the last nonzero digit in the decimal expansion of \\( qzxwvtnp! \\). For instance, \\( hjgrksla(5)=2 \\).\n(a) Show that if \\( nlskdver, gjtkwpza, \\ldots, wqmxzptb \\) are distinct nonnegative integers, then \\( hjgrksla\\left(5^{nlskdver}+5^{gjtkwpza}+\\cdots+5^{wqmxzptb}\\right) \\) depends only on the sum \\( nlskdver+gjtkwpza+\\cdots+wqmxzptb \\).\n(b) Assuming part (a), we can define\n\\[\nzfyodmpr(rjsnwevo)=hjgrksla\\left(5^{nlskdver}+5^{gjtkwpza}+\\cdots+5^{wqmxzptb}\\right)\n\\]\nwhere \\( rjsnwevo=nlskdver+gjtkwpza+\\cdots+wqmxzptb \\). Find the least positive integer \\( dktvhwlo \\) for which\n\\[\nzfyodmpr(rjsnwevo)=zfyodmpr(rjsnwevo+dktvhwlo), \\quad \\text { for all } rjsnwevo \\geqslant 1\n\\]\nor else show that no such \\( dktvhwlo \\) exists.",
      "solution": "A-6.\n(a) All congruences are modulo 10 .\n\nLemma. \\( hjgrksla(5 qzxwvtnp) \\equiv 2^{qzxwvtnp} hjgrksla(qzxwvtnp) \\).\nProof. We have\n(*)\n\\[\n(5 qzxwvtnp)!=10^{qzxwvtnp} qzxwvtnp!\\prod_{xspqgzne=0}^{qzxwvtnp-1} \\frac{(5 xspqgzne+1)(5 xspqgzne+2)(5 xspqgzne+3)(5 xspqgzne+4)}{2} .\n\\]\n\nIf \\( xspqgzne \\) is even, then\n\\[\n\\frac{1}{2}(5 xspqgzne+1)(5 xspqgzne+2)(5 xspqgzne+3)(5 xspqgzne+4) \\equiv \\frac{1}{2}(1 \\cdot 2 \\cdot 3 \\cdot 4) \\equiv 2,\n\\]\nand if \\( xspqgzne \\) is odd, then\n\\[\n\\frac{1}{2}(5 xspqgzne+1)(5 xspqgzne+2)(5 xspqgzne+3)(5 xspqgzne+4) \\equiv \\frac{1}{2}(6 \\cdot 7 \\cdot 8 \\cdot 9) \\equiv 2\n\\]\n\nThus the entire product above is congruent to \\( 2^{qzxwvtnp} \\). From (*) it is clear that the largest power of 10 dividing ( \\( 5 qzxwvtnp \\) )! is the same as the largest power of 10 dividing \\( 10^{qzxwvtnp} qzxwvtnp \\) !, and the proof follows.\n\nWe now show by induction on \\( 5^{nlskdver}+\\cdots+5^{wqmxzptb} \\) that\n\\[\nhjgrksla\\left(5^{nlskdver}+\\cdots+5^{wqmxzptb}\\right) \\equiv 2^{nlskdver+\\cdots+wqmxzptb}\n\\]\n(which depends only on \\( nlskdver+\\cdots+wqmxzptb \\) as desired).\nThis is true for \\( 5^{nlskdver}+\\cdots+5^{wqmxzptb}=1 \\), since \\( hjgrksla\\left(5^{0}\\right) \\equiv 2^{0} \\equiv 1 \\).\n\nCase 1. All \\( nlskdver, gjtkwpza, \\ldots, wqmxzptb>0 \\). By the lemma and induction,\n\\[\n\\begin{aligned}\nhjgrksla\\left(5^{nlskdver}+\\cdots+5^{wqmxzptb}\\right) & \\equiv 2^{5^{nlskdver-1}+\\cdots+5^{wqmxzptb-1}} hjgrksla\\left(5^{nlskdver-1}+\\cdots+5^{wqmxzptb-1}\\right) \\\n& \\equiv 2^{vblmsgqf} \\cdot 2^{\\left(nlskdver-1\\right)+\\cdots+\\left(wqmxzptb-1\\right)} \\quad\\left(\\text { since } 2^{5^{xspqgzne}} \\equiv 2 \\text { for } xspqgzne \\geqslant 0\\right) \\\\\n& \\equiv 2^{nlskdver+\\cdots+wqmxzptb} .\n\\end{aligned}\n\\]\n\nCase 2. Some \\( nlskdver=0 \\), say \\( nlskdver=0 \\). Now\n\\[\n(1+5 uytahvrl)!=(1+5 uytahvrl)(5 uytahvrl)!,\n\\]\nso \\( hjgrksla(1+5 uytahvrl) \\equiv(1+5 uytahvrl) hjgrksla(5 uytahvrl) \\). But \\( hjgrksla(5 uytahvrl) \\) is even for \\( uytahvrl \\geqslant 1 \\) since \\( (5 uytahvrl)! \\) is divisible by a higher power of 2 than of 5 . But\n\\[\n(1+5 uytahvrl) \\cdot(2 cfrdkenu) \\equiv 2 cfrdkenu,\n\\]\nso \\( hjgrksla(1+5 uytahvrl) \\equiv hjgrksla(5 uytahvrl) \\). Letting \\( uytahvrl=5^{gjtkwpza-1}+\\cdots+5^{wqmxzptb-1} \\), the proof follows by induction.\n\n(b) The least \\( dktvhwlo \\geqslant 1 \\) for which \\( 2^{rjsnwevo+dktvhwlo} \\equiv 2^{2} \\) for all \\( rjsnwevo \\geqslant 1 \\) is \\( dktvhwlo=4 \\)."
    },
    "kernel_variant": {
      "question": "Fix the modulus \\(5^{\\,3}=125\\).  \nFor every positive integer \\(n\\) set  \n\\[\n\\lambda(n)=\\frac{n!}{5^{\\,v_{5}(n!)}}\\pmod{125},\\qquad\nv_{5}(n!)=\\left\\lfloor\\frac{n}{5}\\right\\rfloor+\n          \\left\\lfloor\\frac{n}{5^{2}}\\right\\rfloor+\n          \\left\\lfloor\\frac{n}{5^{3}}\\right\\rfloor+\\cdots .\n\\]\n\nA non–negative integer is called a \\(0/1\\)-number if every digit in its\nbase–\\(5\\) expansion equals \\(0\\) or \\(1\\).\nFor a finite set \\(A\\subset\\mathbf Z_{\\ge 0}\\) put  \n\\[\n\\begin{aligned}\nN(A)&=\\sum_{a\\in A}5^{a}\\quad(\\text{so }N(A)\\text{ is a }0/1\\text{–number}),\\\\[2pt]\nw(A)&=\\lvert A\\rvert ,\\qquad\ns(A)=\\sum_{a\\in A}a,\\\\[2pt]\nT_{A}(a)&=\\sum_{\\substack{b\\in A\\\\ b>a}}5^{\\,b-a-1}\\qquad (a\\in A),\\\\[2pt]\n\\theta(A)&=\\dfrac{N(A)-w(A)}{4}.\n\\end{aligned}\n\\]\n(The quotient \\(\\theta(A)\\) is an integer since each power \\(5^{a}\\)\nsatisfies \\(5^{a}\\equiv1\\pmod4\\).)\n\n(a)  Prove the following \\(125\\)-adic factorisation\n\\[\n\\boxed{\\;\n      \\lambda\\!\\bigl(N(A)\\bigr)\\equiv\n      24^{\\,\\theta(A)}\n      \\prod_{a\\in A}\\bigl(1+5\\,T_{A}(a)\\bigr)\n      \\pmod{125}\n     \\;}.\n\\]\n\n(b)  Reduce the congruence in part (a) modulo \\(5\\) and show that the\nresidue of \\(\\lambda\\!\\bigl(N(A)\\bigr)\\) modulo \\(5\\) depends only on\n\\(s(A)\\).  Determine the least positive period \\(p\\) of the function\n\\[\ns\\;\\longmapsto\\;\\lambda\\!\\bigl(N(A)\\bigr)\\pmod 5\n\\qquad\\bigl(s=s(A)\\bigr).\n\\]\n\n(c)  Give explicit sets \\(A,B\\) with  \n\\(w(A)=w(B)\\) and \\(s(A)=s(B)\\) but  \n\\[\n\\lambda\\!\\bigl(N(A)\\bigr)\\not\\equiv\\lambda\\!\\bigl(N(B)\\bigr)\\pmod{125},\n\\]\nthus showing that the pair \\((w,s)\\) no longer determines\n\\(\\lambda(n)\\) once one works modulo \\(125\\).\n\n-------------------------------------------------------------",
      "solution": "Throughout, congruences are understood modulo \\(125\\); the\nsymbol \\(v_{5}(\\,\\cdot\\,)\\) denotes the usual \\(5\\)-adic valuation.\n\n1.  A basic \\(5\\)-adic recursion for \\(\\lambda(n)\\)\n\nWrite \\(n=5q+d\\) with \\(0\\le d\\le4\\).  Separating the last block of five\nfactors,\n\\[\nn!=(5q)!\\prod_{j=1}^{d}(5q+j).\n\\]\nInside \\((5q)!\\) split every complete \\(5\\)-block:\n\\[\n(5q)!=5^{\\,q}\\,q!\\prod_{i=1}^{q}(5i-1)(5i-2)(5i-3)(5i-4).\n\\]\nBecause \\(v_{5}\\bigl((5q)!\\bigr)=q+v_{5}(q!)\\), division by\n\\(5^{\\,v_{5}(n!)}\\) yields\n\\[\n\\boxed{\\;\n   \\lambda(5q+d)=\\lambda(q)\\,C(q)\\,D(q,d)\n \\;}\n\\]\nwith\n\\[\nC(q)=\\prod_{i=1}^{q}(5i-1)(5i-2)(5i-3)(5i-4),\\qquad\nD(q,d)=\\prod_{j=1}^{d}(5q+j).\n\\]\n\n2.  Evaluation of \\(C(q)\\) modulo \\(125\\)\n\nLemma 1. For every \\(q\\ge0\\) one has  \n\\[\n\\boxed{\\;C(q)\\equiv24^{\\,q}\\pmod{125}\\;}.\n\\]\n\nProof.  Put \\(P(x)=(x-1)(x-2)(x-3)(x-4)=x^{4}-10x^{3}+35x^{2}-50x+24\\).\nSubstituting \\(x=5i\\) shows \\(P(5i)\\equiv24\\pmod{125}\\); consequently\nevery factor in \\(C(q)\\) is congruent to \\(24\\), and the lemma follows.\n\\(\\square\\)\n\n3.  Iterating the recursion along the base–\\(5\\) digits of\n\\(N(A)\\)\n\nLet  \n\\[\nN_{0}=N(A),\\qquad\nN_{j+1}=\\left\\lfloor\\frac{N_{j}}{5}\\right\\rfloor ,\\qquad\nd_{j}=N_{j}-5N_{j+1}\\in\\{0,1\\}.\n\\]\nBecause the base–\\(5\\) expansion of \\(N(A)\\) uses only the digits\n\\(0,1\\), we have \\(d_{j}=\\mathbf1_{(j\\in A)}\\) and \\(N_{j}=0\\) for\n\\(j\\gg0\\).\n\nUsing the recursion and Lemma 1,\n\\[\n\\lambda(N_{j})\\equiv\n\\lambda(N_{j+1})\\,24^{\\,N_{j+1}}\\,(5N_{j+1}+1)^{\\,d_{j}}\n\\pmod{125}.\n\\]\nIterating until \\(N_{m}=0\\) gives\n\\[\n\\lambda(N(A))\\equiv\n24^{\\,S}\\,\n\\prod_{j\\ge0}(5N_{j+1}+1)^{\\,d_{j}}\n\\pmod{125},\n\\qquad\nS=\\sum_{j\\ge0}N_{j+1}.\n\\]\n\n4.  Evaluation of the exponent \\(S\\)\n\nBecause \\(N_{j}=\\sum_{k\\ge j}d_{k}5^{\\,k-j}\\),\n\\[\nS=\\sum_{j\\ge0}\\;\\sum_{k\\ge j+1}d_{k}5^{\\,k-j-1}\n   =\\sum_{k\\ge1}d_{k}\\frac{5^{\\,k}-1}{4}\n   =\\frac{1}{4}\\Bigl(N(A)-d_{0}-\\bigl(w(A)-d_{0}\\bigr)\\Bigr)\n   =\\frac{N(A)-w(A)}{4}.\n\\]\nHence \\(S=\\theta(A)\\).\n\n5.  Identification of the factors \\((5N_{j+1}+1)\\)\n\nFor an index \\(a\\) with \\(d_{a}=1\\) we have\n\\[\nN_{a+1}=\\sum_{\\substack{b\\in A\\\\ b>a}}5^{\\,b-a-1}=T_{A}(a),\n\\]\nso \\(5N_{a+1}+1=1+5T_{A}(a)\\).  Splitting the product over those indices\nyields\n\\[\n\\boxed{\\;\n      \\lambda\\!\\bigl(N(A)\\bigr)\\equiv\n      24^{\\,\\theta(A)}\n      \\prod_{a\\in A}\\bigl(1+5\\,T_{A}(a)\\bigr)\n      \\pmod{125}\n     \\;},\n\\]\nestablishing part (a).\n\n6.  Consequences modulo \\(5\\)\n\nReducing the boxed formula modulo \\(5\\) gives  \n\\[\n\\lambda\\!\\bigl(N(A)\\bigr)\\equiv(-1)^{\\,\\theta(A)}\\pmod5\n\\]\nbecause \\(24\\equiv-1\\pmod5\\) and \\(1+5T_{A}(a)\\equiv1\\pmod5\\).\n\nParity connection.  For \\(a\\ge1\\)\n\\[\n\\frac{5^{\\,a}-1}{4}\\equiv\n\\begin{cases}\n0\\pmod2,& a\\text{ even},\\\\\n1\\pmod2,& a\\text{ odd},\n\\end{cases}\n\\]\nso\n\\[\n\\theta(A)\\equiv\\sum_{a\\in A}a\\equiv s(A)\\pmod2.\n\\]\nTherefore\n\\[\n\\boxed{\\;\n   \\lambda\\!\\bigl(N(A)\\bigr)\\equiv(-1)^{\\,s(A)}\\pmod5\n \\;}.\n\\]\n\nSince \\((-1)^{\\,s+2}=(-1)^{\\,s}\\), the least positive period of the map\n\\(s\\mapsto\\lambda\\!\\bigl(N(A)\\bigr)\\pmod5\\) equals  \n\\[\n\\boxed{\\,p=2\\,}.\n\\]\n\n7.  Non–uniqueness modulo \\(125\\)\n\nChoose  \n\\[\nA=\\{0,3\\},\\qquad B=\\{1,2\\}.\n\\]\nBoth sets satisfy  \n\\[\nw(A)=w(B)=2,\\qquad s(A)=s(B)=3,\n\\]\nyet they behave differently modulo \\(125\\).\n\nSet \\(N_{A}=N(A)=1+125=126\\) and \\(N_{B}=N(B)=5+25=30\\).\n\nFirst set \\(A\\).\n\\[\n\\theta(A)=\\frac{126-2}{4}=31,\\quad\n\\prod_{a\\in A}\\bigl(1+5T_{A}(a)\\bigr)\\equiv1,\n\\]\nwhence\n\\[\n\\lambda(126)\\equiv24^{31}\\equiv24\\pmod{125}.\n\\]\n\nSecond set \\(B\\).\n\\[\n\\theta(B)=\\frac{30-2}{4}=7,\\quad\n\\prod_{a\\in B}\\bigl(1+5T_{B}(a)\\bigr)\\equiv(1+5\\cdot1)=6,\n\\]\nso\n\\[\n\\lambda(30)\\equiv24^{7}\\cdot6\\equiv49\\cdot6\\equiv44\\pmod{125}.\n\\]\n\nThus  \n\\[\n\\boxed{\\;\n  \\lambda\\!\\bigl(N(A)\\bigr)=24\\not\\equiv44=\\lambda\\!\\bigl(N(B)\\bigr)\\pmod{125}\n \\;},\n\\]\nalthough \\(w(A)=w(B)\\) and \\(s(A)=s(B)\\).  Part (c) is proved.\n\n-------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.676914",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher modulus.  \nThe original problem works modulo 10; the variant demands work modulo 10^r for arbitrary r.  This forces simultaneous control of congruences mod 2^r and mod 5^r and introduces primitive-root arguments, Euler totients, and Chinese-Remainder synthesis that were absent before.\n\n2.  Explicit 10-adic factorisation.  \nTo establish the key lemma ℓ_r(5n)=12^{n}ℓ_r(n) one must handle the entire block  \n(5j+1)(5j+2)(5j+3)(5j+4)/2  modulo 10^r, rather than merely modulo 10; keeping track of the 2-adic valuation inside this product is considerably subtler.\n\n3.  Two-parameter dependence.  \nUnlike the original problem (where the answer depends solely on s), here the answer depends on both s and the cardinality k.  Proving minimality (part (c)) requires a delicate contradiction using the coprimality of C_k with 5 and shows that no purely one-dimensional period exists.\n\n4.  Period computation for all r.  \nPinpointing p_r relies on the fact that 2 is a primitive root modulo 5^r, an advanced number-theoretic fact; verifying minimality requires combining this with 2-adic considerations.\n\nOverall, the enhanced variant compels the solver to blend 10-adic lifting, primitive-root theory, valuations, and the Chinese Remainder Theorem—elements far beyond the scope of the original problem, and impossible to dispatch by simple pattern-matching."
      }
    },
    "original_kernel_variant": {
      "question": "Fix the modulus \\(5^{\\,3}=125\\).  \nFor every positive integer \\(n\\) set  \n\\[\n\\lambda(n)=\\frac{n!}{5^{\\,v_{5}(n!)}}\\pmod{125},\\qquad\nv_{5}(n!)=\\left\\lfloor\\frac{n}{5}\\right\\rfloor+\n          \\left\\lfloor\\frac{n}{5^{2}}\\right\\rfloor+\n          \\left\\lfloor\\frac{n}{5^{3}}\\right\\rfloor+\\cdots .\n\\]\n\nA non–negative integer is called a \\(0/1\\)-number if every digit in its\nbase–\\(5\\) expansion equals \\(0\\) or \\(1\\).\nFor a finite set \\(A\\subset\\mathbf Z_{\\ge 0}\\) put  \n\\[\n\\begin{aligned}\nN(A)&=\\sum_{a\\in A}5^{a}\\quad(\\text{so }N(A)\\text{ is a }0/1\\text{–number}),\\\\[2pt]\nw(A)&=\\lvert A\\rvert ,\\qquad\ns(A)=\\sum_{a\\in A}a,\\\\[2pt]\nT_{A}(a)&=\\sum_{\\substack{b\\in A\\\\ b>a}}5^{\\,b-a-1}\\qquad (a\\in A),\\\\[2pt]\n\\theta(A)&=\\dfrac{N(A)-w(A)}{4}.\n\\end{aligned}\n\\]\n(The quotient \\(\\theta(A)\\) is an integer since each power \\(5^{a}\\)\nsatisfies \\(5^{a}\\equiv1\\pmod4\\).)\n\n(a)  Prove the following \\(125\\)-adic factorisation\n\\[\n\\boxed{\\;\n      \\lambda\\!\\bigl(N(A)\\bigr)\\equiv\n      24^{\\,\\theta(A)}\n      \\prod_{a\\in A}\\bigl(1+5\\,T_{A}(a)\\bigr)\n      \\pmod{125}\n     \\;}.\n\\]\n\n(b)  Reduce the congruence in part (a) modulo \\(5\\) and show that the\nresidue of \\(\\lambda\\!\\bigl(N(A)\\bigr)\\) modulo \\(5\\) depends only on\n\\(s(A)\\).  Determine the least positive period \\(p\\) of the function\n\\[\ns\\;\\longmapsto\\;\\lambda\\!\\bigl(N(A)\\bigr)\\pmod 5\n\\qquad\\bigl(s=s(A)\\bigr).\n\\]\n\n(c)  Give explicit sets \\(A,B\\) with  \n\\(w(A)=w(B)\\) and \\(s(A)=s(B)\\) but  \n\\[\n\\lambda\\!\\bigl(N(A)\\bigr)\\not\\equiv\\lambda\\!\\bigl(N(B)\\bigr)\\pmod{125},\n\\]\nthus showing that the pair \\((w,s)\\) no longer determines\n\\(\\lambda(n)\\) once one works modulo \\(125\\).\n\n-------------------------------------------------------------",
      "solution": "Throughout, congruences are understood modulo \\(125\\); the\nsymbol \\(v_{5}(\\,\\cdot\\,)\\) denotes the usual \\(5\\)-adic valuation.\n\n1.  A basic \\(5\\)-adic recursion for \\(\\lambda(n)\\)\n\nWrite \\(n=5q+d\\) with \\(0\\le d\\le4\\).  Separating the last block of five\nfactors,\n\\[\nn!=(5q)!\\prod_{j=1}^{d}(5q+j).\n\\]\nInside \\((5q)!\\) split every complete \\(5\\)-block:\n\\[\n(5q)!=5^{\\,q}\\,q!\\prod_{i=1}^{q}(5i-1)(5i-2)(5i-3)(5i-4).\n\\]\nBecause \\(v_{5}\\bigl((5q)!\\bigr)=q+v_{5}(q!)\\), division by\n\\(5^{\\,v_{5}(n!)}\\) yields\n\\[\n\\boxed{\\;\n   \\lambda(5q+d)=\\lambda(q)\\,C(q)\\,D(q,d)\n \\;}\n\\]\nwith\n\\[\nC(q)=\\prod_{i=1}^{q}(5i-1)(5i-2)(5i-3)(5i-4),\\qquad\nD(q,d)=\\prod_{j=1}^{d}(5q+j).\n\\]\n\n2.  Evaluation of \\(C(q)\\) modulo \\(125\\)\n\nLemma 1. For every \\(q\\ge0\\) one has  \n\\[\n\\boxed{\\;C(q)\\equiv24^{\\,q}\\pmod{125}\\;}.\n\\]\n\nProof.  Put \\(P(x)=(x-1)(x-2)(x-3)(x-4)=x^{4}-10x^{3}+35x^{2}-50x+24\\).\nSubstituting \\(x=5i\\) shows \\(P(5i)\\equiv24\\pmod{125}\\); consequently\nevery factor in \\(C(q)\\) is congruent to \\(24\\), and the lemma follows.\n\\(\\square\\)\n\n3.  Iterating the recursion along the base–\\(5\\) digits of\n\\(N(A)\\)\n\nLet  \n\\[\nN_{0}=N(A),\\qquad\nN_{j+1}=\\left\\lfloor\\frac{N_{j}}{5}\\right\\rfloor ,\\qquad\nd_{j}=N_{j}-5N_{j+1}\\in\\{0,1\\}.\n\\]\nBecause the base–\\(5\\) expansion of \\(N(A)\\) uses only the digits\n\\(0,1\\), we have \\(d_{j}=\\mathbf1_{(j\\in A)}\\) and \\(N_{j}=0\\) for\n\\(j\\gg0\\).\n\nUsing the recursion and Lemma 1,\n\\[\n\\lambda(N_{j})\\equiv\n\\lambda(N_{j+1})\\,24^{\\,N_{j+1}}\\,(5N_{j+1}+1)^{\\,d_{j}}\n\\pmod{125}.\n\\]\nIterating until \\(N_{m}=0\\) gives\n\\[\n\\lambda(N(A))\\equiv\n24^{\\,S}\\,\n\\prod_{j\\ge0}(5N_{j+1}+1)^{\\,d_{j}}\n\\pmod{125},\n\\qquad\nS=\\sum_{j\\ge0}N_{j+1}.\n\\]\n\n4.  Evaluation of the exponent \\(S\\)\n\nBecause \\(N_{j}=\\sum_{k\\ge j}d_{k}5^{\\,k-j}\\),\n\\[\nS=\\sum_{j\\ge0}\\;\\sum_{k\\ge j+1}d_{k}5^{\\,k-j-1}\n   =\\sum_{k\\ge1}d_{k}\\frac{5^{\\,k}-1}{4}\n   =\\frac{1}{4}\\Bigl(N(A)-d_{0}-\\bigl(w(A)-d_{0}\\bigr)\\Bigr)\n   =\\frac{N(A)-w(A)}{4}.\n\\]\nHence \\(S=\\theta(A)\\).\n\n5.  Identification of the factors \\((5N_{j+1}+1)\\)\n\nFor an index \\(a\\) with \\(d_{a}=1\\) we have\n\\[\nN_{a+1}=\\sum_{\\substack{b\\in A\\\\ b>a}}5^{\\,b-a-1}=T_{A}(a),\n\\]\nso \\(5N_{a+1}+1=1+5T_{A}(a)\\).  Splitting the product over those indices\nyields\n\\[\n\\boxed{\\;\n      \\lambda\\!\\bigl(N(A)\\bigr)\\equiv\n      24^{\\,\\theta(A)}\n      \\prod_{a\\in A}\\bigl(1+5\\,T_{A}(a)\\bigr)\n      \\pmod{125}\n     \\;},\n\\]\nestablishing part (a).\n\n6.  Consequences modulo \\(5\\)\n\nReducing the boxed formula modulo \\(5\\) gives  \n\\[\n\\lambda\\!\\bigl(N(A)\\bigr)\\equiv(-1)^{\\,\\theta(A)}\\pmod5\n\\]\nbecause \\(24\\equiv-1\\pmod5\\) and \\(1+5T_{A}(a)\\equiv1\\pmod5\\).\n\nParity connection.  For \\(a\\ge1\\)\n\\[\n\\frac{5^{\\,a}-1}{4}\\equiv\n\\begin{cases}\n0\\pmod2,& a\\text{ even},\\\\\n1\\pmod2,& a\\text{ odd},\n\\end{cases}\n\\]\nso\n\\[\n\\theta(A)\\equiv\\sum_{a\\in A}a\\equiv s(A)\\pmod2.\n\\]\nTherefore\n\\[\n\\boxed{\\;\n   \\lambda\\!\\bigl(N(A)\\bigr)\\equiv(-1)^{\\,s(A)}\\pmod5\n \\;}.\n\\]\n\nSince \\((-1)^{\\,s+2}=(-1)^{\\,s}\\), the least positive period of the map\n\\(s\\mapsto\\lambda\\!\\bigl(N(A)\\bigr)\\pmod5\\) equals  \n\\[\n\\boxed{\\,p=2\\,}.\n\\]\n\n7.  Non–uniqueness modulo \\(125\\)\n\nChoose  \n\\[\nA=\\{0,3\\},\\qquad B=\\{1,2\\}.\n\\]\nBoth sets satisfy  \n\\[\nw(A)=w(B)=2,\\qquad s(A)=s(B)=3,\n\\]\nyet they behave differently modulo \\(125\\).\n\nSet \\(N_{A}=N(A)=1+125=126\\) and \\(N_{B}=N(B)=5+25=30\\).\n\nFirst set \\(A\\).\n\\[\n\\theta(A)=\\frac{126-2}{4}=31,\\quad\n\\prod_{a\\in A}\\bigl(1+5T_{A}(a)\\bigr)\\equiv1,\n\\]\nwhence\n\\[\n\\lambda(126)\\equiv24^{31}\\equiv24\\pmod{125}.\n\\]\n\nSecond set \\(B\\).\n\\[\n\\theta(B)=\\frac{30-2}{4}=7,\\quad\n\\prod_{a\\in B}\\bigl(1+5T_{B}(a)\\bigr)\\equiv(1+5\\cdot1)=6,\n\\]\nso\n\\[\n\\lambda(30)\\equiv24^{7}\\cdot6\\equiv49\\cdot6\\equiv44\\pmod{125}.\n\\]\n\nThus  \n\\[\n\\boxed{\\;\n  \\lambda\\!\\bigl(N(A)\\bigr)=24\\not\\equiv44=\\lambda\\!\\bigl(N(B)\\bigr)\\pmod{125}\n \\;},\n\\]\nalthough \\(w(A)=w(B)\\) and \\(s(A)=s(B)\\).  Part (c) is proved.\n\n-------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.530546",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher modulus.  \nThe original problem works modulo 10; the variant demands work modulo 10^r for arbitrary r.  This forces simultaneous control of congruences mod 2^r and mod 5^r and introduces primitive-root arguments, Euler totients, and Chinese-Remainder synthesis that were absent before.\n\n2.  Explicit 10-adic factorisation.  \nTo establish the key lemma ℓ_r(5n)=12^{n}ℓ_r(n) one must handle the entire block  \n(5j+1)(5j+2)(5j+3)(5j+4)/2  modulo 10^r, rather than merely modulo 10; keeping track of the 2-adic valuation inside this product is considerably subtler.\n\n3.  Two-parameter dependence.  \nUnlike the original problem (where the answer depends solely on s), here the answer depends on both s and the cardinality k.  Proving minimality (part (c)) requires a delicate contradiction using the coprimality of C_k with 5 and shows that no purely one-dimensional period exists.\n\n4.  Period computation for all r.  \nPinpointing p_r relies on the fact that 2 is a primitive root modulo 5^r, an advanced number-theoretic fact; verifying minimality requires combining this with 2-adic considerations.\n\nOverall, the enhanced variant compels the solver to blend 10-adic lifting, primitive-root theory, valuations, and the Chinese Remainder Theorem—elements far beyond the scope of the original problem, and impossible to dispatch by simple pattern-matching."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}