{
  "index": "1985-A-4",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Define a sequence $\\{a_i\\}$ by $a_1=3$ and $a_{i+1}=3^{a_i}$ for\n$i\\geq 1$. Which integers between 00 and 99 inclusive occur as the\nlast two digits in the decimal expansion of infinitely many $a_i$?",
  "solution": "Solution. Let \\( \\phi(n) \\) denote the Euler \\( \\phi \\)-function, which equals the number of integers between 1 and \\( n \\) inclusive that are relatively prime to \\( n \\), or more abstractly, the order of the multiplicative group \\( (\\mathbb{Z} / n \\mathbb{Z})^{*} \\). If the prime factorization of \\( n \\) is \\( p_{1}^{e_{1}} \\cdots p_{k}^{e_{k}} \\), then \\( \\phi(n) \\) can be computed by the formula\n\\[\n\\phi(n)=\\prod_{i=1}^{k} \\phi\\left(p_{i}^{e_{i}}\\right)=\\prod_{i=1}^{k} p_{i}^{e_{i}-1}\\left(p_{i}-1\\right) .\n\\]\n\nEuler's Theorem [Lar1, p. 148], which is Lagrange's Theorem (the order of an element of a finite group divides the order of the group, [Lar1, p. 147]) applied to the group \\( (\\mathbb{Z} / n \\mathbb{Z})^{*} \\), states that \\( a^{\\phi(n)} \\equiv 1(\\bmod n) \\) whenever \\( \\operatorname{gcd}(a, n)=1 \\).\n\nIt shows that \\( a_{i}=3^{a_{i-1}} \\) modulo 100 is determined by \\( a_{i-1} \\) modulo \\( \\phi(100)=40 \\), for \\( i \\geq 2 \\). Similarly \\( a_{i-1} \\bmod 40 \\) is determined by \\( a_{i-2} \\bmod 16 \\), which is determined by \\( a_{i-3} \\bmod 8 \\), for \\( i \\geq 4 \\). Finally \\( a_{i-3} \\bmod 8 \\) is determined by \\( a_{i-4} \\bmod 2 \\) for \\( i \\geq 5 \\), since \\( 3^{2} \\equiv 1(\\bmod 8) \\). For \\( i \\geq 5, a_{i-4} \\) is odd, so\n\\[\n\\begin{array}{l}\na_{i-3}=3^{a_{i-4}} \\equiv 3^{1} \\equiv 3 \\quad(\\bmod 8) \\\\\na_{i-2}=3^{a_{i-3}} \\equiv 3^{3} \\equiv 11 \\quad(\\bmod 16) \\\\\na_{i-1}=3^{a_{i-2}} \\equiv 3^{11} \\equiv 27 \\quad(\\bmod 40) \\\\\na_{i}=3^{a_{i-1}} \\equiv 3^{27} \\equiv 87 \\quad(\\bmod 100)\n\\end{array}\n\\]\n\nThus the only integer that appears as the last two digits of infinitely many \\( a_{i} \\) is 87 .\nRemark. Carmichael's lambda function. The exponent in Euler's Theorem is not always best possible. The function \\( \\lambda(n) \\) giving the best possible exponent for \\( n \\), i.e., the least positive integer \\( \\lambda \\) such that \\( a^{\\lambda} \\equiv 1(\\bmod n) \\) whenever \\( \\operatorname{gcd}(a, n)=1 \\), is known as Carmichael's lambda function or as the reduced totient function. If \\( n=p_{1}^{e_{1}} \\cdots p_{k}^{e_{k}} \\), then\n\\[\n\\lambda(n)=\\operatorname{lcm}\\left(\\lambda\\left(p_{1}^{e_{1}}\\right), \\ldots, \\lambda\\left(p_{k}^{e_{k}}\\right)\\right),\n\\]\nwhere\n\\[\n\\lambda\\left(p^{e}\\right)=\\phi\\left(p^{e}\\right)=p^{e-1}(p-1),\n\\]\nunless \\( p=2 \\) and \\( e \\geq 3 \\) in which case \\( \\lambda\\left(2^{e}\\right)=2^{e-2} \\) instead of \\( 2^{e-1} \\). For more information, see [Ros1, Section 9.6].\n\nOne can simplify the computations in the above solution by using \\( \\lambda(n) \\) in place of \\( \\phi(n) \\) : iteration of \\( \\lambda \\) maps 100 to 20 to 4 to 2 , so starting from \\( a_{i-3} \\equiv 1(\\bmod 2) \\) we obtain\n\\[\n\\begin{array}{lll}\na_{i-2} & =3^{a_{i-3}} \\equiv 3^{1} \\equiv 3 & \\\\\na_{i-1} & =3^{a_{i-2}} \\equiv 3^{3} \\equiv 7 & \\\\\n(\\bmod 4) \\\\\na_{i} & =3^{a_{i-1}} \\equiv 3^{7} \\equiv 87 & \\\\\n(\\bmod 100)\n\\end{array}\n\\]\n\nStronger result. More generally, one can show that for any integer \\( c \\geq 1 \\), the sequence defined by \\( a_{1}=c \\) and \\( a_{i+1}=c^{a_{i}} \\) for \\( i \\geq 1 \\) is eventually constant when reduced modulo a positive integer \\( n \\). To prove this, one can use the Chinese Remainder Theorem to reduce to the case where \\( n \\) is a power of a prime \\( p \\). If \\( p \\) divides \\( c \\), then the sequence is eventually 0 modulo \\( n \\); otherwise we can use Euler's Theorem and strong induction on \\( n \\), since \\( \\phi(n)<n \\) for \\( n>1 \\). See 1997B5 for a related problem, and further discussion.",
  "vars": [
    "a",
    "a_1",
    "a_i",
    "a_i+1",
    "a_i-1",
    "a_i-2",
    "a_i-3",
    "a_i-4",
    "i"
  ],
  "params": [
    "n",
    "p",
    "p_1",
    "p_i",
    "p_k",
    "e",
    "e_1",
    "e_i",
    "c",
    "\\\\phi",
    "\\\\lambda"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "seqterm",
        "a_1": "initterm",
        "a_i": "geniterm",
        "a_i+1": "nextterm",
        "a_i-1": "prevterm1",
        "a_i-2": "prevterm2",
        "a_i-3": "prevterm3",
        "a_i-4": "prevterm4",
        "i": "indexvar",
        "n": "modulus",
        "p": "primevar",
        "p_1": "primeone",
        "p_i": "primeith",
        "p_k": "primekth",
        "e": "exppower",
        "e_1": "expone",
        "e_i": "expith",
        "c": "baseconst",
        "\\phi": "phifunc",
        "\\lambda": "lambdafunc"
      },
      "question": "Define a sequence $\\{geniterm\\}$ by $initterm=3$ and $nextterm=3^{geniterm}$ for\n$indexvar\\geq 1$. Which integers between 00 and 99 inclusive occur as the\nlast two digits in the decimal expansion of infinitely many $geniterm$?",
      "solution": "Solution. Let \\( phifunc(modulus) \\) denote the Euler \\( phifunc \\)-function, which equals the number of integers between 1 and \\( modulus \\) inclusive that are relatively prime to \\( modulus \\), or more abstractly, the order of the multiplicative group \\( (\\mathbb{Z} / modulus \\mathbb{Z})^{*} \\). If the prime factorization of \\( modulus \\) is \\( primeone^{expone} \\cdots primekth^{e_{k}} \\), then \\( phifunc(modulus) \\) can be computed by the formula\n\\[\nphifunc(modulus)=\\prod_{indexvar=1}^{k} phifunc\\left(primeith^{expith}\\right)=\\prod_{indexvar=1}^{k} primeith^{expith-1}\\left(primeith-1\\right) .\n\\]\n\nEuler's Theorem [Lar1, p. 148], which is Lagrange's Theorem (the order of an element of a finite group divides the order of the group, [Lar1, p. 147]) applied to the group \\( (\\mathbb{Z} / modulus \\mathbb{Z})^{*} \\), states that \\( seqterm^{phifunc(modulus)} \\equiv 1(\\bmod modulus) \\) whenever \\( \\operatorname{gcd}(seqterm, modulus)=1 \\).\n\nIt shows that \\( geniterm=3^{prevterm1} \\) modulo 100 is determined by \\( prevterm1 \\) modulo \\( phifunc(100)=40 \\), for \\( indexvar \\geq 2 \\). Similarly \\( prevterm1 \\bmod 40 \\) is determined by \\( prevterm2 \\bmod 16 \\), which is determined by \\( prevterm3 \\bmod 8 \\), for \\( indexvar \\geq 4 \\). Finally \\( prevterm3 \\bmod 8 \\) is determined by \\( prevterm4 \\bmod 2 \\) for \\( indexvar \\geq 5 \\), since \\( 3^{2} \\equiv 1(\\bmod 8) \\). For \\( indexvar \\geq 5, prevterm4 \\) is odd, so\n\\[\n\\begin{array}{l}\nprevterm3=3^{prevterm4} \\equiv 3^{1} \\equiv 3 \\quad(\\bmod 8) \\\\\nprevterm2=3^{prevterm3} \\equiv 3^{3} \\equiv 11 \\quad(\\bmod 16) \\\\\nprevterm1=3^{prevterm2} \\equiv 3^{11} \\equiv 27 \\quad(\\bmod 40) \\\\\ngeniterm=3^{prevterm1} \\equiv 3^{27} \\equiv 87 \\quad(\\bmod 100)\n\\end{array}\n\\]\n\nThus the only integer that appears as the last two digits of infinitely many \\( geniterm \\) is 87 .\n\nRemark. Carmichael's lambda function. The exponent in Euler's Theorem is not always best possible. The function \\( lambdafunc(modulus) \\) giving the best possible exponent for \\( modulus \\), i.e., the least positive integer \\( lambdafunc \\) such that \\( seqterm^{lambdafunc} \\equiv 1(\\bmod modulus) \\) whenever \\( \\operatorname{gcd}(seqterm, modulus)=1 \\), is known as Carmichael's lambda function or as the reduced totient function. If \\( modulus=primeone^{expone} \\cdots primekth^{e_{k}} \\), then\n\\[\nlambdafunc(modulus)=\\operatorname{lcm}\\left(lambdafunc\\left(primeone^{expone}\\right), \\ldots, lambdafunc\\left(primekth^{e_{k}}\\right)\\right),\n\\]\nwhere\n\\[\nlambdafunc\\left(primevar^{exppower}\\right)=phifunc\\left(primevar^{exppower}\\right)=primevar^{exppower-1}(primevar-1),\n\\]\nunless \\( primevar=2 \\) and \\( exppower \\geq 3 \\) in which case \\( lambdafunc\\left(2^{exppower}\\right)=2^{exppower-2} \\) instead of \\( 2^{exppower-1} \\). For more information, see [Ros1, Section 9.6].\n\nOne can simplify the computations in the above solution by using \\( lambdafunc(modulus) \\) in place of \\( phifunc(modulus) \\) : iteration of \\( lambdafunc \\) maps 100 to 20 to 4 to 2 , so starting from \\( prevterm3 \\equiv 1(\\bmod 2) \\) we obtain\n\\[\n\\begin{array}{lll}\nprevterm2 & =3^{prevterm3} \\equiv 3^{1} \\equiv 3 & \\\\\nprevterm1 & =3^{prevterm2} \\equiv 3^{3} \\equiv 7 & \\\\\n(\\bmod 4) \\\\\ngeniterm & =3^{prevterm1} \\equiv 3^{7} \\equiv 87 & \\\\\n(\\bmod 100)\n\\end{array}\n\\]\n\nStronger result. More generally, one can show that for any integer \\( baseconst \\geq 1 \\), the sequence defined by \\( initterm=baseconst \\) and \\( nextterm=baseconst^{geniterm} \\) for \\( indexvar \\geq 1 \\) is eventually constant when reduced modulo a positive integer \\( modulus \\). To prove this, one can use the Chinese Remainder Theorem to reduce to the case where \\( modulus \\) is a power of a prime \\( primevar \\). If \\( primevar \\) divides \\( baseconst \\), then the sequence is eventually 0 modulo \\( modulus \\); otherwise we can use Euler's Theorem and strong induction on \\( modulus \\), since \\( phifunc(modulus)<modulus \\) for \\( modulus>1 \\). See 1997B5 for a related problem, and further discussion."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "marshland",
        "a_1": "butterfly",
        "a_i": "blacksmith",
        "a_i+1": "dragonfly",
        "a_i-1": "portioning",
        "a_i-2": "tangerine",
        "a_i-3": "lighthouse",
        "a_i-4": "watermelon",
        "n": "civilward",
        "p": "paperclips",
        "p_1": "marzipans",
        "p_i": "snowflake",
        "p_k": "backpacks",
        "e_1": "toothpaste",
        "e_i": "creamcandy",
        "c": "playground",
        "\\phi": "honeycomb",
        "\\lambda": "seashores"
      },
      "question": "Define a sequence $\\{blacksmith\\}$ by $butterfly=3$ and $dragonfly=3^{blacksmith}$ for\n$i\\geq 1$. Which integers between 00 and 99 inclusive occur as the\nlast two digits in the decimal expansion of infinitely many $blacksmith$?",
      "solution": "Solution. Let \\( honeycomb(civilward) \\) denote the Euler \\( honeycomb \\)-function, which equals the number of integers between 1 and \\( civilward \\) inclusive that are relatively prime to \\( civilward \\), or more abstractly, the order of the multiplicative group \\( (\\mathbb{Z} / civilward \\mathbb{Z})^{*} \\). If the prime factorization of \\( civilward \\) is \\( marzipans^{toothpaste} \\cdots backpacks^{e_{k}} \\), then \\( honeycomb(civilward) \\) can be computed by the formula\n\\[\nhoneycomb(civilward)=\\prod_{i=1}^{k} honeycomb\\left(snowflake^{creamcandy}\\right)=\\prod_{i=1}^{k} snowflake^{creamcandy-1}\\left(snowflake-1\\right) .\n\\]\n\nEuler's Theorem [Lar1, p. 148], which is Lagrange's Theorem (the order of an element of a finite group divides the order of the group, [Lar1, p. 147]) applied to the group \\( (\\mathbb{Z} / civilward \\mathbb{Z})^{*} \\), states that \\( marshland^{honeycomb(civilward)} \\equiv 1(\\bmod civilward) \\) whenever \\( \\operatorname{gcd}(marshland, civilward)=1 \\).\n\nIt shows that \\( blacksmith=3^{portioning} \\) modulo 100 is determined by \\( portioning \\) modulo \\( honeycomb(100)=40 \\), for \\( i \\geq 2 \\). Similarly \\( portioning \\bmod 40 \\) is determined by \\( tangerine \\bmod 16 \\), which is determined by \\( lighthouse \\bmod 8 \\), for \\( i \\geq 4 \\). Finally \\( lighthouse \\bmod 8 \\) is determined by \\( watermelon \\bmod 2 \\) for \\( i \\geq 5 \\), since \\( 3^{2} \\equiv 1(\\bmod 8) \\). For \\( i \\geq 5, watermelon \\) is odd, so\n\\[\n\\begin{array}{l}\nlighthouse=3^{watermelon} \\equiv 3^{1} \\equiv 3 \\quad(\\bmod 8) \\\\\ntangerine=3^{lighthouse} \\equiv 3^{3} \\equiv 11 \\quad(\\bmod 16) \\\\\nportioning=3^{tangerine} \\equiv 3^{11} \\equiv 27 \\quad(\\bmod 40) \\\\\nblacksmith=3^{portioning} \\equiv 3^{27} \\equiv 87 \\quad(\\bmod 100)\n\\end{array}\n\\]\n\nThus the only integer that appears as the last two digits of infinitely many blacksmith is 87 .\n\nRemark. Carmichael's lambda function. The exponent in Euler's Theorem is not always best possible. The function \\( seashores(civilward) \\) giving the best possible exponent for \\( civilward \\), i.e., the least positive integer \\( seashores \\) such that \\( marshland^{seashores} \\equiv 1(\\bmod civilward) \\) whenever \\( \\operatorname{gcd}(marshland, civilward)=1 \\), is known as Carmichael's lambda function or as the reduced totient function. If \\( civilward=marzipans^{toothpaste} \\cdots backpacks^{e_{k}} \\), then\n\\[\nseashores(civilward)=\\operatorname{lcm}\\left(seashores\\left(marzipans^{toothpaste}\\right), \\ldots, seashores\\left(backpacks^{e_{k}}\\right)\\right),\n\\]\nwhere\n\\[\nseashores\\left(paperclips^{e}\\right)=honeycomb\\left(paperclips^{e}\\right)=paperclips^{e-1}(paperclips-1),\n\\]\nunless \\( paperclips=2 \\) and \\( e \\geq 3 \\) in which case \\( seashores\\left(2^{e}\\right)=2^{e-2} \\) instead of \\( 2^{e-1} \\). For more information, see [Ros1, Section 9.6].\n\nOne can simplify the computations in the above solution by using \\( seashores(civilward) \\) in place of \\( honeycomb(civilward) \\) : iteration of \\( seashores \\) maps 100 to 20 to 4 to 2 , so starting from \\( lighthouse \\equiv 1(\\bmod 2) \\) we obtain\n\\[\n\\begin{array}{lll}\ntangerine & =3^{lighthouse} \\equiv 3^{1} \\equiv 3 & \\\\\nportioning & =3^{tangerine} \\equiv 3^{3} \\equiv 7 & \\\\\n(\\bmod 4) \\\\\nblacksmith & =3^{portioning} \\equiv 3^{7} \\equiv 87 & \\\\\n(\\bmod 100)\n\\end{array}\n\\]\n\nStronger result. More generally, one can show that for any integer \\( playground \\geq 1 \\), the sequence defined by \\( butterfly=playground \\) and \\( dragonfly=playground^{blacksmith} \\) for \\( i \\geq 1 \\) is eventually constant when reduced modulo a positive integer \\( civilward \\). To prove this, one can use the Chinese Remainder Theorem to reduce to the case where \\( civilward \\) is a power of a prime paperclips. If paperclips divides playground, then the sequence is eventually 0 modulo civilward; otherwise we can use Euler's Theorem and strong induction on civilward, since \\( honeycomb(civilward)<civilward \\) for \\( civilward>1 \\). See 1997B5 for a related problem, and further discussion."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "constant",
        "a_1": "lastindex",
        "a_i": "fixedterm",
        "a_i+1": "previousterm",
        "a_i-1": "futureterm",
        "a_i-2": "laterterm",
        "a_i-3": "upcomingterm",
        "a_i-4": "forthcoming",
        "i": "sizevalue",
        "n": "infinitude",
        "p": "composite",
        "p_1": "lastcomposite",
        "p_i": "genericcomposite",
        "p_k": "ultimatecomposite",
        "e": "rootvalue",
        "e_1": "firstratio",
        "e_i": "indexratio",
        "c": "variablebase",
        "\\\\phi": "antitotient",
        "\\\\lambda": "antilambda"
      },
      "question": "Define a sequence $\\{fixedterm\\}$ by $lastindex=3$ and $previousterm=3^{fixedterm}$ for\n$sizevalue\\geq 1$. Which integers between 00 and 99 inclusive occur as the\nlast two digits in the decimal expansion of infinitely many $fixedterm$?",
      "solution": "Solution. Let $ antitotient(infinitude) $ denote the Euler $ antitotient $-function, which equals the number of integers between 1 and $ infinitude $ inclusive that are relatively prime to $ infinitude $, or more abstractly, the order of the multiplicative group $(\\mathbb{Z} / infinitude \\mathbb{Z})^{*}$. If the prime factorization of $ infinitude $ is $ lastcomposite^{firstratio} \\cdots ultimatecomposite^{e_{k}} $, then\n\\[\nantitotient(infinitude)=\\prod_{sizevalue=1}^{k} antitotient\\left(genericcomposite^{indexratio}\\right)=\\prod_{sizevalue=1}^{k} genericcomposite^{indexratio-1}\\left(genericcomposite-1\\right) .\n\\]\n\nEuler's Theorem [Lar1, p. 148], which is Lagrange's Theorem (the order of an element of a finite group divides the order of the group, [Lar1, p. 147]) applied to the group $(\\mathbb{Z} / infinitude \\mathbb{Z})^{*}$, states that $ constant^{antitotient(infinitude)} \\equiv 1(\\bmod infinitude) $ whenever $ \\operatorname{gcd}(constant, infinitude)=1 $.\n\nIt shows that $ fixedterm=3^{futureterm} $ modulo 100 is determined by $ futureterm $ modulo $ antitotient(100)=40 $, for $ sizevalue \\geq 2 $. Similarly $ futureterm \\bmod 40 $ is determined by $ laterterm \\bmod 16 $, which is determined by $ upcomingterm \\bmod 8 $, for $ sizevalue \\geq 4 $. Finally $ upcomingterm \\bmod 8 $ is determined by $ forthcoming \\bmod 2 $ for $ sizevalue \\geq 5 $, since $ 3^{2} \\equiv 1(\\bmod 8) $. For $ sizevalue \\geq 5, forthcoming $ is odd, so\n\\[\n\\begin{array}{l}\nupcomingterm=3^{forthcoming} \\equiv 3^{1} \\equiv 3 \\quad(\\bmod 8) \\\\\nlaterterm=3^{upcomingterm} \\equiv 3^{3} \\equiv 11 \\quad(\\bmod 16) \\\\\nfutureterm=3^{laterterm} \\equiv 3^{11} \\equiv 27 \\quad(\\bmod 40) \\\\\nfixedterm=3^{futureterm} \\equiv 3^{27} \\equiv 87 \\quad(\\bmod 100)\n\\end{array}\n\\]\n\nThus the only integer that appears as the last two digits of infinitely many $ fixedterm $ is 87 .\n\nRemark. Carmichael's antilambda function. The exponent in Euler's Theorem is not always best possible. The function $ antilambda(infinitude) $ giving the best possible exponent for $ infinitude $, i.e., the least positive integer $ antilambda $ such that $ constant^{antilambda} \\equiv 1(\\bmod infinitude) $ whenever $ \\operatorname{gcd}(constant, infinitude)=1 $, is known as Carmichael's antilambda function or as the reduced totient function. If $ infinitude=lastcomposite^{firstratio} \\cdots ultimatecomposite^{e_{k}} $, then\n\\[\nantilambda(infinitude)=\\operatorname{lcm}\\left(antilambda\\left(lastcomposite^{firstratio}\\right), \\ldots, antilambda\\left(ultimatecomposite^{e_{k}}\\right)\\right),\n\\]\nwhere\n\\[\nantilambda\\left(composite^{rootvalue}\\right)=antitotient\\left(composite^{rootvalue}\\right)=composite^{rootvalue-1}(composite-1),\n\\]\nunless $ composite=2 $ and $ rootvalue \\geq 3 $ in which case $ antilambda\\left(2^{rootvalue}\\right)=2^{rootvalue-2} $ instead of $ 2^{rootvalue-1} $. For more information, see [Ros1, Section 9.6].\n\nOne can simplify the computations in the above solution by using $ antilambda(infinitude) $ in place of $ antitotient(infinitude) $ : iteration of $ antilambda $ maps 100 to 20 to 4 to 2 , so starting from $ upcomingterm \\equiv 1(\\bmod 2) $ we obtain\n\\[\n\\begin{array}{lll}\nlaterterm & =3^{upcomingterm} \\equiv 3^{1} \\equiv 3 & \\\\\nfutureterm & =3^{laterterm} \\equiv 3^{3} \\equiv 7 & \\\\\n(\\bmod 4) \\\\\nfixedterm & =3^{futureterm} \\equiv 3^{7} \\equiv 87 & \\\\\n(\\bmod 100)\n\\end{array}\n\\]\n\nStronger result. More generally, one can show that for any integer $ variablebase \\geq 1 $, the sequence defined by $ lastindex=variablebase $ and $ previousterm=variablebase^{fixedterm} $ for $ sizevalue \\geq 1 $ is eventually constant when reduced modulo a positive integer $ infinitude $. To prove this, one can use the Chinese Remainder Theorem to reduce to the case where $ infinitude $ is a power of a prime $ composite $. If $ composite $ divides $ variablebase $, then the sequence is eventually 0 modulo $ infinitude $; otherwise we can use Euler's Theorem and strong induction on $ infinitude $, since $ antitotient(infinitude)<infinitude $ for $ infinitude>1 $. See 1997B5 for a related problem, and further discussion."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "a_1": "hjgrksla",
        "a_i": "wkmbrtqn",
        "a_i+1": "cvprndsl",
        "a_i-1": "lxkzefor",
        "a_i-2": "smjupdal",
        "a_i-3": "pturbvqa",
        "a_i-4": "znxqlweh",
        "i": "rplozkmi",
        "n": "vpqswkzn",
        "p": "rghmzkqu",
        "p_1": "cszlpmar",
        "p_i": "fxlkrdwn",
        "p_k": "hvbeqjon",
        "e": "bxdwljrs",
        "e_1": "mgvcralt",
        "e_i": "tjvopksu",
        "c": "quxntfao",
        "\\phi": "\\ytrencua",
        "\\lambda": "\\xkwoarjf"
      },
      "question": "Define a sequence $\\{wkmbrtqn\\}$ by $hjgrksla=3$ and $cvprndsl=3^{wkmbrtqn}$ for $rplozkmi\\geq 1$. Which integers between 00 and 99 inclusive occur as the last two digits in the decimal expansion of infinitely many $wkmbrtqn$?",
      "solution": "Solution. Let \\( \\ytrencua(vpqswkzn) \\) denote the Euler \\( \\ytrencua \\)-function, which equals the number of integers between 1 and \\( vpqswkzn \\) inclusive that are relatively prime to \\( vpqswkzn \\), or more abstractly, the order of the multiplicative group \\( (\\mathbb{Z} / vpqswkzn \\mathbb{Z})^{*} \\). If the prime factorization of \\( vpqswkzn \\) is \\( cszlpmar^{mgvcralt} \\cdots hvbeqjon^{e_{k}} \\), then \\( \\ytrencua(vpqswkzn) \\) can be computed by the formula\n\\[\n\\ytrencua(vpqswkzn)=\\prod_{rplozkmi=1}^{k} \\ytrencua\\left(fxlkrdwn^{tjvopksu}\\right)=\\prod_{rplozkmi=1}^{k} fxlkrdwn^{tjvopksu-1}\\left(fxlkrdwn-1\\right) .\n\\]\n\nEuler's Theorem [Lar1, p. 148], which is Lagrange's Theorem (the order of an element of a finite group divides the order of the group, [Lar1, p. 147]) applied to the group \\( (\\mathbb{Z} / vpqswkzn \\mathbb{Z})^{*} \\), states that \\( qzxwvtnp^{\\ytrencua(vpqswkzn)} \\equiv 1(\\bmod vpqswkzn) \\) whenever \\( \\operatorname{gcd}(qzxwvtnp, vpqswkzn)=1 \\).\n\nIt shows that \\( wkmbrtqn=3^{lxkzefor} \\) modulo 100 is determined by \\( lxkzefor \\) modulo \\( \\ytrencua(100)=40 \\), for \\( rplozkmi \\geq 2 \\). Similarly \\( lxkzefor \\bmod 40 \\) is determined by \\( smjupdal \\bmod 16 \\), which is determined by \\( pturbvqa \\bmod 8 \\), for \\( rplozkmi \\geq 4 \\). Finally \\( pturbvqa \\bmod 8 \\) is determined by \\( znxqlweh \\bmod 2 \\) for \\( rplozkmi \\geq 5 \\), since \\( 3^{2} \\equiv 1(\\bmod 8) \\). For \\( rplozkmi \\geq 5, znxqlweh \\) is odd, so\n\\[\n\\begin{array}{l}\npturbvqa=3^{znxqlweh} \\equiv 3^{1} \\equiv 3 \\quad(\\bmod 8) \\\\\nsmjupdal=3^{pturbvqa} \\equiv 3^{3} \\equiv 11 \\quad(\\bmod 16) \\\\\nlxkzefor=3^{smjupdal} \\equiv 3^{11} \\equiv 27 \\quad(\\bmod 40) \\\\\nwkmbrtqn=3^{lxkzefor} \\equiv 3^{27} \\equiv 87 \\quad(\\bmod 100)\n\\end{array}\n\\]\n\nThus the only integer that appears as the last two digits of infinitely many \\( wkmbrtqn \\) is 87 .\n\nRemark. Carmichael's lambda function. The exponent in Euler's Theorem is not always best possible. The function \\( \\xkwoarjf(vpqswkzn) \\) giving the best possible exponent for \\( vpqswkzn \\), i.e., the least positive integer \\( \\xkwoarjf \\) such that \\( qzxwvtnp^{\\xkwoarjf} \\equiv 1(\\bmod vpqswkzn) \\) whenever \\( \\operatorname{gcd}(qzxwvtnp, vpqswkzn)=1 \\), is known as Carmichael's lambda function or as the reduced totient function. If \\( vpqswkzn= cszlpmar^{mgvcralt} \\cdots hvbeqjon^{e_{k}} \\), then\n\\[\n\\xkwoarjf(vpqswkzn)=\\operatorname{lcm}\\left(\\xkwoarjf\\left(cszlpmar^{mgvcralt}\\right), \\ldots, \\xkwoarjf\\left(hvbeqjon^{e_{k}}\\right)\\right),\n\\]\nwhere\n\\[\n\\xkwoarjf\\left(rghmzkqu^{bxdwljrs}\\right)=\\ytrencua\\left(rghmzkqu^{bxdwljrs}\\right)=rghmzkqu^{bxdwljrs-1}(rghmzkqu-1),\n\\]\nunless \\( rghmzkqu=2 \\) and \\( bxdwljrs \\geq 3 \\) in which case \\( \\xkwoarjf\\left(2^{bxdwljrs}\\right)=2^{bxdwljrs-2} \\) instead of \\( 2^{bxdwljrs-1} \\). For more information, see [Ros1, Section 9.6].\n\nOne can simplify the computations in the above solution by using \\( \\xkwoarjf \\) in place of \\( \\ytrencua \\) : iteration of \\( \\xkwoarjf \\) maps 100 to 20 to 4 to 2 , so starting from \\( pturbvqa \\equiv 1(\\bmod 2) \\) we obtain\n\\[\n\\begin{array}{lll}\nsmjupdal & =3^{pturbvqa} \\equiv 3^{1} \\equiv 3 & \\\\\nlxkzefor & =3^{smjupdal} \\equiv 3^{3} \\equiv 7 & \\\\\n(\\bmod 4) \\\\\nwkmbrtqn & =3^{lxkzefor} \\equiv 3^{7} \\equiv 87 & \\\\\n(\\bmod 100)\n\\end{array}\n\\]\n\nStronger result. More generally, one can show that for any integer \\( quxntfao \\geq 1 \\), the sequence defined by \\( hjgrksla=quxntfao \\) and \\( cvprndsl=quxntfao^{wkmbrtqn} \\) for \\( rplozkmi \\geq 1 \\) is eventually constant when reduced modulo a positive integer \\( vpqswkzn \\). To prove this, one can use the Chinese Remainder Theorem to reduce to the case where \\( vpqswkzn \\) is a power of a prime \\( rghmzkqu \\). If \\( rghmzkqu \\) divides \\( quxntfao \\), then the sequence is eventually 0 modulo \\( vpqswkzn \\); otherwise we can use Euler's Theorem and strong induction on \\( vpqswkzn \\), since \\( \\ytrencua(vpqswkzn)<vpqswkzn \\) for \\( vpqswkzn>1 \\). See 1997B5 for a related problem, and further discussion."
    },
    "kernel_variant": {
      "question": "Let the sequence $\\bigl(a_n\\bigr)_{n\\ge 1}$ be defined recursively by  \n\\[\na_1=3 ,\\qquad a_{n+1}=3^{\\,a_n}\\qquad(n\\ge 1).\n\\]\n\n(a)\\;  Define  \n\\[\nf:\\mathbf Z_{10}\\longrightarrow\\mathbf Z_{10},\\qquad f(x)=3^{x}.\n\\]\nShow that $f$ admits a (unique) $10$–adic analytic continuation to all of $\\mathbf Z_{10}$ and prove that the functional equation  \n\\[\n\\alpha \\;=\\;3^{\\alpha}\\qquad\\bigl(\\alpha\\in\\mathbf Z_{10}\\bigr)\n\\tag{$\\star$}\n\\]\nhas one and only one solution $\\alpha\\in\\mathbf Z_{10}$.  \nMoreover, verify that the whole orbit $\\bigl(a_n\\bigr)_{n\\ge 1}$ is\n$10$–adically attracted to $\\alpha$.\n\n(b)\\;  Determine the residue class of $\\alpha$ modulo $10^{5}$; that is,\nfind the last five decimal digits of $\\alpha$.\n\n(c)\\;  Prove that the $5$–digit integer obtained in part\\,(b) is the only\nresidue modulo $10^{5}$ that appears infinitely often as the last five\ndigits of the members of the sequence $\\bigl(a_n\\bigr)$.\n\n────────────────────────────────────────────────────────",
      "solution": "Throughout the solution we write $\\mathbf Z_p$ for the ring of $p$-adic\nintegers, $v_p(\\cdot)$ for the $p$-adic valuation and\n$|\\cdot|_p=p^{-v_p(\\cdot)}$ for the associated absolute value.  On the\nintersection\n$\\mathbf Z_{10}:=\\mathbf Z_{2}\\cap\\mathbf Z_{5}$ we set  \n\n\\[\n|x|_{10}:=\\max\\bigl\\{|x|_2,\\,|x|_5\\bigr\\},\\qquad\nv_{10}(x):=\\min\\bigl\\{v_2(x),\\,v_5(x)\\bigr\\}.\n\\]\n\n--------------------------------------------------------------------\n(a)  Analytic continuation and attraction of the fixed point\n--------------------------------------------------------------------\n1.\\;Analytic continuation of $x\\mapsto 3^{x}$.\n\n$\\bullet$ $2$–adic factor.  \nBecause $3\\equiv1\\pmod2$, $\\log3$ is defined in $\\mathbf Z_{2}$ and\nsatisfies $\\log3\\in2\\mathbf Z_{2}$; hence for every $x\\in\\mathbf Z_{2}$  \n\n\\[\n3^{x}=\\exp\\!\\bigl(x\\log3\\bigr),\n\\]\nso the map $f$ is $2$-adically analytic on $\\mathbf Z_{2}$.\n\n$\\bullet$ $5$–adic factor.  \nHere $3\\not\\equiv1\\pmod5$, so we split $x\\in\\mathbf Z_{5}$ uniquely as\n$x=r+4q$ with $r\\in\\{0,1,2,3\\}$ and $q\\in\\mathbf Z_{5}$.  Then  \n\n\\[\n3^{x}=3^{r}\\exp\\!\\bigl(q\\log81\\bigr),\\qquad\n\\log81\\in5\\mathbf Z_{5},\n\\]\nand the exponential series converges for every $q\\in\\mathbf Z_{5}$.\nHence $f$ is analytic on $\\mathbf Z_{5}$, and therefore on their\nintersection $\\mathbf Z_{10}$.\n\n2.\\;Construction and uniqueness of the fixed point.\n\nPut $g(x):=3^{x}-x\\in\\mathbf Z_{5}\\{x\\}$.  Choose $x_0=7$; a direct\ncheck gives $g(7)=3^{7}-7=2180\\equiv0\\pmod5$.  Because  \n\n\\[\ng'(x)=3^{x}\\cdot\\frac{\\log81}{4}-1\n\\qquad\\text{and}\\qquad \\frac{\\log81}{4}\\in5\\mathbf Z_{5},\n\\]\nwe have $g'(7)\\equiv-1\\pmod5$, a $5$-adic unit.  Hensel’s lemma yields a\nunique root $\\alpha_{5}\\in\\mathbf Z_{5}$ with\n$\\alpha_{5}\\equiv7\\pmod5$.\n\nExactly the same argument works $2$-adically with the starting value\n$x_0=1$, giving a unique root $\\alpha_{2}\\in\\mathbf Z_{2}$.\nBy the Chinese Remainder Theorem there is a unique  \n\n\\[\n\\alpha\\in\\mathbf Z_{10}\n\\quad\\text{satisfying}\\quad\n\\alpha\\equiv\\alpha_{2}\\pmod{2^{\\infty}},\\;\n\\alpha\\equiv\\alpha_{5}\\pmod{5^{\\infty}},\n\\]\nand this $\\alpha$ is the sole solution of $(\\star)$.\n\n3.\\;Attraction in the $2$–adic component.\n\nFor $x\\in\\mathbf Z_{2}$ we have $f'(x)=3^{x}\\log3$ and\n$|f'(x)|_2=|\\log3|_2=\\tfrac12$.  The non-Archimedean mean-value\nestimate gives  \n\n\\[\n|f(x)-f(y)|_2\\le \\tfrac12\\,|x-y|_2\\qquad(x,y\\in\\mathbf Z_{2}),\n\\]\nso $f$ is a global $2$-adic contraction with constant $\\tfrac12$ and\nhence $a_n\\to\\alpha$ in $\\mathbf Z_{2}$.\n\n4.\\;The orbit enters $\\alpha_{5}+20\\mathbf Z_{5}$.\n\nModulo $20$ we have  \n\n\\[\na_1=3,\\;\na_2=3^{3}=27\\equiv7,\\;\na_3=3^{27}\\equiv3^{7}\\equiv7,\\qquad\n3^{20}\\equiv1\\pmod{20},\n\\]\nso $a_n\\equiv7\\pmod{20}$ for all $n\\ge2$.  Because\n$\\alpha\\equiv7\\pmod{20}$ as well,  \n\n\\[\ne_n:=a_n-\\alpha\\;\\in\\;20\\mathbf Z_{5}\\qquad(n\\ge2).\n\\]\n\n5.\\;Exact growth of the $5$-adic valuations.\n\nWrite $e_n=20k$ with $k\\in\\mathbf Z_{5}$.  Then  \n\n\\[\n3^{e_n}=3^{4\\cdot5k}=81^{5k}.\n\\]\nUsing the lifting-the-exponent lemma,\n\n\\[\nv_5\\!\\bigl(81^{t}-1\\bigr)=v_5(81-1)+v_5(t)=1+v_5(t),\n\\]\nwe obtain  \n\n\\[\nv_5\\!\\bigl(3^{e_n}-1\\bigr)=1+v_5(e_n).\n\\]\nSince $3^{\\alpha}$ is a $5$-adic unit,\n\n\\[\nv_5(e_{n+1})=v_5\\!\\bigl(3^{a_n}-3^{\\alpha}\\bigr)\n            =v_5\\!\\bigl(3^{e_n}-1\\bigr)\n            =1+v_5(e_n).\n\\tag{1}\n\\]\nThus $v_5(e_n)$ increases by exactly one at each step, which is\nequivalent to $|e_{n+1}|_5=\\tfrac15|e_n|_5$.  As $|e_2|_5\\le\\tfrac15$,\niteration of (1) shows $|e_n|_5\\le 5^{-(n-1)}\\to0$, i.e.\\\n$a_n\\to\\alpha$ in $\\mathbf Z_{5}$.\n\n6.\\;Convergence in $\\mathbf Z_{10}$.\n\nBecause\n$|x|_{10}=\\max\\{|x|_2,|x|_5\\}$ and convergence holds in both factors,\nwe have  \n\n\\[\n\\boxed{\\displaystyle\\lim_{n\\to\\infty}a_n=\\alpha\\quad\n        \\text{in }\\mathbf Z_{10}.}\n\\]\n\n--------------------------------------------------------------------\n(b)  Determining $\\alpha\\bmod10^{5}$\n--------------------------------------------------------------------\nWe lift the congruence $\\alpha=3^{\\alpha}$ successively through the\nchain  \n\n\\[\n2 \\;\\to\\;4 \\;\\to\\;20 \\;\\to\\;100 \\;\\to\\;1000 \\;\\to\\;10^{4} \\;\\to\\;10^{5},\n\\]\nusing at each stage the fact that $g'(\\alpha)\\equiv-1\\pmod{10}$, which\nguarantees uniqueness.\n\n\\begin{itemize}\n\\item \\textbf{Step 1:} $\\pmod2$.\n      Since $3^{x}\\equiv1\\pmod2$, we have $\\alpha\\equiv1\\pmod2$.\n\n\\item \\textbf{Step 2:} $\\pmod4$.\n      Among the odd residues only $x\\equiv3$ works, so\n      $\\alpha\\equiv3\\pmod4$.\n\n\\item \\textbf{Step 3:} $\\pmod{20}$.\n      Checking $x\\equiv3,7,11,15\\pmod{20}$ shows\n      $\\alpha\\equiv7\\pmod{20}$.\n\n\\item \\textbf{Step 4:} $\\pmod{100}$.\n      Write $x=7+20t$, $0\\le t\\le4$.  As $3^{20}\\equiv1\\pmod{100}$,\n\n      \\[\n      3^{x}\\equiv3^{7}\\equiv87\\pmod{100},\n      \\]\n      which forces $t=4$ and therefore $\\alpha\\equiv87\\pmod{100}$.\n\n\\item \\textbf{Step 5:} $\\pmod{1000}$.\n      Carmichael’s value $\\lambda(1000)=100$ allows the reduction\n      $3^{y}\\equiv3^{\\,y\\bmod100}\\pmod{1000}$.  Taking $y\\equiv87$\n      yields  \n\n      \\[\n      \\alpha\\equiv3^{87}\\equiv387\\pmod{1000}.\n      \\]\n\n\\item \\textbf{Step 6:} $\\pmod{10^{4}}$.\n      Since $\\lambda(10^{4})=\\operatorname{lcm}\\!\\bigl(4,500\\bigr)=500$\n      and $387\\equiv387\\pmod{500}$,\n\n      \\[\n      \\alpha\\equiv3^{387}\\equiv5387\\pmod{10^{4}}.\n      \\]\n\n\\item \\textbf{Step 7:} $\\pmod{10^{5}}$.\n      Write $x=5387+10^{4}t$ with $0\\le t\\le9$.  Because\n      $\\lambda(10^{5})=5000$ we have $3^{10^{4}}\\equiv1\\pmod{10^{5}}$,\n      whence  \n\n      \\[\n      3^{x}\\equiv3^{5387}\\pmod{10^{5}}.\n      \\]\n      A Chinese Remainder computation gives  \n\n      \\[\n      3^{5387}\\equiv27\\pmod{32},\\qquad\n      3^{5387}\\equiv1637\\pmod{3125},\n      \\]\n      and combining the two congruences yields  \n\n      \\[\n      3^{5387}\\equiv95\\,387\\pmod{10^{5}}.\n      \\]\n      Therefore  \n\n      \\[\n      5387+10^{4}t\\equiv95\\,387\\pmod{10^{5}}\n      \\;\\Longrightarrow\\;\n      10^{4}t\\equiv90\\,000\\pmod{10^{5}}\n      \\;\\Longrightarrow\\;\n      t\\equiv9\\pmod{10},\n      \\]\n      so $t=9$ and finally  \n\n      \\[\n      \\boxed{\\alpha\\equiv95\\,387\\pmod{100\\,000}.}\n      \\]\n\\end{itemize}\n\n--------------------------------------------------------------------\n(c)  Uniqueness of the five-digit tail\n--------------------------------------------------------------------\nSet $r:=95\\,387$.\n\n(i)\\; By part\\,(a) the sequence $(a_n)$ converges to $\\alpha$ in\n$\\mathbf Z_{10}$; hence $a_n\\equiv r\\pmod{10^{5}}$ for all sufficiently\nlarge $n$, so $r$ occurs infinitely often.\n\n(ii)\\; Conversely, suppose a residue $s\\pmod{10^{5}}$ occurs for\ninfinitely many indices.  The corresponding subsequence is then\n$10$-adically Cauchy and must converge to the unique limit $\\alpha$.\nThus $s\\equiv\\alpha\\equiv r\\pmod{10^{5}}$.\n\nHence  \n\n\\[\n\\boxed{\\text{The residue }95\\,387\\text{ is the \\emph{only} five-digit\ntail that appears infinitely often.}}\\qquad\\square\n\n────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.682240",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional setting: the problem is lifted from ordinary\n  modular arithmetic to the full $10$-adic number system, requiring\n  simultaneous control of the $2$-adic and $5$-adic components.\n\n• Deeper theory: tools from $p$-adic analysis (non-Archimedean\n  metrics, contractions, Hensel-type lifting, the $p$-adic Banach\n  fixed-point theorem) are indispensable, far beyond the elementary\n  Euler-totient considerations of the original problem.\n\n• Longer chain of reductions: to reach the modulus $10^{5}$ we must\n  iterate Carmichael’s $\\lambda$-function five times and perform\n  several non-trivial modular exponentiations.\n\n• Uniqueness argument: part (c) needs an understanding of convergence\n  in $\\Bbb Z_{10}$ to rule out any other residue, adding a conceptual\n  layer absent from the original variant.\n\n• Convergence proof: demonstrating that $(a_n)$ converges $10$-adically\n  involves local-Lipschitz estimates derived from the Lifting-the-Exponent\n  lemma, again markedly more sophisticated than the finite-modulus\n  periodicity used before.\n\nThese additions make the enhanced kernel variant substantially more\ntechnical and demanding than both the original contest problem and the\nexisting three-digit kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let the sequence $\\bigl(a_n\\bigr)_{n\\ge 1}$ be defined by \n\\[\na_1=3 ,\\qquad a_{n+1}=3^{\\,a_n}\\qquad(n\\ge 1).\n\\]\n\n(a)\\; Show that the assignment \n\\[\nf:\\mathbf Z_{10}\\longrightarrow\\mathbf Z_{10},\\qquad f(x)=3^{x},\n\\]\nadmits a (unique) $10$–adic analytic continuation and prove that the\nfunctional equation \n\\[\n\\alpha=3^{\\alpha}\\qquad\\bigl(\\alpha\\in\\mathbf Z_{10}\\bigr)\n\\tag{$\\star$}\n\\]\npossesses one and only one solution $\\alpha\\in\\mathbf Z_{10}$.  \nMoreover, verify that the entire orbit $\\bigl(a_n\\bigr)_{n\\ge 1}$ is\n$10$–adically attracted to $\\alpha$.\n\n(b)\\; Determine the residue class of $\\alpha$ modulo $10^{5}$, i.e.\\\nits last five decimal digits.\n\n(c)\\; Prove that the $5$–digit integer found in part (b) is the sole\nresidue modulo $10^{5}$ that occurs infinitely often as the last five\ndigits of the elements of the sequence $\\bigl(a_n\\bigr)$.",
      "solution": "Throughout, $\\mathbf Z_p$ denotes the ring of $p$–adic integers,\n$v_p(\\cdot)$ the $p$–adic valuation, and $|\\cdot|_p=p^{-v_p(\\cdot)}$ the\nassociated absolute value.  On \n$\\mathbf Z_{10}=\\mathbf Z_2\\cap\\mathbf Z_5$ we use\n\\[\n|x|_{10}:=\\max\\bigl\\{|x|_2,|x|_5\\bigr\\},\\qquad\nv_{10}(x):=\\min\\bigl\\{v_2(x),v_5(x)\\bigr\\}.\n\\]\n\n--------------------------------------------------------------------\n(a)  Analytic continuation, location and attractiveness of the fixed point\n--------------------------------------------------------------------\n1.  Analytic extension of $x\\longmapsto3^{x}$.\n\nBecause $3\\equiv1\\pmod2$, one may write\n\\[\n3^{x}=\\exp\\!\\bigl(x\\log3\\bigr),\\qquad x\\in\\mathbf Z_2 ,\n\\]\nwhere the $2$–adic exponential converges on $2\\mathbf Z_2$ and\n$\\log3\\in2\\mathbf Z_2$; hence the right hand side converges for all\n$x\\in\\mathbf Z_2$.  The same argument works in $\\mathbf Z_5$ after\nsplitting off a fourth power:\nwrite every $x\\in\\mathbf Z_5$ uniquely as $x=r+4q$ with\n$r\\in\\{0,1,2,3\\}$ and $q\\in\\mathbf Z_5$.\nSince $3^{4}=81\\equiv1+5\\cdot16$, set\n\\[\n3^{x}=3^{r}\\exp\\!\\bigl(q\\log81\\bigr)\\qquad\\bigl(x=r+4q\\bigr),\n\\]\nand $\\log81\\in5\\mathbf Z_5$ guarantees convergence for all\n$q\\in\\mathbf Z_5$.\nHence $f(x):=3^{x}$ is a well–defined analytic self–map of\n$\\mathbf Z_p$ for $p=2$ and $p=5$, therefore also of their intersection\n$\\mathbf Z_{10}$.\n\n2.  Construction of the fixed point by Hensel’s lemma.\n\nPut $g(x):=3^{x}-x\\in\\mathbf Z_5\\{x\\}$.\nThe integer $x_0:=7$ satisfies\n\\[\ng(7)=3^{7}-7=2187-7=2180\\equiv0\\pmod5 ,\\qquad\ng'(7)=3^{7}\\log3-1\\equiv-1\\pmod5 ,\n\\]\nbecause $3^{7}\\equiv2\\pmod5$ and\n$v_5\\!\\bigl(\\log3\\bigr)=1$.  Thus $g'(7)$ is a $5$–adic unit, and\nHensel’s lemma produces a unique root\n$\\alpha_5\\in\\mathbf Z_5$ with $\\alpha_5\\equiv7\\pmod5$.\n\nThe same device works $2$–adically: with $x_0:=1$ we have\n\\[\ng(1)=3-1=2\\equiv0\\pmod2,\\qquad\ng'(1)=3\\log3-1\\equiv-1\\pmod2 ,\n\\]\nso a unique fixed point $\\alpha_2\\in\\mathbf Z_2$ exists.\nThe Chinese remainder theorem gives a unique\n\\[\n\\alpha\\in\\mathbf Z_{10}\\quad\\text{such that}\\quad\n\\alpha\\equiv\\alpha_2\\pmod{2^{\\infty}},\\;\n\\alpha\\equiv\\alpha_5\\pmod{5^{\\infty}},\n\\]\nand obviously $\\alpha$ fulfills $(\\star)$.\n\nUniqueness in $\\mathbf Z_{10}$:  \nif $\\beta$ satisfies $3^{\\beta}=\\beta$, then\n$\\beta\\equiv\\alpha\\pmod{2^{\\infty}}$ and\n$\\beta\\equiv\\alpha\\pmod{5^{\\infty}}$, hence $\\beta=\\alpha$.\n\n3.  Dynamics in $\\mathbf Z_2$.\n\nFor $x,y\\in\\mathbf Z_2$ put $h=x-y$.\nThe binomial expansion\n$3^{h}=1+h\\log3+\\dfrac{h^2(\\log3)^2}{2!}+\\cdots$\ngives\n\\[\nv_2\\!\\bigl(3^{h}-1\\bigr)=v_2(h)+1,\\qquad\n|3^{h}-1|_2=\\tfrac12\\,|h|_2 .\n\\]\nHence\n$|f(x)-f(y)|_2=\\dfrac12\\,|x-y|_2$, i.e. $f$ is a global\ncontraction on $\\mathbf Z_2$ with constant $\\tfrac12$.\nConsequently $a_n\\longrightarrow\\alpha$ in $\\mathbf Z_2$.\n\n4.  Local contraction in $\\mathbf Z_5$ and entry of the orbit.\n\nIf $h\\in5\\mathbf Z_5$, the same computation as above yields\n\\[\n|3^{h}-1|_5=\\tfrac15\\,|h|_5 .\n\\tag{1}\n\\]\nTherefore $f$ is a strict contraction on every coset\n$x+5\\mathbf Z_5$.  In particular, on the closed ball\n\\[\nB:=\\alpha_5+5\\mathbf Z_5\n\\]\nthe Lipschitz constant is exactly $\\tfrac15$.\n\nIt remains to check that the sequence $(a_n)$\nenters $B$ once and for all.  We compute modulo $20$:\n\\[\na_1=3,\\;\na_2=3^{3}=27\\equiv7\\pmod{20},\\;\na_3=3^{27}\\equiv3^{7}\\equiv7\\pmod{20},\n\\]\nand the same congruence persists forever because\n$3^{20}\\equiv1\\pmod{20}$.  Hence $a_n\\equiv7\\pmod{20}$ for every\n$n\\ge2$.  Since $\\alpha_5\\equiv7\\pmod5$, it follows that\n\\[\na_n-\\alpha\\;\\in\\;5\\mathbf Z_5\\qquad(n\\ge2),\n\\]\nso $a_n\\in B$ for all $n\\ge2$.\n\n5.  Convergence in $\\mathbf Z_5$.\n\nLet $n\\ge2$ and write $h_n:=a_n-\\alpha\\in5\\mathbf Z_5$.\n Using $(\\star)$ and (1),\n\\[\na_{n+1}-\\alpha\n   =3^{a_n}-3^{\\alpha}=3^{\\alpha}\\bigl(3^{h_n}-1\\bigr),\n\\quad\n|a_{n+1}-\\alpha|_5\n   =|\\alpha|_5\\cdot|3^{h_n}-1|_5\n   =\\tfrac15\\,|h_n|_5 .\n\\]\nHence the error shrinks by a factor $\\tfrac15$ at every step,\nso $a_n\\longrightarrow\\alpha$ in $\\mathbf Z_5$.\n\n6.  Convergence in $\\mathbf Z_{10}$.\n\nBecause $|\\cdot|_{10}=\\max\\{|\\cdot|_2,|\\cdot|_5\\}$ and we already have\nconvergence in each component, we conclude\n\\[\n\\boxed{\\displaystyle\n\\lim_{n\\to\\infty}a_n=\\alpha\\quad\\text{in }\\mathbf Z_{10}}.\n\\]\n\n--------------------------------------------------------------------\n(b)  Determining $\\alpha\\bmod10^{5}$\n--------------------------------------------------------------------\nWe successively lift the congruence\n$\\alpha=3^{\\alpha}$ modulo $10^{k}$ for $k=1,\\dots,5$; at\neach step uniqueness is guaranteed by the invertibility of\n$g'(\\alpha)=-1+3^{\\alpha}\\log3$.\n\n$\\boldsymbol{k=1}$ (mod $2$).  \nBecause $3^{x}\\equiv1\\pmod2$ for all $x$, necessarily\n$\\alpha\\equiv1\\pmod2$.\n\n$\\boldsymbol{k=2}$ (mod $4$).  \nOdd candidates are $1,3$; computing\n$3^{1}\\equiv3\\pmod4,\\;3^{3}\\equiv27\\equiv3\\pmod4$ selects\n$\\alpha\\equiv3\\pmod4$.\n\n$\\boldsymbol{k=3}$ (mod $20$).  \nAmong $3,7,11,15$ (all $\\equiv3\\pmod4$) we need\n$3^{x}\\equiv x\\pmod{20}$.  \nA direct check isolates $x\\equiv7\\pmod{20}$.\n\n$\\boldsymbol{k=4}$ (mod $100$).  \nWrite $x=7+20t$ with $t\\in\\{0,\\dots,4\\}$.  Since\n$3^{20}\\equiv1\\pmod{100}$, we have\n$3^{x}\\equiv3^{7}\\equiv87\\pmod{100}$, hence $t=4$ and\n$\\alpha\\equiv87\\pmod{100}$.\n\n$\\boldsymbol{k=5}$ (mod $1000$).  \nBecause $\\lambda(1000)=100$,  \n$3^{y}\\equiv3^{\\,y\\bmod100}\\pmod{1000}$.  Using $y\\equiv87$ we find\n\\[\n\\alpha\\equiv3^{87}\\equiv387\\pmod{1000}.\n\\]\n\n$\\boldsymbol{k=6}$ (mod $10^{4}$).  \n$\\lambda(10^{4})=\\operatorname{lcm}\\bigl(\\lambda(16),\\lambda(5^{4})\\bigr)\n =\\operatorname{lcm}(4,500)=500$, and $387\\equiv387\\pmod{500}$, whence\n\\[\n\\alpha\\equiv3^{387}\\equiv5387\\pmod{10^{4}}.\n\\]\n\n$\\boldsymbol{k=7}$ (mod $10^{5}$).  \nWith $\\lambda(10^{5})=\\operatorname{lcm}(8,2500)=5000$ and again\n$387\\equiv387\\pmod{5000}$, a square–and–multiply calculation yields\n\\[\n3^{387}\\equiv95387\\pmod{10^{5}}.\n\\]\nTherefore\n\\[\n\\boxed{\\alpha\\equiv95\\,387\\pmod{100\\,000}}.\n\\]\n\n--------------------------------------------------------------------\n(c)  Uniqueness of the five–digit tail\n--------------------------------------------------------------------\nLet $r:=95\\,387$.\n\n(i)\\; Convergence proved in (a) shows that\n$a_n\\equiv r\\pmod{10^{5}}$ for all sufficiently large $n$, hence $r$\nindeed appears infinitely often.\n\n(ii)\\; Conversely, suppose a residue $s\\pmod{10^{5}}$ occurs for\ninfinitely many indices.  The corresponding subsequence is\n$10$–adically Cauchy and must converge to the same limit $\\alpha$ as the\nfull sequence; thus $s\\equiv\\alpha\\equiv r\\pmod{10^{5}}$.\n\nTherefore\n\\[\n\\boxed{\\text{the residue }95\\,387\\text{ is the \\emph{only} five–digit\ntail that appears infinitely many times.}}\\qquad\\square",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.535038",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional setting: the problem is lifted from ordinary\n  modular arithmetic to the full $10$-adic number system, requiring\n  simultaneous control of the $2$-adic and $5$-adic components.\n\n• Deeper theory: tools from $p$-adic analysis (non-Archimedean\n  metrics, contractions, Hensel-type lifting, the $p$-adic Banach\n  fixed-point theorem) are indispensable, far beyond the elementary\n  Euler-totient considerations of the original problem.\n\n• Longer chain of reductions: to reach the modulus $10^{5}$ we must\n  iterate Carmichael’s $\\lambda$-function five times and perform\n  several non-trivial modular exponentiations.\n\n• Uniqueness argument: part (c) needs an understanding of convergence\n  in $\\Bbb Z_{10}$ to rule out any other residue, adding a conceptual\n  layer absent from the original variant.\n\n• Convergence proof: demonstrating that $(a_n)$ converges $10$-adically\n  involves local-Lipschitz estimates derived from the Lifting-the-Exponent\n  lemma, again markedly more sophisticated than the finite-modulus\n  periodicity used before.\n\nThese additions make the enhanced kernel variant substantially more\ntechnical and demanding than both the original contest problem and the\nexisting three-digit kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}