{
  "index": "2003-B-3",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Show that for each positive integer  n,\n\\[\n    n! = \\prod_{i=1}^n  \\mathrm{lcm}\\{1, 2, \\dots, \\lfloor n/i\\rfloor\\} .\n\\]\n(Here $\\mathrm{lcm}$ denotes the least common multiple, and\n$\\lfloor x \\rfloor$ denotes the greatest integer $\\leq x$.)",
  "solution": "\\textbf{First solution:}\nIt is enough to show that for each prime $p$, the exponent of $p$ in\nthe prime factorization of both sides is the same.\nOn the left side, it is well-known that the exponent of $p$ in the\nprime factorization of $n!$ is\n\\[\n\\sum_{i =1}^n \\left\\lfloor \\frac{n}{p^i} \\right\\rfloor.\n\\]\n(To see this, note that the $i$-th term counts the multiples of $p^i$ among\n$1, \\dots, n$, so that a number divisible exactly by $p^i$ gets counted\nexactly $i$ times.)  This number can be reinterpreted as the cardinality\nof the set $S$ of points in the plane with positive integer coordinates\nlying on or under the curve $y = np^{-x}$: namely, each summand is the\nnumber of points of $S$ with $x=i$.\n\nOn the right side, the exponent of $p$ in the prime factorization of $\\lcm(1,\n\\dots, \\lfloor n/i \\rfloor)$ is $\\lfloor \\log_p \\lfloor n/i \\rfloor \\rfloor\n= \\lfloor \\log_p (n/i) \\rfloor$.  However, this is precisely the number\nof points of $S$ with $y=i$. Thus\n\\[\n\\sum_{i=1}^n\n\\lfloor \\log_p \\lfloor n/i \\rfloor \\rfloor\n= \\sum_{i =1}^n \\left\\lfloor \\frac{n}{p^i} \\right\\rfloor,\n\\]\nand the desired result follows.\n\n\\textbf{Second solution:}\nWe prove the result by induction on $n$, the case $n=1$ being obvious.\nWhat we actually show is that going from $n-1$ to $n$ changes both\nsides by the same multiplicative factor, that is,\n\\[\nn = \\prod_{i=1}^{n-1} \\frac{\\lcm\\{1, 2, \\dots, \\lfloor n/i \\rfloor\\}}{\\lcm\n\\{1, 2, \\dots, \\lfloor (n-1)/i \\rfloor\\}}.\n\\]\nNote that the $i$-th term in the product is equal to 1 if $n/i$ is not\nan integer, i.e., if $n/i$ is not a divisor of $n$.\nIt is also equal to 1 if $n/i$ is a divisor of $n$ but not a prime power,\nsince any composite number divides the lcm of all smaller numbers.\nHowever, if $n/i$ is a power of $p$, then the $i$-th term is equal to $p$.\n\nSince $n/i$ runs over all proper divisors of $n$, the product on the right\nside includes one factor of the prime $p$\nfor each factor of $p$ in the prime factorization of $n$. Thus the whole\nproduct is indeed equal to $n$, completing the induction.",
  "vars": [
    "i",
    "x",
    "y",
    "S"
  ],
  "params": [
    "n",
    "p"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "i": "counter",
        "x": "abscissa",
        "y": "ordinate",
        "S": "pointset",
        "n": "posintn",
        "p": "primep"
      },
      "question": "Show that for each positive integer posintn,\n\\[\n    \\posintn! = \\prod_{\\counter=1}^{\\posintn}  \\mathrm{lcm}\\{1, 2, \\dots, \\lfloor \\posintn/\\counter\\rfloor\\} .\n\\]\n(Here $\\mathrm{lcm}$ denotes the least common multiple, and\n$\\lfloor abscissa \\rfloor$ denotes the greatest integer $\\leq abscissa$.)",
      "solution": "\\textbf{First solution:}\nIt is enough to show that for each prime $primep$, the exponent of $primep$ in\nthe prime factorization of both sides is the same.\nOn the left side, it is well-known that the exponent of $primep$ in the\nprime factorization of $\\posintn!$ is\n\\[\n\\sum_{\\counter =1}^{\\posintn} \\left\\lfloor \\frac{\\posintn}{primep^{\\counter}} \\right\\rfloor.\n\\]\n(To see this, note that the $\\counter$-th term counts the multiples of $primep^{\\counter}$ among\n$1, \\dots, \\posintn$, so that a number divisible exactly by $primep^{\\counter}$ gets counted\nexactly $\\counter$ times.)  This number can be reinterpreted as the cardinality\nof the set $pointset$ of points in the plane with positive integer coordinates\nlying on or under the curve $ordinate = \\posintn primep^{-\\abscissa}$: namely, each summand is the\nnumber of points of $pointset$ with $\\abscissa=\\counter$.\n\nOn the right side, the exponent of $primep$ in the prime factorization of $\\lcm(1,\n\\dots, \\lfloor \\posintn/\\counter \\rfloor)$ is $\\lfloor \\log_{primep} \\lfloor \\posintn/\\counter \\rfloor \\rfloor\n= \\lfloor \\log_{primep} (\\posintn/\\counter) \\rfloor$.  However, this is precisely the number\nof points of $pointset$ with $ordinate=\\counter$. Thus\n\\[\n\\sum_{\\counter=1}^{\\posintn}\n\\lfloor \\log_{primep} \\lfloor \\posintn/\\counter \\rfloor \\rfloor\n= \\sum_{\\counter =1}^{\\posintn} \\left\\lfloor \\frac{\\posintn}{primep^{\\counter}} \\right\\rfloor,\n\\]\nand the desired result follows.\n\n\\textbf{Second solution:}\nWe prove the result by induction on $\\posintn$, the case $\\posintn=1$ being obvious.\nWhat we actually show is that going from $\\posintn-1$ to $\\posintn$ changes both\nsides by the same multiplicative factor, that is,\n\\[\n\\posintn = \\prod_{\\counter=1}^{\\posintn-1} \\frac{\\lcm\\{1, 2, \\dots, \\lfloor \\posintn/\\counter \\rfloor\\}}{\\lcm\n\\{1, 2, \\dots, \\lfloor (\\posintn-1)/\\counter \\rfloor\\}}.\n\\]\nNote that the $\\counter$-th term in the product is equal to 1 if $\\posintn/\\counter$ is not\nan integer, i.e., if $\\posintn/\\counter$ is not a divisor of $\\posintn$.\nIt is also equal to 1 if $\\posintn/\\counter$ is a divisor of $\\posintn$ but not a prime power,\nsince any composite number divides the lcm of all smaller numbers.\nHowever, if $\\posintn/\\counter$ is a power of $primep$, then the $\\counter$-th term is equal to $primep$.\n\nSince $\\posintn/\\counter$ runs over all proper divisors of $\\posintn$, the product on the right\nside includes one factor of the prime $primep$\nfor each factor of $primep$ in the prime factorization of $\\posintn$. Thus the whole\nproduct is indeed equal to $\\posintn$, completing the induction."
    },
    "descriptive_long_confusing": {
      "map": {
        "i": "buttercup",
        "x": "marshmallow",
        "y": "hurricane",
        "S": "platypus",
        "n": "rhinoceros",
        "p": "chandelier"
      },
      "question": "Show that for each positive integer  rhinoceros,\n\\[\n    rhinoceros! = \\prod_{buttercup=1}^{rhinoceros}  \\mathrm{lcm}\\{1, 2, \\dots, \\lfloor rhinoceros/buttercup\\rfloor\\} .\n\\]\n(Here $\\mathrm{lcm}$ denotes the least common multiple, and\n$\\lfloor marshmallow \\rfloor$ denotes the greatest integer $\\leq marshmallow$.)",
      "solution": "\\textbf{First solution:}\nIt is enough to show that for each prime chandelier, the exponent of chandelier in\nthe prime factorization of both sides is the same.\nOn the left side, it is well-known that the exponent of chandelier in the\nprime factorization of rhinoceros! is\n\\[\n\\sum_{buttercup =1}^{rhinoceros} \\left\\lfloor \\frac{rhinoceros}{chandelier^{buttercup}} \\right\\rfloor.\n\\]\n(To see this, note that the buttercup-th term counts the multiples of $chandelier^{buttercup}$ among\n$1, \\dots, rhinoceros$, so that a number divisible exactly by $chandelier^{buttercup}$ gets counted\nexactly buttercup times.)  This number can be reinterpreted as the cardinality\nof the set platypus of points in the plane with positive integer coordinates\nlying on or under the curve $hurricane = rhinoceros chandelier^{-marshmallow}$: namely, each summand is the\nnumber of points of platypus with $marshmallow=buttercup$.\n\nOn the right side, the exponent of chandelier in the prime factorization of $\\lcm(1,\n\\dots, \\lfloor rhinoceros/buttercup \\rfloor)$ is $\\lfloor \\log_{chandelier} \\lfloor rhinoceros/buttercup \\rfloor \\rfloor\n= \\lfloor \\log_{chandelier} (rhinoceros/buttercup) \\rfloor$.  However, this is precisely the number\nof points of platypus with $hurricane=buttercup$. Thus\n\\[\n\\sum_{buttercup=1}^{rhinoceros}\n\\lfloor \\log_{chandelier} \\lfloor rhinoceros/buttercup \\rfloor \\rfloor\n= \\sum_{buttercup =1}^{rhinoceros} \\left\\lfloor \\frac{rhinoceros}{chandelier^{buttercup}} \\right\\rfloor,\n\\]\nand the desired result follows.\n\n\\textbf{Second solution:}\nWe prove the result by induction on rhinoceros, the case $rhinoceros=1$ being obvious.\nWhat we actually show is that going from $rhinoceros-1$ to $rhinoceros$ changes both\nsides by the same multiplicative factor, that is,\n\\[\nrhinoceros = \\prod_{buttercup=1}^{rhinoceros-1} \\frac{\\lcm\\{1, 2, \\dots, \\lfloor rhinoceros/buttercup \\rfloor\\}}{\\lcm\n\\{1, 2, \\dots, \\lfloor (rhinoceros-1)/buttercup \\rfloor\\}}.\n\\]\nNote that the buttercup-th term in the product is equal to 1 if $rhinoceros/buttercup$ is not\nan integer, i.e., if $rhinoceros/buttercup$ is not a divisor of rhinoceros.\nIt is also equal to 1 if $rhinoceros/buttercup$ is a divisor of rhinoceros but not a prime power,\nsince any composite number divides the lcm of all smaller numbers.\nHowever, if $rhinoceros/buttercup$ is a power of chandelier, then the buttercup-th term is equal to chandelier.\n\nSince $rhinoceros/buttercup$ runs over all proper divisors of rhinoceros, the product on the right\nside includes one factor of the prime chandelier\nfor each factor of chandelier in the prime factorization of rhinoceros. Thus the whole\nproduct is indeed equal to rhinoceros, completing the induction."
    },
    "descriptive_long_misleading": {
      "map": {
        "i": "aggregate",
        "x": "ordinate",
        "y": "abscissa",
        "S": "complement",
        "n": "boundless",
        "p": "composite"
      },
      "question": "Show that for each positive integer  boundless,\n\\[\n    boundless! = \\prod_{aggregate=1}^{boundless}  \\mathrm{lcm}\\{1, 2, \\dots, \\lfloor boundless/aggregate\\rfloor\\} .\n\\]\n(Here $\\mathrm{lcm}$ denotes the least common multiple, and\n$\\lfloor ordinate \\rfloor$ denotes the greatest integer $\\leq ordinate$.)",
      "solution": "\\textbf{First solution:}\nIt is enough to show that for each prime $composite$, the exponent of $composite$ in\nthe prime factorization of both sides is the same.\nOn the left side, it is well-known that the exponent of $composite$ in the\nprime factorization of $boundless!$ is\n\\[\n\\sum_{aggregate =1}^{boundless} \\left\\lfloor \\frac{boundless}{composite^{aggregate}} \\right\\rfloor.\n\\]\n(To see this, note that the $aggregate$-th term counts the multiples of $composite^{aggregate}$ among\n$1, \\dots, boundless$, so that a number divisible exactly by $composite^{aggregate}$ gets counted\nexactly $aggregate$ times.)  This number can be reinterpreted as the cardinality\nof the set $complement$ of points in the plane with positive integer coordinates\nlying on or under the curve $abscissa = boundless composite^{-ordinate}$: namely, each summand is the\nnumber of points of $complement$ with $ordinate=aggregate$.\n\nOn the right side, the exponent of $composite$ in the prime factorization of $\\lcm(1,\n\\dots, \\lfloor boundless/aggregate \\rfloor)$ is $\\lfloor \\log_{composite} \\lfloor boundless/aggregate \\rfloor \\rfloor\n= \\lfloor \\log_{composite} (boundless/aggregate) \\rfloor$.  However, this is precisely the number\nof points of $complement$ with $abscissa=aggregate$. Thus\n\\[\n\\sum_{aggregate=1}^{boundless}\n\\lfloor \\log_{composite} \\lfloor boundless/aggregate \\rfloor \\rfloor\n= \\sum_{aggregate =1}^{boundless} \\left\\lfloor \\frac{boundless}{composite^{aggregate}} \\right\\rfloor,\n\\]\nand the desired result follows.\n\n\\textbf{Second solution:}\nWe prove the result by induction on $boundless$, the case $boundless=1$ being obvious.\nWhat we actually show is that going from $boundless-1$ to $boundless$ changes both\nsides by the same multiplicative factor, that is,\n\\[\nboundless = \\prod_{aggregate=1}^{boundless-1} \\frac{\\lcm\\{1, 2, \\dots, \\lfloor boundless/aggregate \\rfloor\\}}{\\lcm\n\\{1, 2, \\dots, \\lfloor (boundless-1)/aggregate \\rfloor\\}}.\n\\]\nNote that the $aggregate$-th term in the product is equal to 1 if $boundless/aggregate$ is not\nan integer, i.e., if $boundless/aggregate$ is not a divisor of $boundless$.\nIt is also equal to 1 if $boundless/aggregate$ is a divisor of $boundless$ but not a prime power,\nsince any composite number divides the lcm of all smaller numbers.\nHowever, if $boundless/aggregate$ is a power of $composite$, then the $aggregate$-th term is equal to $composite$.\n\nSince $boundless/aggregate$ runs over all proper divisors of $boundless$, the product on the right\nside includes one factor of the prime $composite$\nfor each factor of $composite$ in the prime factorization of $boundless$. Thus the whole\nproduct is indeed equal to $boundless$, completing the induction."
    },
    "garbled_string": {
      "map": {
        "i": "mdfplrqe",
        "x": "tcnvrbsa",
        "y": "gzkqfmno",
        "S": "wrvjhbkd",
        "n": "qzxwvtnp",
        "p": "hjgrksla"
      },
      "question": "Show that for each positive integer  qzxwvtnp,\n\\[\n    qzxwvtnp! = \\prod_{mdfplrqe=1}^{qzxwvtnp}  \\mathrm{lcm}\\{1, 2, \\dots, \\lfloor qzxwvtnp/mdfplrqe\\rfloor\\} .\n\\]\n(Here $\\mathrm{lcm}$ denotes the least common multiple, and\n$\\lfloor tcnvrbsa \\rfloor$ denotes the greatest integer $\\leq tcnvrbsa$.)",
      "solution": "\\textbf{First solution:}\nIt is enough to show that for each prime hjgrksla, the exponent of hjgrksla in\nthe prime factorization of both sides is the same.\nOn the left side, it is well-known that the exponent of hjgrksla in the\nprime factorization of qzxwvtnp! is\n\\[\n\\sum_{mdfplrqe =1}^{qzxwvtnp} \\left\\lfloor \\frac{qzxwvtnp}{hjgrksla^{mdfplrqe}} \\right\\rfloor.\n\\]\n(To see this, note that the mdfplrqe-th term counts the multiples of hjgrksla^{mdfplrqe} among\n1, \\dots, qzxwvtnp, so that a number divisible exactly by hjgrksla^{mdfplrqe} gets counted\nexactly mdfplrqe times.)  This number can be reinterpreted as the cardinality\nof the set wrvjhbkd of points in the plane with positive integer coordinates\nlying on or under the curve gzkqfmno = qzxwvtnp hjgrksla^{-tcnvrbsa}: namely, each summand is the\nnumber of points of wrvjhbkd with tcnvrbsa=mdfplrqe.\n\nOn the right side, the exponent of hjgrksla in the prime factorization of $\\lcm(1,\n\\dots, \\lfloor qzxwvtnp/mdfplrqe \\rfloor)$ is $\\lfloor \\log_{hjgrksla} \\lfloor qzxwvtnp/mdfplrqe \\rfloor \\rfloor\n= \\lfloor \\log_{hjgrksla} (qzxwvtnp/mdfplrqe) \\rfloor$.  However, this is precisely the number\nof points of wrvjhbkd with gzkqfmno=mdfplrqe. Thus\n\\[\n\\sum_{mdfplrqe=1}^{qzxwvtnp}\n\\lfloor \\log_{hjgrksla} \\lfloor qzxwvtnp/mdfplrqe \\rfloor \\rfloor\n= \\sum_{mdfplrqe =1}^{qzxwvtnp} \\left\\lfloor \\frac{qzxwvtnp}{hjgrksla^{mdfplrqe}} \\right\\rfloor,\n\\]\nand the desired result follows.\n\n\\textbf{Second solution:}\nWe prove the result by induction on qzxwvtnp, the case qzxwvtnp=1 being obvious.\nWhat we actually show is that going from qzxwvtnp-1 to qzxwvtnp changes both\nsides by the same multiplicative factor, that is,\n\\[\nqzxwvtnp = \\prod_{mdfplrqe=1}^{qzxwvtnp-1} \\frac{\\lcm\\{1, 2, \\dots, \\lfloor qzxwvtnp/mdfplrqe \\rfloor\\}}{\\lcm\n\\{1, 2, \\dots, \\lfloor (qzxwvtnp-1)/mdfplrqe \\rfloor\\}}.\n\\]\nNote that the mdfplrqe-th term in the product is equal to 1 if qzxwvtnp/mdfplrqe is not\nan integer, i.e., if qzxwvtnp/mdfplrqe is not a divisor of qzxwvtnp.\nIt is also equal to 1 if qzxwvtnp/mdfplrqe is a divisor of qzxwvtnp but not a prime power,\nsince any composite number divides the lcm of all smaller numbers.\nHowever, if qzxwvtnp/mdfplrqe is a power of hjgrksla, then the mdfplrqe-th term is equal to hjgrksla.\n\nSince qzxwvtnp/mdfplrqe runs over all proper divisors of qzxwvtnp, the product on the right\nside includes one factor of the prime hjgrksla\nfor each factor of hjgrksla in the prime factorization of qzxwvtnp. Thus the whole\nproduct is indeed equal to qzxwvtnp, completing the induction."
    },
    "kernel_variant": {
      "question": "Let K be a number field with ring of integers 𝒪_K.  \nFor a finite non-empty subset S ⊂ 𝒪_K \\ {0} define the ideal–theoretic least common multiple by  \n\n  lcm_{𝒪_K}(S) := ⋂_{β∈S} β 𝒪_K  \n        = ⋂_{𝔭} 𝔭^{\\,max_{β∈S} v_{𝔭}(β)},  \n\nand put lcm_{𝒪_K}(∅)=𝒪_K.\n\nProve that for every positive integer n one has the following equality of (principal) ideals of 𝒪_K:  \n\n           (n!) 𝒪_K = ∏_{i=1}^{n} lcm_{𝒪_K}{1,2,…,⌊n/i⌋}.  (★)\n\n(Here ⌊x⌋ denotes the greatest integer ≤ x.  For indices i with ⌊n/i⌋=0 the empty lcm is interpreted as 𝒪_K and hence contributes the unit ideal.)",
      "solution": "We work in the Dedekind domain 𝒪_K.  An ideal is uniquely determined by its valuations at all non-zero prime ideals, so it suffices to compare the 𝔭-adic valuations on both sides of (★) for every prime ideal 𝔭⊂𝒪_K.\n\nNotation  \n• p := 𝔭 ∩ ℤ is the rational prime below 𝔭;  \n• e_𝔭 and f_𝔭 are the ramification index and residue degree of 𝔭 over p;  \n• v_p(·) is the usual p-adic valuation on ℤ.  \nBecause ℤ → 𝒪_K is integral, for any m ∈ ℤ we have  \n\n  v_{𝔭}(m)=e_{𝔭}·v_p(m).                                        (1)\n\nStep 1. The 𝔭-adic valuation of the left–hand side  \nApplying (1) together with Legendre’s formula gives  \n\n v_{𝔭}(n!)=e_{𝔭}·v_p(n!)=e_{𝔭}·∑_{k≥1}⌊n/p^{k}⌋.             (2)\n\nStep 2. The 𝔭-adic valuation of each factor on the right  \nFix i with 1 ≤ i ≤ n and set m_i:=⌊n/i⌋.  In the list 1,2,…,m_i the largest power of p dividing any entry is p^{⌊log_p m_i⌋}; hence  \n\n v_p(lcm_{ℤ}{1,…,m_i}) = ⌊log_p m_i⌋.                         (3)\n\nUsing (1) we get  \n\n v_{𝔭}(lcm_{𝒪_K}{1,…,m_i}) = e_{𝔭}·⌊log_p m_i⌋.             (4)\n\nStep 3. The 𝔭-adic valuation of the complete product  \nSumming (4) over i gives  \n\n v_{𝔭} (RHS of (★)) = ∑_{i=1}^{n} e_{𝔭}·⌊log_p ⌊n/i⌋⌋  \n           = e_{𝔭}·∑_{i=1}^{n}⌊log_p ⌊n/i⌋⌋.             (5)\n\nStep 4. A combinatorial identity  \nExactly as in the classical integer proof, one shows  \n\n  ∑_{i=1}^{n}⌊log_p ⌊n/i⌋⌋ = ∑_{k≥1}⌊n/p^{k}⌋.                (6)\n\n(A convenient interpretation is that both sides count the lattice points with positive integral coordinates lying on or beneath the curve y = n p^{–x}.)\n\nSubstituting (6) into (5) yields  \n\n v_{𝔭}(RHS) = e_{𝔭}·∑_{k≥1}⌊n/p^{k}⌋.                        (7)\n\nStep 5. Equality of valuations  \nComparing (2) and (7) we have v_{𝔭}(LHS)=v_{𝔭}(RHS) for every 𝔭, hence the ideals agree:\n\n  (n!) 𝒪_K = ∏_{i=1}^{n} lcm_{𝒪_K}{1,2,…,⌊n/i⌋}. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.781703",
        "was_fixed": false,
        "difficulty_analysis": "• The original statement concerns ordinary integers; the enhanced variant moves to the ring of integers of an arbitrary number field, forcing the solver to work simultaneously with all prime ideals lying over each rational prime.  \n• One must understand ideal–theoretic lcms, ramification indices, and how valuations behave on principal ideals generated by rational integers (formula (1)).  \n• The Legendre formula has to be lifted to every 𝔭, introducing the ramification factor e_𝔭.  \n• The combinatorial lattice-point argument survives but now has to be carefully interwoven with algebraic-number-theoretic data.  \n• All of these ingredients—Dedekind domains, prime ideal decomposition, ramification, and valuation theory—are far beyond the elementary techniques needed for the original identity, making this kernel variant substantially more intricate and conceptually deeper."
      }
    },
    "original_kernel_variant": {
      "question": "Let K be a number field with ring of integers 𝒪_K.  \nFor a finite non-empty subset S ⊂ 𝒪_K \\ {0} define the ideal–theoretic least common multiple by  \n\n  lcm_{𝒪_K}(S) := ⋂_{β∈S} β 𝒪_K  \n        = ⋂_{𝔭} 𝔭^{\\,max_{β∈S} v_{𝔭}(β)},  \n\nand put lcm_{𝒪_K}(∅)=𝒪_K.\n\nProve that for every positive integer n one has the following equality of (principal) ideals of 𝒪_K:  \n\n           (n!) 𝒪_K = ∏_{i=1}^{n} lcm_{𝒪_K}{1,2,…,⌊n/i⌋}.  (★)\n\n(Here ⌊x⌋ denotes the greatest integer ≤ x.  For indices i with ⌊n/i⌋=0 the empty lcm is interpreted as 𝒪_K and hence contributes the unit ideal.)",
      "solution": "We work in the Dedekind domain 𝒪_K.  An ideal is uniquely determined by its valuations at all non-zero prime ideals, so it suffices to compare the 𝔭-adic valuations on both sides of (★) for every prime ideal 𝔭⊂𝒪_K.\n\nNotation  \n• p := 𝔭 ∩ ℤ is the rational prime below 𝔭;  \n• e_𝔭 and f_𝔭 are the ramification index and residue degree of 𝔭 over p;  \n• v_p(·) is the usual p-adic valuation on ℤ.  \nBecause ℤ → 𝒪_K is integral, for any m ∈ ℤ we have  \n\n  v_{𝔭}(m)=e_{𝔭}·v_p(m).                                        (1)\n\nStep 1. The 𝔭-adic valuation of the left–hand side  \nApplying (1) together with Legendre’s formula gives  \n\n v_{𝔭}(n!)=e_{𝔭}·v_p(n!)=e_{𝔭}·∑_{k≥1}⌊n/p^{k}⌋.             (2)\n\nStep 2. The 𝔭-adic valuation of each factor on the right  \nFix i with 1 ≤ i ≤ n and set m_i:=⌊n/i⌋.  In the list 1,2,…,m_i the largest power of p dividing any entry is p^{⌊log_p m_i⌋}; hence  \n\n v_p(lcm_{ℤ}{1,…,m_i}) = ⌊log_p m_i⌋.                         (3)\n\nUsing (1) we get  \n\n v_{𝔭}(lcm_{𝒪_K}{1,…,m_i}) = e_{𝔭}·⌊log_p m_i⌋.             (4)\n\nStep 3. The 𝔭-adic valuation of the complete product  \nSumming (4) over i gives  \n\n v_{𝔭} (RHS of (★)) = ∑_{i=1}^{n} e_{𝔭}·⌊log_p ⌊n/i⌋⌋  \n           = e_{𝔭}·∑_{i=1}^{n}⌊log_p ⌊n/i⌋⌋.             (5)\n\nStep 4. A combinatorial identity  \nExactly as in the classical integer proof, one shows  \n\n  ∑_{i=1}^{n}⌊log_p ⌊n/i⌋⌋ = ∑_{k≥1}⌊n/p^{k}⌋.                (6)\n\n(A convenient interpretation is that both sides count the lattice points with positive integral coordinates lying on or beneath the curve y = n p^{–x}.)\n\nSubstituting (6) into (5) yields  \n\n v_{𝔭}(RHS) = e_{𝔭}·∑_{k≥1}⌊n/p^{k}⌋.                        (7)\n\nStep 5. Equality of valuations  \nComparing (2) and (7) we have v_{𝔭}(LHS)=v_{𝔭}(RHS) for every 𝔭, hence the ideals agree:\n\n  (n!) 𝒪_K = ∏_{i=1}^{n} lcm_{𝒪_K}{1,2,…,⌊n/i⌋}. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.598829",
        "was_fixed": false,
        "difficulty_analysis": "• The original statement concerns ordinary integers; the enhanced variant moves to the ring of integers of an arbitrary number field, forcing the solver to work simultaneously with all prime ideals lying over each rational prime.  \n• One must understand ideal–theoretic lcms, ramification indices, and how valuations behave on principal ideals generated by rational integers (formula (1)).  \n• The Legendre formula has to be lifted to every 𝔭, introducing the ramification factor e_𝔭.  \n• The combinatorial lattice-point argument survives but now has to be carefully interwoven with algebraic-number-theoretic data.  \n• All of these ingredients—Dedekind domains, prime ideal decomposition, ramification, and valuation theory—are far beyond the elementary techniques needed for the original identity, making this kernel variant substantially more intricate and conceptually deeper."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}