{
  "index": "1974-A-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "A-6. It is well known that the value of the polynomial \\( (x+1)(x+2) \\cdots(x+n) \\) is exactly divisible by \\( n \\) for every integer \\( x \\). Given \\( n \\), let \\( \\boldsymbol{k}=\\boldsymbol{k}(\\boldsymbol{n}) \\) be the minimal degree of any monic integral polynomial\n\\[\nf(x)=x^{k}+a_{1} x^{k-1}+\\cdots+a_{k}\n\\]\n(with integer coefficients and leading coefficient 1) such that the value of \\( f(x) \\) is exactly divisible by \\( n \\) for every integer \\( \\boldsymbol{x} \\).\n\nFind the relationship between \\( n \\) and \\( k=k(n) \\). In particular, find the value of \\( k \\) corresponding to \\( n=1000000 \\).",
  "solution": "A-6.\nLet \\( p(k, x) \\) be the monic polynomial \\( (x+1)(x+2) \\cdots(x+k) \\) and let \\( m \\) be an integer. Then \\( p(k, m) \\) is exactly divisible by \\( k \\) ! since the absolute value of the quotient is a binomial coefficient (even when \\( m \\) is negative). Hence, if \\( n \\mid k! \\) there is a monic integral polynomial \\( f(x) \\) of degree \\( k \\) with \\( n \\mid f(m) \\) for all integers \\( m \\). Conversely, the condition \\( n \\mid k! \\) is necessary since the \\( k \\)-th difference \\( k \\) ! of a monic integral polynomial of degree \\( k \\) is divisible by any common divisor of all the values \\( f(m) \\).\n\nIn particular, \\( k\\left(10^{6}\\right)=k\\left(5^{6} 2^{6}\\right)=25 \\) since the smallest \\( s \\) with \\( 5^{6} \\mid s! \\) is \\( s=25 \\).",
  "vars": [
    "x",
    "m"
  ],
  "params": [
    "n",
    "k",
    "p",
    "a_1",
    "a_k-1",
    "a_k",
    "s"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "inputvar",
        "m": "intvalue",
        "n": "modulusval",
        "k": "mindegree",
        "p": "polyprod",
        "a_1": "coeffone",
        "a_k-1": "coeffprev",
        "a_k": "coefflast",
        "s": "smallestn"
      },
      "question": "A-6. It is well known that the value of the polynomial \\( (inputvar+1)(inputvar+2) \\cdots(inputvar+modulusval) \\) is exactly divisible by \\( modulusval \\) for every integer \\( inputvar \\). Given \\( modulusval \\), let \\( \\boldsymbol{mindegree}=\\boldsymbol{mindegree}(\\boldsymbol{modulusval}) \\) be the minimal degree of any monic integral polynomial\n\\[\nf(inputvar)=inputvar^{mindegree}+coeffone\\, inputvar^{mindegree-1}+\\cdots+coefflast\n\\]\n(with integer coefficients and leading coefficient 1) such that the value of \\( f(inputvar) \\) is exactly divisible by \\( modulusval \\) for every integer \\( \\boldsymbol{inputvar} \\).\n\nFind the relationship between \\( modulusval \\) and \\( mindegree=mindegree(modulusval) \\). In particular, find the value of \\( mindegree \\) corresponding to \\( modulusval=1000000 \\).",
      "solution": "A-6.\nLet \\( polyprod(mindegree, inputvar) \\) be the monic polynomial \\( (inputvar+1)(inputvar+2) \\cdots(inputvar+mindegree) \\) and let \\( intvalue \\) be an integer. Then \\( polyprod(mindegree, intvalue) \\) is exactly divisible by \\( mindegree! \\) since the absolute value of the quotient is a binomial coefficient (even when \\( intvalue \\) is negative). Hence, if \\( modulusval \\mid mindegree! \\) there is a monic integral polynomial \\( f(inputvar) \\) of degree \\( mindegree \\) with \\( modulusval \\mid f(intvalue) \\) for all integers \\( intvalue \\). Conversely, the condition \\( modulusval \\mid mindegree! \\) is necessary since the \\( mindegree \\)-th difference \\( mindegree! \\) of a monic integral polynomial of degree \\( mindegree \\) is divisible by any common divisor of all the values \\( f(intvalue) \\).\n\nIn particular, \\( mindegree\\left(10^{6}\\right)=mindegree\\left(5^{6} 2^{6}\\right)=25 \\) since the smallest \\( smallestn \\) with \\( 5^{6} \\mid smallestn! \\) is \\( smallestn=25 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "windswept",
        "m": "blackthorn",
        "n": "sailcloth",
        "k": "stonework",
        "p": "daffodils",
        "a_1": "marigolds",
        "a_k-1": "lighthouse",
        "a_k": "cherrywine",
        "s": "dragonfly"
      },
      "question": "A-6. It is well known that the value of the polynomial \\( (windswept+1)(windswept+2) \\cdots(windswept+sailcloth) \\) is exactly divisible by \\( sailcloth \\) for every integer \\( windswept \\). Given \\( sailcloth \\), let \\( \\boldsymbol{stonework}=\\boldsymbol{stonework}(\\boldsymbol{sailcloth}) \\) be the minimal degree of any monic integral polynomial\n\\[\nf(windswept)=windswept^{stonework}+marigolds\\, windswept^{stonework-1}+\\cdots+cherrywine\n\\]\n(with integer coefficients and leading coefficient 1) such that the value of \\( f(windswept) \\) is exactly divisible by \\( sailcloth \\) for every integer \\( \\boldsymbol{windswept} \\).\n\nFind the relationship between \\( sailcloth \\) and \\( stonework=stonework(sailcloth) \\). In particular, find the value of \\( stonework \\) corresponding to \\( sailcloth=1000000 \\).",
      "solution": "A-6.\nLet \\( daffodils(stonework, windswept) \\) be the monic polynomial \\( (windswept+1)(windswept+2) \\cdots(windswept+stonework) \\) and let \\( blackthorn \\) be an integer. Then \\( daffodils(stonework, blackthorn) \\) is exactly divisible by \\( stonework ! \\) since the absolute value of the quotient is a binomial coefficient (even when \\( blackthorn \\) is negative). Hence, if \\( sailcloth \\mid stonework! \\) there is a monic integral polynomial \\( f(windswept) \\) of degree \\( stonework \\) with \\( sailcloth \\mid f(blackthorn) \\) for all integers \\( blackthorn \\). Conversely, the condition \\( sailcloth \\mid stonework! \\) is necessary since the \\( stonework \\)-th difference \\( stonework ! \\) of a monic integral polynomial of degree \\( stonework \\) is divisible by any common divisor of all the values \\( f(blackthorn) \\).\n\nIn particular, \\( stonework\\left(10^{6}\\right)=stonework\\left(5^{6} 2^{6}\\right)=25 \\) since the smallest \\( dragonfly \\) with \\( 5^{6} \\mid dragonfly! \\) is \\( dragonfly=25 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "m": "fractionalnumber",
        "n": "boundless",
        "k": "maximumdegree",
        "p": "monomial",
        "a_1": "variableone",
        "a_k-1": "variabletwo",
        "a_k": "variablethree",
        "s": "finishvalue"
      },
      "question": "A-6. It is well known that the value of the polynomial \\( (constantvalue+1)(constantvalue+2) \\cdots(constantvalue+boundless) \\) is exactly divisible by \\( boundless \\) for every integer \\( constantvalue \\). Given \\( boundless \\), let \\( \\boldsymbol{maximumdegree}=\\boldsymbol{maximumdegree}(\\boldsymbol{boundless}) \\) be the minimal degree of any monic integral polynomial\n\\[\nf(constantvalue)=constantvalue^{maximumdegree}+variableone\\, constantvalue^{maximumdegree-1}+\\cdots+variablethree\n\\]\n(with integer coefficients and leading coefficient 1) such that the value of \\( f(constantvalue) \\) is exactly divisible by \\( boundless \\) for every integer \\( \\boldsymbol{constantvalue} \\).\n\nFind the relationship between \\( boundless \\) and \\( maximumdegree=maximumdegree(boundless) \\). In particular, find the value of \\( maximumdegree \\) corresponding to \\( boundless=1000000 \\).",
      "solution": "A-6.\nLet \\( monomial(maximumdegree, constantvalue) \\) be the monic polynomial \\( (constantvalue+1)(constantvalue+2) \\cdots(constantvalue+maximumdegree) \\) and let \\( fractionalnumber \\) be an integer. Then \\( monomial(maximumdegree, fractionalnumber) \\) is exactly divisible by \\( maximumdegree \\) ! since the absolute value of the quotient is a binomial coefficient (even when \\( fractionalnumber \\) is negative). Hence, if \\( boundless \\mid maximumdegree! \\) there is a monic integral polynomial \\( f(constantvalue) \\) of degree \\( maximumdegree \\) with \\( boundless \\mid f(fractionalnumber) \\) for all integers \\( fractionalnumber \\). Conversely, the condition \\( boundless \\mid maximumdegree! \\) is necessary since the \\( maximumdegree \\)-th difference \\( maximumdegree \\) ! of a monic integral polynomial of degree \\( maximumdegree \\) is divisible by any common divisor of all the values \\( f(fractionalnumber) \\).\n\nIn particular, \\( maximumdegree\\left(10^{6}\\right)=maximumdegree\\left(5^{6} 2^{6}\\right)=25 \\) since the smallest \\( finishvalue \\) with \\( 5^{6} \\mid finishvalue! \\) is \\( finishvalue=25 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "m": "hjgrksla",
        "n": "vckezfub",
        "k": "rbyiqdmo",
        "p": "lxshcenq",
        "a_1": "ujtoarhf",
        "a_k-1": "ginwlexb",
        "a_k": "perfqasu",
        "s": "owimznky"
      },
      "question": "A-6. It is well known that the value of the polynomial \\( (qzxwvtnp+1)(qzxwvtnp+2) \\cdots(qzxwvtnp+vckezfub) \\) is exactly divisible by \\( vckezfub \\) for every integer \\( qzxwvtnp \\). Given \\( vckezfub \\), let \\( \\boldsymbol{rbyiqdmo}=\\boldsymbol{rbyiqdmo}(\\boldsymbol{vckezfub}) \\) be the minimal degree of any monic integral polynomial\n\\[\nf(qzxwvtnp)=qzxwvtnp^{rbyiqdmo}+ujtoarhf\\, qzxwvtnp^{rbyiqdmo-1}+\\cdots+perfqasu\n\\]\n(with integer coefficients and leading coefficient 1) such that the value of \\( f(qzxwvtnp) \\) is exactly divisible by \\( vckezfub \\) for every integer \\( \\boldsymbol{qzxwvtnp} \\).\n\nFind the relationship between \\( vckezfub \\) and \\( rbyiqdmo=rbyiqdmo(vckezfub) \\). In particular, find the value of \\( rbyiqdmo \\) corresponding to \\( vckezfub=1000000 \\).",
      "solution": "A-6.\nLet \\( lxshcenq(rbyiqdmo, qzxwvtnp) \\) be the monic polynomial \\( (qzxwvtnp+1)(qzxwvtnp+2) \\cdots(qzxwvtnp+rbyiqdmo) \\) and let \\( hjgrksla \\) be an integer. Then \\( lxshcenq(rbyiqdmo, hjgrksla) \\) is exactly divisible by \\( rbyiqdmo! \\) since the absolute value of the quotient is a binomial coefficient (even when \\( hjgrksla \\) is negative). Hence, if \\( vckezfub \\mid rbyiqdmo! \\) there is a monic integral polynomial \\( f(qzxwvtnp) \\) of degree \\( rbyiqdmo \\) with \\( vckezfub \\mid f(hjgrksla) \\) for all integers \\( hjgrksla \\). Conversely, the condition \\( vckezfub \\mid rbyiqdmo! \\) is necessary since the \\( rbyiqdmo \\)-th difference \\( rbyiqdmo! \\) of a monic integral polynomial of degree \\( rbyiqdmo \\) is divisible by any common divisor of all the values \\( f(hjgrksla) \\).\n\nIn particular, \\( rbyiqdmo\\left(10^{6}\\right)=rbyiqdmo\\left(5^{6} 2^{6}\\right)=25 \\) since the smallest \\( owimznky \\) with \\( 5^{6} \\mid owimznky! \\) is \\( owimznky=25 \\)."
    },
    "kernel_variant": {
      "question": "Fix integers n ≥ 1 and r ≥ 2 and, for k ∈ ℕ, let  \n\n  P_k (n , r) = { monic f(x)=x^{k}+a_{1}x^{k-1}+⋯+a_{k}∈ℤ[x] : n^{r} divides f(m) for every m∈ℤ }.  \n\nPut  \n\n  k_r (n)=min{ k∈ℕ : P_k (n , r)≠∅ }.\n\na)  Prove that\n  k_r (n)=min{ k∈ℕ : n^{r} | k! }.\n\nb)  Compute k_2 (27720).\n\nc)  (Fine structure of the extremal polynomials)\n   Write k:=k_r (n) and, for 0≤j≤k−1, put  \n\n  d_j :=gcd(j!, n^{r}).  \n\n   Show that every polynomial f∈P_k (n , r) can be written uniquely in the form  \n\n  f(x)=k!·C(x,k)+n^{r}·g(x)  (∗)  \n\n   where C(x,k)=\\(\\binom{x}{k}\\) and  \n\n  g(x)=∑_{j=0}^{k-1} c_j C(x,j)  (c_j ∈ℤ)          (†)  \n\n   with the additional integrality conditions  \n\n  j! divides n^{r}·c_j  for every 0≤j≤k−1.          (‡)  \n\n   Conversely, any choice of integers c_0 ,…,c_{k-1} satisfying (‡) produces, via (∗), a polynomial that belongs to P_k (n , r).\n\nd)  (Counting residue classes)\n   On P_k (n , r) consider the two equivalence relations  \n\n  f ≡_c g  ⇔  every coefficient of f−g lies in n^{r}ℤ,  \n  f ≡_v g  ⇔  f(m)≡g(m) (mod n^{r}) for all m∈ℤ.  \n\n   Determine  \n\n  |P_k (n , r)/≡_c | and |P_k (n , r)/≡_v |  \n\n   in terms of the numbers d_0 ,…,d_{k-1}.",
      "solution": "Throughout we work with the falling–factorial (binomial) basis  \n\n  C(x,j)=\\(\\binom{x}{j}\\)=x^{\\underline{j}}/j!  (0≤j≤k),  \n\nand with the forward–difference operator Δp(x)=p(x+1)−p(x).  \nTwo facts will be used repeatedly:\n\n(1)  For j≥1 we have ΔC(x,j)=C(x,j−1).\n\n(2)  If p(x)=∑_{j=0}^{m} b_j C(x,j) with b_j ∈ℤ, then the j-fold difference Δ^{j}p(x) is the constant polynomial b_j .\n\nFrom (2) it follows immediately that  \n\n  n^{r} | p(m) ∀m∈ℤ ⇔ n^{r} | b_j for every j. (∗∗)\n\n\n\na)  Determination of k_r (n)\n\nNecessity.  \nLet f(x)=∑_{j=0}^{k} b_j C(x,j) be any element of P_k (n , r).  \nBy (1) and induction, Δ^{k}f(x)=b_k =k!.  Since all k-fold differences of f\nare divisible by n^{r}, so is k!; hence n^{r} | k!.\n\nSufficiency.  \nIf n^{r} | k!, consider f(x)=k!·C(x,k).  Its leading term is x^{k}, so f is monic; and f(m)=k!·\\(\\binom{m}{k}\\) is always a multiple of k!, hence of n^{r}.  Thus f∈P_k (n , r).\n\nTherefore  \n\n  k_r (n)=min{ k : n^{r} | k! }. ∎\n\n\n\nb)  The value k_2 (27720)\n\nFactor n=27720=2³·3²·5·7·11; hence n²=2⁶·3⁴·5²·7²·11².  \nUsing Legendre’s formula v_p (k!)=∑_{t≥1}⌊k/p^{t}⌋ we look for the least k with\n\n v₂(k!)≥6, v₃(k!)≥4, v₅(k!)≥2, v₇(k!)≥2, v₁₁(k!)≥2.\n\nA routine check gives the first k satisfying these simultaneously:\n\n  k=22.\n\nConsequently k_2 (27720)=22. ∎\n\n\n\nc)  Fine structure\n\nFix k=k_r (n) and write every f∈P_k (n , r) in the binomial basis:\n\n  f(x)=∑_{j=0}^{k} b_j C(x,j),  b_k =k!.\n\nBecause n^{r} | f(m) for all m, property (∗∗) yields n^{r} | b_j for j<k.  \nWrite b_j =n^{r} c_j  (j<k) with c_j ∈ℤ.  Then\n\n  f(x)=k!·C(x,k)+n^{r}·g(x) with g(x)=∑_{j=0}^{k-1} c_j C(x,j).\n\nThis proves the existence of a decomposition (∗).  \nFor (∗) to define a polynomial with integral coefficients we must have\n j! | n^{r} c_j   (0≤j≤k−1),  \nbecause j!·C(x,j) has integral coefficients.  Conditions (‡) are therefore necessary.\n\nConversely, take arbitrary integers c_0 ,…,c_{k-1} satisfying (‡).  \nSet g(x)=∑_{j} c_j C(x,j) and define f by (∗).  Then f is monic of degree k and clearly lies in ℤ[x].  Moreover n^{r}|f(m) for every integer m because the second summand in (∗) is already divisible by n^{r}.  Hence f∈P_k (n , r).\n\nUniqueness of the representation (∗) follows from the ℚ-linear independence of the functions C(x,0),…,C(x,k) on ℤ. ∎\n\n\n\nd)  Counting the residue classes\n\nWrite again L_j := j!/d_j = j! / gcd(j!, n^{r}).  \nEvery f∈P_k (n , r) corresponds to a unique k-tuple (s_0 ,…,s_{k-1})∈ℤ^{k} via\n\n  c_j = L_j · s_j  (0≤j≤k−1),\n\nand therefore  \n\n  f(x)=k!·C(x,k)+∑_{j=0}^{k-1} n^{r}L_j s_j C(x,j).  (♣)\n\n1.  Functional reduction (≡_v).  \nFor f,g described by (♣) the difference is\n\n  f(x)−g(x)=n^{r}·h(x) with h(x)=∑_{j=0}^{k-1} L_j (s_j−t_j) C(x,j)∈Int(ℤ).\n\nHence n^{r} divides (f−g)(m) for every integer m, i.e. f≡_v g.  \nThus all extremal polynomials are functionally congruent and\n\n  |P_k (n , r)/≡_v | = 1. (8)\n\n\n2.  Coefficientwise reduction (≡_c).  \nTwo polynomials f,g (with parameters s and t) are ≡_c  \niff all coefficients of f−g are multiples of n^{r}.  \nLooking at the coefficient of x^{j} with the largest index j\nfor which s_j≠t_j shows (because the matrix transforming the basis\n{ C(x,j) } to the monomial basis is unit-triangular) that this happens exactly when\n\n  s_j ≡ t_j (mod d_j)  for every j.\n\nHence the map  \n\n  ψ : P_k (n , r) → ⨁_{j=0}^{k-1} ℤ/d_j ℤ,  ψ(f)=(s_0 (mod d_0),…,s_{k-1} (mod d_{k-1}))\n\ninduces a bijection between ≡_c-classes and the finite product ring on the\nright.  Therefore  \n\n  |P_k (n , r)/≡_c | = ∏_{j=0}^{k-1} d_j = ∏_{j=0}^{k-1} gcd(j!, n^{r}). (9)\n\nEquations (8) and (9) complete the answer. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.613978",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional structural layer – forward differences up to order s introduce the falling-factorial (k)_{s} instead of the single factorial k!, forcing the solver to work with several consecutive integers rather than one cumulative product.  \n\n2. Interaction of many constraints – each prime factor of n behaves differently inside a short block of consecutive integers; analysing which primes must coexist in the *same* term and which can be split is considerably subtler than checking valuations in k!.  \n\n3. Congruence synthesis – the necessity of making one designated integer simultaneously satisfy several high prime-power divisibilities while controlling its residues modulo 64 demands non-trivial modular arithmetic.  \n\n4. Enormous search space – the minimal k is around 1.8·10^{8}, far outside the range approachable by naïve brute force; the argument must be purely theoretical.  \n\n5. Layered reasoning – the solution combines valuation theory, combinatorial number theory (distribution of multiples in short intervals), Newton series, and Chinese-remainder-type congruence solving, going well beyond the original factorial divisibility argument."
      }
    },
    "original_kernel_variant": {
      "question": "Fix integers n ≥ 1 and r ≥ 2 and, for k ∈ ℕ, let  \n\n  P_k (n , r) = { monic f(x)=x^{k}+a_{1}x^{k-1}+⋯+a_{k}∈ℤ[x] : n^{r} divides f(m) for every m∈ℤ }.  \n\nPut  \n\n  k_r (n)=min{ k∈ℕ : P_k (n , r)≠∅ }.\n\na)  Prove that\n  k_r (n)=min{ k∈ℕ : n^{r} | k! }.\n\nb)  Compute k_2 (27720).\n\nc)  (Fine structure of the extremal polynomials)\n   Write k:=k_r (n) and, for 0≤j≤k−1, put  \n\n  d_j :=gcd(j!, n^{r}).  \n\n   Show that every polynomial f∈P_k (n , r) can be written uniquely in the form  \n\n  f(x)=k!·C(x,k)+n^{r}·g(x)  (∗)  \n\n   where C(x,k)=\\(\\binom{x}{k}\\) and  \n\n  g(x)=∑_{j=0}^{k-1} c_j C(x,j)  (c_j ∈ℤ)          (†)  \n\n   with the additional integrality conditions  \n\n  j! divides n^{r}·c_j  for every 0≤j≤k−1.          (‡)  \n\n   Conversely, any choice of integers c_0 ,…,c_{k-1} satisfying (‡) produces, via (∗), a polynomial that belongs to P_k (n , r).\n\nd)  (Counting residue classes)\n   On P_k (n , r) consider the two equivalence relations  \n\n  f ≡_c g  ⇔  every coefficient of f−g lies in n^{r}ℤ,  \n  f ≡_v g  ⇔  f(m)≡g(m) (mod n^{r}) for all m∈ℤ.  \n\n   Determine  \n\n  |P_k (n , r)/≡_c | and |P_k (n , r)/≡_v |  \n\n   in terms of the numbers d_0 ,…,d_{k-1}.",
      "solution": "Throughout we work with the falling–factorial (binomial) basis  \n\n  C(x,j)=\\(\\binom{x}{j}\\)=x^{\\underline{j}}/j!  (0≤j≤k),  \n\nand with the forward–difference operator Δp(x)=p(x+1)−p(x).  \nTwo facts will be used repeatedly:\n\n(1)  For j≥1 we have ΔC(x,j)=C(x,j−1).\n\n(2)  If p(x)=∑_{j=0}^{m} b_j C(x,j) with b_j ∈ℤ, then the j-fold difference Δ^{j}p(x) is the constant polynomial b_j .\n\nFrom (2) it follows immediately that  \n\n  n^{r} | p(m) ∀m∈ℤ ⇔ n^{r} | b_j for every j. (∗∗)\n\n\n\na)  Determination of k_r (n)\n\nNecessity.  \nLet f(x)=∑_{j=0}^{k} b_j C(x,j) be any element of P_k (n , r).  \nBy (1) and induction, Δ^{k}f(x)=b_k =k!.  Since all k-fold differences of f\nare divisible by n^{r}, so is k!; hence n^{r} | k!.\n\nSufficiency.  \nIf n^{r} | k!, consider f(x)=k!·C(x,k).  Its leading term is x^{k}, so f is monic; and f(m)=k!·\\(\\binom{m}{k}\\) is always a multiple of k!, hence of n^{r}.  Thus f∈P_k (n , r).\n\nTherefore  \n\n  k_r (n)=min{ k : n^{r} | k! }. ∎\n\n\n\nb)  The value k_2 (27720)\n\nFactor n=27720=2³·3²·5·7·11; hence n²=2⁶·3⁴·5²·7²·11².  \nUsing Legendre’s formula v_p (k!)=∑_{t≥1}⌊k/p^{t}⌋ we look for the least k with\n\n v₂(k!)≥6, v₃(k!)≥4, v₅(k!)≥2, v₇(k!)≥2, v₁₁(k!)≥2.\n\nA routine check gives the first k satisfying these simultaneously:\n\n  k=22.\n\nConsequently k_2 (27720)=22. ∎\n\n\n\nc)  Fine structure\n\nFix k=k_r (n) and write every f∈P_k (n , r) in the binomial basis:\n\n  f(x)=∑_{j=0}^{k} b_j C(x,j),  b_k =k!.\n\nBecause n^{r} | f(m) for all m, property (∗∗) yields n^{r} | b_j for j<k.  \nWrite b_j =n^{r} c_j  (j<k) with c_j ∈ℤ.  Then\n\n  f(x)=k!·C(x,k)+n^{r}·g(x) with g(x)=∑_{j=0}^{k-1} c_j C(x,j).\n\nThis proves the existence of a decomposition (∗).  \nFor (∗) to define a polynomial with integral coefficients we must have\n j! | n^{r} c_j   (0≤j≤k−1),  \nbecause j!·C(x,j) has integral coefficients.  Conditions (‡) are therefore necessary.\n\nConversely, take arbitrary integers c_0 ,…,c_{k-1} satisfying (‡).  \nSet g(x)=∑_{j} c_j C(x,j) and define f by (∗).  Then f is monic of degree k and clearly lies in ℤ[x].  Moreover n^{r}|f(m) for every integer m because the second summand in (∗) is already divisible by n^{r}.  Hence f∈P_k (n , r).\n\nUniqueness of the representation (∗) follows from the ℚ-linear independence of the functions C(x,0),…,C(x,k) on ℤ. ∎\n\n\n\nd)  Counting the residue classes\n\nWrite again L_j := j!/d_j = j! / gcd(j!, n^{r}).  \nEvery f∈P_k (n , r) corresponds to a unique k-tuple (s_0 ,…,s_{k-1})∈ℤ^{k} via\n\n  c_j = L_j · s_j  (0≤j≤k−1),\n\nand therefore  \n\n  f(x)=k!·C(x,k)+∑_{j=0}^{k-1} n^{r}L_j s_j C(x,j).  (♣)\n\n1.  Functional reduction (≡_v).  \nFor f,g described by (♣) the difference is\n\n  f(x)−g(x)=n^{r}·h(x) with h(x)=∑_{j=0}^{k-1} L_j (s_j−t_j) C(x,j)∈Int(ℤ).\n\nHence n^{r} divides (f−g)(m) for every integer m, i.e. f≡_v g.  \nThus all extremal polynomials are functionally congruent and\n\n  |P_k (n , r)/≡_v | = 1. (8)\n\n\n2.  Coefficientwise reduction (≡_c).  \nTwo polynomials f,g (with parameters s and t) are ≡_c  \niff all coefficients of f−g are multiples of n^{r}.  \nLooking at the coefficient of x^{j} with the largest index j\nfor which s_j≠t_j shows (because the matrix transforming the basis\n{ C(x,j) } to the monomial basis is unit-triangular) that this happens exactly when\n\n  s_j ≡ t_j (mod d_j)  for every j.\n\nHence the map  \n\n  ψ : P_k (n , r) → ⨁_{j=0}^{k-1} ℤ/d_j ℤ,  ψ(f)=(s_0 (mod d_0),…,s_{k-1} (mod d_{k-1}))\n\ninduces a bijection between ≡_c-classes and the finite product ring on the\nright.  Therefore  \n\n  |P_k (n , r)/≡_c | = ∏_{j=0}^{k-1} d_j = ∏_{j=0}^{k-1} gcd(j!, n^{r}). (9)\n\nEquations (8) and (9) complete the answer. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.491330",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional structural layer – forward differences up to order s introduce the falling-factorial (k)_{s} instead of the single factorial k!, forcing the solver to work with several consecutive integers rather than one cumulative product.  \n\n2. Interaction of many constraints – each prime factor of n behaves differently inside a short block of consecutive integers; analysing which primes must coexist in the *same* term and which can be split is considerably subtler than checking valuations in k!.  \n\n3. Congruence synthesis – the necessity of making one designated integer simultaneously satisfy several high prime-power divisibilities while controlling its residues modulo 64 demands non-trivial modular arithmetic.  \n\n4. Enormous search space – the minimal k is around 1.8·10^{8}, far outside the range approachable by naïve brute force; the argument must be purely theoretical.  \n\n5. Layered reasoning – the solution combines valuation theory, combinatorial number theory (distribution of multiples in short intervals), Newton series, and Chinese-remainder-type congruence solving, going well beyond the original factorial divisibility argument."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}