{
  "index": "1958-2-B-2",
  "type": "NT",
  "tag": [
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "2. Given a set of \\( n+1 \\) positive integers, none of which exceeds \\( 2 n \\), show that at least one member of the set must divide another member of the set.",
  "solution": "Solution. Every positive integer can be written uniquely in the form \\( 2^{p} q \\) where \\( p \\) is a non-negative integer and \\( q \\) is a positive odd integer, the odd part of \\( n \\). The odd part of an integer in the set \\( S=\\{1,2,3, \\ldots, 2 n\\} \\) must be one of the \\( n \\) integers \\( 1,3,5, \\ldots,(2 n-1) \\). Given \\( (n+1) \\) integers in \\( S \\), at least two must have the same odd part, that is they must be of the form\n\\[\n2^{p_{1}} q \\quad \\text { and } \\quad 2^{p_{2}} q\n\\]\nwhere \\( p_{1} \\neq p_{2} \\). We can choose the notation so that \\( p_{1}<p_{2} \\); then \\( 2^{p_{1}} q \\) divides \\( \\mathbf{2}^{\\boldsymbol{p}_{\\mathbf{2}} \\boldsymbol{q}} \\).",
  "vars": [
    "p",
    "p_1",
    "p_2",
    "q"
  ],
  "params": [
    "n",
    "S"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "p": "twopowerindex",
        "p_1": "firsttwopower",
        "p_2": "secondtwopower",
        "q": "oddcomponent",
        "n": "countvalue",
        "S": "integerset"
      },
      "question": "2. Given a set of \\( countvalue+1 \\) positive integers, none of which exceeds \\( 2 countvalue \\), show that at least one member of the set must divide another member of the set.",
      "solution": "Solution. Every positive integer can be written uniquely in the form \\( 2^{twopowerindex} oddcomponent \\) where \\( twopowerindex \\) is a non-negative integer and \\( oddcomponent \\) is a positive odd integer, the odd part of \\( countvalue \\). The odd part of an integer in the set \\( integerset=\\{1,2,3, \\ldots, 2 countvalue\\} \\) must be one of the \\( countvalue \\) integers \\( 1,3,5, \\ldots,(2 countvalue-1) \\). Given \\( (countvalue+1) \\) integers in \\( integerset \\), at least two must have the same odd part, that is they must be of the form\n\\[\n2^{firsttwopower} oddcomponent \\quad \\text { and } \\quad 2^{secondtwopower} oddcomponent\n\\]\nwhere \\( firsttwopower \\neq secondtwopower \\). We can choose the notation so that \\( firsttwopower<secondtwopower \\); then \\( 2^{firsttwopower} oddcomponent \\) divides \\( \\mathbf{2}^{\\boldsymbol{secondtwopower} \\boldsymbol{oddcomponent}} \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "p": "labyrinth",
        "p_1": "buttercup",
        "p_2": "wheelhouse",
        "q": "chandelier",
        "n": "sandcastle",
        "S": "raincloud"
      },
      "question": "2. Given a set of \\( sandcastle+1 \\) positive integers, none of which exceeds \\( 2 sandcastle \\), show that at least one member of the set must divide another member of the set.",
      "solution": "Solution. Every positive integer can be written uniquely in the form \\( 2^{labyrinth} chandelier \\) where \\( labyrinth \\) is a non-negative integer and \\( chandelier \\) is a positive odd integer, the odd part of \\( sandcastle \\). The odd part of an integer in the set \\( raincloud=\\{1,2,3, \\ldots, 2 sandcastle\\} \\) must be one of the \\( sandcastle \\) integers \\( 1,3,5, \\ldots,(2 sandcastle-1) \\). Given \\( (sandcastle+1) \\) integers in \\( raincloud \\), at least two must have the same odd part, that is they must be of the form\n\\[\n2^{buttercup} chandelier \\quad \\text { and } \\quad 2^{wheelhouse} chandelier\n\\]\nwhere \\( buttercup \\neq wheelhouse \\). We can choose the notation so that \\( buttercup<wheelhouse \\); then \\( 2^{buttercup} chandelier \\) divides \\( \\mathbf{2}^{\\boldsymbol{wheelhouse} \\boldsymbol{chandelier}} \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "p": "rootnumber",
        "p_1": "rootnumberone",
        "p_2": "rootnumbertwo",
        "q": "evensegment",
        "n": "nothingness",
        "S": "chaosgroup"
      },
      "question": "2. Given a set of \\( nothingness+1 \\) positive integers, none of which exceeds \\( 2\\,nothingness \\), show that at least one member of the set must divide another member of the set.",
      "solution": "Solution. Every positive integer can be written uniquely in the form \\( 2^{rootnumber} evensegment \\) where \\( rootnumber \\) is a non-negative integer and \\( evensegment \\) is a positive odd integer, the odd part of \\( nothingness \\). The odd part of an integer in the set \\( chaosgroup=\\{1,2,3, \\ldots, 2 nothingness\\} \\) must be one of the \\( nothingness \\) integers \\( 1,3,5, \\ldots,(2 nothingness-1) \\). Given \\( (nothingness+1) \\) integers in \\( chaosgroup \\), at least two must have the same odd part, that is they must be of the form\\n\\[\\n2^{rootnumberone} evensegment \\quad \\text { and } \\quad 2^{rootnumbertwo} evensegment\\n\\]\\nwhere \\( rootnumberone \\neq rootnumbertwo \\). We can choose the notation so that \\( rootnumberone<rootnumbertwo \\); then \\( 2^{rootnumberone} evensegment \\) divides \\( \\mathbf{2}^{\\boldsymbol{rootnumbertwo} \\boldsymbol{evensegment}} \\)."
    },
    "garbled_string": {
      "map": {
        "p": "hjgrksla",
        "p_1": "qzxwvtnp",
        "p_2": "fjdsoier",
        "q": "vbmtyeui",
        "n": "kdlsewpo",
        "S": "mncxropt"
      },
      "question": "2. Given a set of \\( kdlsewpo+1 \\) positive integers, none of which exceeds \\( 2 kdlsewpo \\), show that at least one member of the set must divide another member of the set.",
      "solution": "Solution. Every positive integer can be written uniquely in the form \\( 2^{hjgrksla} vbmtyeui \\) where \\( hjgrksla \\) is a non-negative integer and \\( vbmtyeui \\) is a positive odd integer, the odd part of \\( kdlsewpo \\). The odd part of an integer in the set \\( mncxropt=\\{1,2,3, \\ldots, 2 kdlsewpo\\} \\) must be one of the \\( kdlsewpo \\) integers \\( 1,3,5, \\ldots,(2 kdlsewpo-1) \\). Given \\( (kdlsewpo+1) \\) integers in \\( mncxropt \\), at least two must have the same odd part, that is they must be of the form\n\\[\n2^{qzxwvtnp} vbmtyeui \\quad \\text { and } \\quad 2^{fjdsoier} vbmtyeui\n\\]\nwhere \\( qzxwvtnp \\neq fjdsoier \\). We can choose the notation so that \\( qzxwvtnp<fjdsoier \\); then \\( 2^{qzxwvtnp} vbmtyeui \\) divides \\( \\mathbf{2}^{\\boldsymbol{fjdsoier} \\boldsymbol{vbmtyeui}} \\)."
    },
    "kernel_variant": {
      "question": "Fix a prime number p ≥ 2 and a positive integer m.  \nThroughout, the notation  \n\n  {1,2,… ,p m}\n\nrefers to the first p m positive integers (the product p·m, not the power p^m).\n\nFor two different positive integers a < b write  \n  a ≺ₚ b iff b = p^k a for some integer k ≥ 1.  \nA set of positive integers is called p-power–free if it contains no ordered pair (a,b) with a ≺ₚ b.\n\n1. (Extremal size)  \n   Determine the largest possible cardinality of a p-power–free subset of  \n   {1,2,… ,p m}.\n\n2. (Structure and enumeration)  \n   For every integer u with 1 ≤ u ≤ p m and p ∤ u set  \n\n  t(u) := ⌊log_p(p m / u)⌋  and  C_u := {p^k u : 0 ≤ k ≤ t(u)}.\n\n   The sets C_u are called p-chains.\n\n   (a) Prove that the p-chains form a partition of {1,2,… ,p m} into exactly (p−1)m classes.\n\n   (b) Show that a subset A ⊆ {1,… ,p m} is p-power–free and of maximal size iff it contains exactly one element of each chain.\n\n   (c) Let  \n\n    N(p,m) := ∏_{1 ≤ u ≤ p m,\\,p∤u} (t(u)+1).\n\n   Prove that N(p,m) equals the number of distinct extremal p-power–free sets and that  \n\n    N(p,m) ≥ 2^{\\,m − ⌊m/p⌋},\n\n   with strict inequality whenever m ≥ p.\n\n3. (Forced occurrence)  \n   Let f(p,m) be the least positive integer with the following property:\n\n  Every collection of f(p,m) distinct positive integers not exceeding p m  \n  contains two numbers whose quotient is a positive power of p.\n\n   Prove that  \n\n    f(p,m) = (p−1)m + 1.",
      "solution": "Step 0. Partition of {1,… ,p m} into p-chains  \nEvery n ∈ {1,… ,p m} can be written uniquely as n = p^v u with v ≥ 0 and p ∤ u.  Fix such a u.  Put  \n\n  t(u) := ⌊log_p(p m/u)⌋  (so p^{t(u)}u ≤ p m < p^{t(u)+1}u)\n\nand define  \n\n  C_u := {p^k u : 0 ≤ k ≤ t(u)}.                                (1)\n\nBecause the representation n = p^v u with p ∤ u is unique, the chains C_u are pairwise disjoint and cover {1,… ,p m}.  Exactly those u with 1 ≤ u ≤ p m and p ∤ u arise; there are (p−1)m such u (in every block of p consecutive integers precisely p−1 are coprime to p).  Hence the ground set is partitioned into (p−1)m disjoint p-chains.\n\n──────────────────────────────────────────────────────────────\n1. Extremal size\n\nLet A ⊆ {1,… ,p m} be p-power–free.  \nIf two distinct elements of A lay in the same chain C_u, their quotient would be a power of p, contradicting p-power–freeness.  Therefore A meets each chain in at most one element, whence\n\n  |A| ≤ number of chains = (p−1)m.                      (2)\n\nConversely, choose one arbitrary element from every chain and collect them in A.  Then |A| = (p−1)m and A is p-power–free by construction.  Hence\n\n  max{|A| : A ⊆ {1,… ,p m} p-power–free} = (p−1)m.      (3)\n\n──────────────────────────────────────────────────────────────\n2. Structure and enumeration of extremal sets\n\nBy inequality (2), a p-power–free set A with |A| = (p−1)m must intersect every chain in exactly one element.  Conversely, any such transversal is p-power–free and of maximal size.  Thus\n\n  A is extremal  ⇔  (∀u with 1 ≤ u ≤ p m, p∤u) |A ∩ C_u| = 1.    (4)\n\nFor a fixed u the chain C_u has length |C_u| = t(u)+1, so there are exactly t(u)+1 possibilities for A ∩ C_u.  The choices for different u are independent, giving\n\n  N(p,m) = ∏_{1≤u≤p m,\\,p∤u} (t(u)+1).                  (5)\n\nLower bound for N(p,m).  \nWrite  \n\n  S₁ := {u : 1 ≤ u ≤ m, p∤u},  S₂ := {u : m<u≤p m, p∤u}.\n\nIf u∈S₂ then u>m implies p m/u< p, hence t(u)=0 and the factor (t(u)+1)=1.  \nIf u∈S₁ we have u ≤ m, so p m/u ≥ p and therefore t(u) ≥ 1; consequently (t(u)+1) ≥ 2.  The number of elements in S₁ is\n\n  |S₁| = m − ⌊m/p⌋                                      (6)\n\n(the m integers 1,… ,m lose precisely ⌊m/p⌋ multiples of p).  Hence\n\n  N(p,m) = ∏_{u∈S₁}(t(u)+1) ≥ 2^{|S₁|} = 2^{m − ⌊m/p⌋}.    (7)\n\nStrict inequality when m ≥ p.  \nIf m ≥ p, then u=1 satisfies p m/u = p m ≥ p², so t(1) ≥ 2 and the corresponding factor equals at least 3.  Thus N(p,m) > 2^{m−⌊m/p⌋}.  This completes Part 2.\n\n──────────────────────────────────────────────────────────────\n3. The minimal forcing number f(p,m)\n\nLet S ⊆ {1,… ,p m} with |S| = (p−1)m + 1 be arbitrary.  \nBecause the chains C_u form a partition into only (p−1)m classes, the pigeon-hole principle forces two distinct elements of S to lie in the same chain.  Their quotient is a positive power of p, so every set of size (p−1)m + 1 already contains the forbidden pair.  Hence\n\n  f(p,m) ≤ (p−1)m + 1.                                  (8)\n\nOn the other hand, Part 1 produced a p-power–free set A of size (p−1)m.  Such a set contains no two elements whose quotient is a power of p, so\n\n  f(p,m) ≥ (p−1)m + 1.                                  (9)\n\nCombining (8) and (9) we obtain\n\n  f(p,m) = (p−1)m + 1.                                  ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.497133",
        "was_fixed": false,
        "difficulty_analysis": "1.  The original problem involves only the prime 2; the enhanced version works for an arbitrary prime p.  This forces the solver to generalise every step (counting, factorisation, and pigeon-hole arguments) to an arbitrary modulus p, demanding a clear understanding of how the argument depends on the arithmetic of 2 and how to replace it by p.\n\n2.  The number of admissible p–free parts is (p – 1)m instead of simply m for the “odd parts’’ when p = 2.  Handling this larger, prime-dependent counting correctly is an extra technical layer.\n\n3.  While the original conclusion is merely “one divides the other,’’ the new conclusion prescribes the precise quotient (a pure power of p).  Establishing this stronger statement requires keeping track of the exact difference in p-adic exponents rather than just their inequality.\n\n4.  All combinatorial bounds (how many numbers guarantee a collision of p–free parts) must be recomputed in terms of p, adding an additional variable and therefore more subtle book-keeping.\n\n5.  Conceptually, the solver must see that the argument is really about a “valuation at a prime’’ and not about parity alone; recognising and exploiting this abstraction represents a deeper theoretical requirement than the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Fix a prime number p ≥ 2 and a positive integer m.  \nThroughout, the notation  \n\n  {1,2,… ,p m}\n\nrefers to the first p m positive integers (the product p·m, not the power p^m).\n\nFor two different positive integers a < b write  \n  a ≺ₚ b iff b = p^k a for some integer k ≥ 1.  \nA set of positive integers is called p-power–free if it contains no ordered pair (a,b) with a ≺ₚ b.\n\n1. (Extremal size)  \n   Determine the largest possible cardinality of a p-power–free subset of  \n   {1,2,… ,p m}.\n\n2. (Structure and enumeration)  \n   For every integer u with 1 ≤ u ≤ p m and p ∤ u set  \n\n  t(u) := ⌊log_p(p m / u)⌋  and  C_u := {p^k u : 0 ≤ k ≤ t(u)}.\n\n   The sets C_u are called p-chains.\n\n   (a) Prove that the p-chains form a partition of {1,2,… ,p m} into exactly (p−1)m classes.\n\n   (b) Show that a subset A ⊆ {1,… ,p m} is p-power–free and of maximal size iff it contains exactly one element of each chain.\n\n   (c) Let  \n\n    N(p,m) := ∏_{1 ≤ u ≤ p m,\\,p∤u} (t(u)+1).\n\n   Prove that N(p,m) equals the number of distinct extremal p-power–free sets and that  \n\n    N(p,m) ≥ 2^{\\,m − ⌊m/p⌋},\n\n   with strict inequality whenever m ≥ p.\n\n3. (Forced occurrence)  \n   Let f(p,m) be the least positive integer with the following property:\n\n  Every collection of f(p,m) distinct positive integers not exceeding p m  \n  contains two numbers whose quotient is a positive power of p.\n\n   Prove that  \n\n    f(p,m) = (p−1)m + 1.",
      "solution": "Step 0. Partition of {1,… ,p m} into p-chains  \nEvery n ∈ {1,… ,p m} can be written uniquely as n = p^v u with v ≥ 0 and p ∤ u.  Fix such a u.  Put  \n\n  t(u) := ⌊log_p(p m/u)⌋  (so p^{t(u)}u ≤ p m < p^{t(u)+1}u)\n\nand define  \n\n  C_u := {p^k u : 0 ≤ k ≤ t(u)}.                                (1)\n\nBecause the representation n = p^v u with p ∤ u is unique, the chains C_u are pairwise disjoint and cover {1,… ,p m}.  Exactly those u with 1 ≤ u ≤ p m and p ∤ u arise; there are (p−1)m such u (in every block of p consecutive integers precisely p−1 are coprime to p).  Hence the ground set is partitioned into (p−1)m disjoint p-chains.\n\n──────────────────────────────────────────────────────────────\n1. Extremal size\n\nLet A ⊆ {1,… ,p m} be p-power–free.  \nIf two distinct elements of A lay in the same chain C_u, their quotient would be a power of p, contradicting p-power–freeness.  Therefore A meets each chain in at most one element, whence\n\n  |A| ≤ number of chains = (p−1)m.                      (2)\n\nConversely, choose one arbitrary element from every chain and collect them in A.  Then |A| = (p−1)m and A is p-power–free by construction.  Hence\n\n  max{|A| : A ⊆ {1,… ,p m} p-power–free} = (p−1)m.      (3)\n\n──────────────────────────────────────────────────────────────\n2. Structure and enumeration of extremal sets\n\nBy inequality (2), a p-power–free set A with |A| = (p−1)m must intersect every chain in exactly one element.  Conversely, any such transversal is p-power–free and of maximal size.  Thus\n\n  A is extremal  ⇔  (∀u with 1 ≤ u ≤ p m, p∤u) |A ∩ C_u| = 1.    (4)\n\nFor a fixed u the chain C_u has length |C_u| = t(u)+1, so there are exactly t(u)+1 possibilities for A ∩ C_u.  The choices for different u are independent, giving\n\n  N(p,m) = ∏_{1≤u≤p m,\\,p∤u} (t(u)+1).                  (5)\n\nLower bound for N(p,m).  \nWrite  \n\n  S₁ := {u : 1 ≤ u ≤ m, p∤u},  S₂ := {u : m<u≤p m, p∤u}.\n\nIf u∈S₂ then u>m implies p m/u< p, hence t(u)=0 and the factor (t(u)+1)=1.  \nIf u∈S₁ we have u ≤ m, so p m/u ≥ p and therefore t(u) ≥ 1; consequently (t(u)+1) ≥ 2.  The number of elements in S₁ is\n\n  |S₁| = m − ⌊m/p⌋                                      (6)\n\n(the m integers 1,… ,m lose precisely ⌊m/p⌋ multiples of p).  Hence\n\n  N(p,m) = ∏_{u∈S₁}(t(u)+1) ≥ 2^{|S₁|} = 2^{m − ⌊m/p⌋}.    (7)\n\nStrict inequality when m ≥ p.  \nIf m ≥ p, then u=1 satisfies p m/u = p m ≥ p², so t(1) ≥ 2 and the corresponding factor equals at least 3.  Thus N(p,m) > 2^{m−⌊m/p⌋}.  This completes Part 2.\n\n──────────────────────────────────────────────────────────────\n3. The minimal forcing number f(p,m)\n\nLet S ⊆ {1,… ,p m} with |S| = (p−1)m + 1 be arbitrary.  \nBecause the chains C_u form a partition into only (p−1)m classes, the pigeon-hole principle forces two distinct elements of S to lie in the same chain.  Their quotient is a positive power of p, so every set of size (p−1)m + 1 already contains the forbidden pair.  Hence\n\n  f(p,m) ≤ (p−1)m + 1.                                  (8)\n\nOn the other hand, Part 1 produced a p-power–free set A of size (p−1)m.  Such a set contains no two elements whose quotient is a power of p, so\n\n  f(p,m) ≥ (p−1)m + 1.                                  (9)\n\nCombining (8) and (9) we obtain\n\n  f(p,m) = (p−1)m + 1.                                  ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.416088",
        "was_fixed": false,
        "difficulty_analysis": "1.  The original problem involves only the prime 2; the enhanced version works for an arbitrary prime p.  This forces the solver to generalise every step (counting, factorisation, and pigeon-hole arguments) to an arbitrary modulus p, demanding a clear understanding of how the argument depends on the arithmetic of 2 and how to replace it by p.\n\n2.  The number of admissible p–free parts is (p – 1)m instead of simply m for the “odd parts’’ when p = 2.  Handling this larger, prime-dependent counting correctly is an extra technical layer.\n\n3.  While the original conclusion is merely “one divides the other,’’ the new conclusion prescribes the precise quotient (a pure power of p).  Establishing this stronger statement requires keeping track of the exact difference in p-adic exponents rather than just their inequality.\n\n4.  All combinatorial bounds (how many numbers guarantee a collision of p–free parts) must be recomputed in terms of p, adding an additional variable and therefore more subtle book-keeping.\n\n5.  Conceptually, the solver must see that the argument is really about a “valuation at a prime’’ and not about parity alone; recognising and exploiting this abstraction represents a deeper theoretical requirement than the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}