{
  "index": "2017-B-2",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Suppose that a positive integer $N$ can be expressed as the sum of $k$ consecutive positive integers\n\\[\nN = a + (a+1) +(a+2) + \\cdots + (a+k-1)\n\\]\nfor $k=2017$ but for no other values of $k>1$. Considering all positive integers $N$ with this property,\nwhat is the smallest positive integer $a$ that occurs in any of these expressions?",
  "solution": "We prove that the smallest value of $a$ is 16.\n\nNote that the expression for $N$ can be rewritten as $k(2a+k-1)/2$,\nso that $2N = k(2a+k-1)$. In this expression, $k>1$ by requirement;\n$k < 2a+k-1$ because $a>1$; and obviously $k$ and $2a+k-1$ have opposite parity. Conversely, for any factorization $2N = mn$ with $1<m<n$ and $m,n$ of opposite parity, we obtain an expression of $N$ in the desired form by taking\n$k = m$, $a = (n+1-m)/2$.\n\nWe now note that $2017$ is prime. (On the exam, solvers would have had to verify this by hand.\nSince $2017 < 45^2$, this can be done by trial division by the primes up to 43.)\nFor $2N = 2017(2a+2016)$ not to have another expression of the specified form, it must be the case that\n$2a+2016$ has no odd divisor greater than 1; that is, $2a+2016$ must be a power of 2.\nThis first occurs for $2a+2016=2048$, yielding the claimed result.\n\n\\textbf{Reinterpretation:}\n(by Karl Mahlburg)\nTo avoid $N$ having another representation, for $k = 2, \\dots, 2016$, we must have\n\\[\nN \\not\\equiv \\begin{cases} k/2 & k \\equiv 0 \\pmod{2} \\\\\n0 & k \\equiv 1 \\pmod{2}.\n\\end{cases}\n\\]\nConsequently, $N \\not\\equiv 0 \\pmod{p}$ for any odd prime $p<2017$\nand $N \\equiv 0 \\pmod{1024}$. Since $N$ must be divisible by 2017, this again yields the claimed value of $a$.",
  "vars": [
    "N",
    "k",
    "a",
    "m",
    "n",
    "p"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "N": "totalnum",
        "k": "consecnum",
        "a": "startint",
        "m": "factora",
        "n": "factorb",
        "p": "primevar"
      },
      "question": "Suppose that a positive integer totalnum can be expressed as the sum of consecnum consecutive positive integers\n\\[\ntotalnum = startint + (startint+1) +(startint+2) + \\cdots + (startint+consecnum-1)\n\\]\nfor consecnum=2017 but for no other values of consecnum>1. Considering all positive integers totalnum with this property,\nwhat is the smallest positive integer startint that occurs in any of these expressions?",
      "solution": "We prove that the smallest value of startint is 16.\n\nNote that the expression for totalnum can be rewritten as consecnum(2startint+consecnum-1)/2,\nso that \\(2totalnum = consecnum(2startint+consecnum-1)\\). In this expression, consecnum>1 by requirement;\nconsecnum < 2startint+consecnum-1 because startint>1; and obviously consecnum and 2startint+consecnum-1 have opposite parity. Conversely, for any factorization \\(2totalnum = factora\\,factorb\\) with \\(1<factora<factorb\\) and factora and factorb of opposite parity, we obtain an expression of totalnum in the desired form by taking\n\\(consecnum = factora\\), \\(startint = (factorb+1-factora)/2\\).\n\nWe now note that 2017 is prime. (On the exam, solvers would have had to verify this by hand.\nSince \\(2017 < 45^2\\), this can be done by trial division by the primes up to 43.)\nFor \\(2totalnum = 2017(2startint+2016)\\) not to have another expression of the specified form, it must be the case that\n\\(2startint+2016\\) has no odd divisor greater than 1; that is, \\(2startint+2016\\) must be a power of 2.\nThis first occurs for \\(2startint+2016=2048\\), yielding the claimed result.\n\n\\textbf{Reinterpretation:}\n(by Karl Mahlburg)\nTo avoid totalnum having another representation, for consecnum = 2, \\dots, 2016, we must have\n\\[\ntotalnum \\not\\equiv \\begin{cases} consecnum/2 & consecnum \\equiv 0 \\pmod{2} \\\\\n0 & consecnum \\equiv 1 \\pmod{2}.\n\\end{cases}\n\\]\nConsequently, \\(totalnum \\not\\equiv 0 \\pmod{primevar}\\) for any odd prime primevar<2017\nand \\(totalnum \\equiv 0 \\pmod{1024}\\). Since totalnum must be divisible by 2017, this again yields the claimed value of startint."
    },
    "descriptive_long_confusing": {
      "map": {
        "N": "moonlight",
        "k": "windsock",
        "a": "pineapple",
        "m": "waterfall",
        "n": "hurricane",
        "p": "butterfly"
      },
      "question": "Suppose that a positive integer $moonlight$ can be expressed as the sum of $windsock$ consecutive positive integers\n\\[\nmoonlight = pineapple + (pineapple+1) +(pineapple+2) + \\cdots + (pineapple+windsock-1)\n\\]\nfor $windsock=2017$ but for no other values of $windsock>1$. Considering all positive integers $moonlight$ with this property,\nwhat is the smallest positive integer $pineapple$ that occurs in any of these expressions?",
      "solution": "We prove that the smallest value of $pineapple$ is 16.\n\nNote that the expression for $moonlight$ can be rewritten as $windsock(2pineapple+windsock-1)/2$,\nso that $2moonlight = windsock(2pineapple+windsock-1)$. In this expression, $windsock>1$ by requirement;\n$windsock < 2pineapple+windsock-1$ because $pineapple>1$; and obviously $windsock$ and $2pineapple+windsock-1$ have opposite parity. Conversely, for any factorization $2moonlight = waterfallhurricane$ with $1<waterfall<hurricane$ and $waterfall,hurricane$ of opposite parity, we obtain an expression of $moonlight$ in the desired form by taking\n$windsock = waterfall$, $pineapple = (hurricane+1-waterfall)/2$.\n\nWe now note that $2017$ is prime. (On the exam, solvers would have had to verify this by hand.\nSince $2017 < 45^2$, this can be done by trial division by the primes up to 43.)\nFor $2moonlight = 2017(2pineapple+2016)$ not to have another expression of the specified form, it must be the case that\n$2pineapple+2016$ has no odd divisor greater than 1; that is, $2pineapple+2016$ must be a power of 2.\nThis first occurs for $2pineapple+2016=2048$, yielding the claimed result.\n\n\\textbf{Reinterpretation:}\n(by Karl Mahlburg)\nTo avoid $moonlight$ having another representation, for $windsock = 2, \\dots, 2016$, we must have\n\\[\nmoonlight \\not\\equiv \\begin{cases} windsock/2 & windsock \\equiv 0 \\pmod{2} \\\\\n0 & windsock \\equiv 1 \\pmod{2}.\n\\end{cases}\n\\]\nConsequently, $moonlight \\not\\equiv 0 \\pmod{butterfly}$ for any odd prime $butterfly<2017$\nand $moonlight \\equiv 0 \\pmod{1024}$. Since $moonlight$ must be divisible by 2017, this again yields the claimed value of $pineapple$.}",
      "params": []
    },
    "descriptive_long_misleading": {
      "map": {
        "N": "nothingness",
        "k": "solitude",
        "a": "terminus",
        "m": "colossus",
        "n": "minuscule",
        "p": "composite"
      },
      "question": "Suppose that a positive integer $nothingness$ can be expressed as the sum of $solitude$ consecutive positive integers\n\\[\nnothingness = terminus + (terminus+1) +(terminus+2) + \\cdots + (terminus+solitude-1)\n\\]\nfor $solitude=2017$ but for no other values of $solitude>1$. Considering all positive integers $nothingness$ with this property,\nwhat is the smallest positive integer $terminus$ that occurs in any of these expressions?",
      "solution": "We prove that the smallest value of $terminus$ is 16.\n\nNote that the expression for $nothingness$ can be rewritten as $solitude(2terminus+solitude-1)/2$, so that $2nothingness = solitude(2terminus+solitude-1)$. In this expression, $solitude>1$ by requirement; $solitude < 2terminus+solitude-1$ because $terminus>1$; and obviously $solitude$ and $2terminus+solitude-1$ have opposite parity. Conversely, for any factorization $2nothingness = colossus\\,minuscule$ with $1<colossus<minuscule$ and $colossus,minuscule$ of opposite parity, we obtain an expression of $nothingness$ in the desired form by taking\n$solitude = colossus$, $terminus = (minuscule+1-colossus)/2$.\n\nWe now note that 2017 is prime. (On the exam, solvers would have had to verify this by hand.\nSince 2017 $< 45^2$, this can be done by trial division by the primes up to 43.)\nFor $2nothingness = 2017(2terminus+2016)$ not to have another expression of the specified form, it must be the case that\n$2terminus+2016$ has no odd divisor greater than 1; that is, $2terminus+2016$ must be a power of 2.\nThis first occurs for $2terminus+2016=2048$, yielding the claimed result.\n\n\\textbf{Reinterpretation:}\n(by Karl Mahlburg)\nTo avoid $nothingness$ having another representation, for $solitude = 2, \\dots, 2016$, we must have\n\\[\nnothingness \\not\\equiv \\begin{cases} solitude/2 & solitude \\equiv 0 \\pmod{2} \\\\ 0 & solitude \\equiv 1 \\pmod{2}. \\end{cases}\n\\]\nConsequently, $nothingness \\not\\equiv 0 \\pmod{composite}$ for any odd prime $composite<2017$\nand $nothingness \\equiv 0 \\pmod{1024}$. Since $nothingness$ must be divisible by 2017, this again yields the claimed value of $terminus$. "
    },
    "garbled_string": {
      "map": {
        "N": "qzxwvtnp",
        "k": "hjgrksla",
        "a": "plmdqrfz",
        "m": "xcbvneuq",
        "n": "trslpgoa",
        "p": "dfhjklqw"
      },
      "question": "Suppose that a positive integer $qzxwvtnp$ can be expressed as the sum of $hjgrksla$ consecutive positive integers\n\\[\nqzxwvtnp = plmdqrfz + (plmdqrfz+1) +(plmdqrfz+2) + \\cdots + (plmdqrfz+hjgrksla-1)\n\\]\nfor $hjgrksla=2017$ but for no other values of $hjgrksla>1$. Considering all positive integers $qzxwvtnp$ with this property,\nwhat is the smallest positive integer $plmdqrfz$ that occurs in any of these expressions?",
      "solution": "We prove that the smallest value of $plmdqrfz$ is 16.\n\nNote that the expression for $qzxwvtnp$ can be rewritten as $hjgrksla(2plmdqrfz+hjgrksla-1)/2$, so that $2qzxwvtnp = hjgrksla(2plmdqrfz+hjgrksla-1)$. In this expression, $hjgrksla>1$ by requirement; $hjgrksla < 2plmdqrfz+hjgrksla-1$ because $plmdqrfz>1$; and obviously $hjgrksla$ and $2plmdqrfz+hjgrksla-1$ have opposite parity. Conversely, for any factorization $2qzxwvtnp = xcbvneuq trslpgoa$ with $1<xcbvneuq<trslpgoa$ and $xcbvneuq,trslpgoa$ of opposite parity, we obtain an expression of $qzxwvtnp$ in the desired form by taking $hjgrksla = xcbvneuq$, $plmdqrfz = (trslpgoa+1-xcbvneuq)/2$.\n\nWe now note that 2017 is prime. (On the exam, solvers would have had to verify this by hand. Since $2017 < 45^2$, this can be done by trial division by the primes up to 43.) For $2qzxwvtnp = 2017(2plmdqrfz+2016)$ not to have another expression of the specified form, it must be the case that $2plmdqrfz+2016$ has no odd divisor greater than 1; that is, $2plmdqrfz+2016$ must be a power of 2. This first occurs for $2plmdqrfz+2016=2048$, yielding the claimed result.\n\n\\textbf{Reinterpretation:} (by Karl Mahlburg) To avoid $qzxwvtnp$ having another representation, for $hjgrksla = 2, \\dots, 2016$, we must have\n\\[\nqzxwvtnp \\not\\equiv \\begin{cases} hjgrksla/2 & hjgrksla \\equiv 0 \\pmod{2} \\\\ 0 & hjgrksla \\equiv 1 \\pmod{2}. \\end{cases}\n\\]\nConsequently, $qzxwvtnp \\not\\equiv 0 \\pmod{dfhjklqw}$ for any odd prime $dfhjklqw<2017$ and $qzxwvtnp \\equiv 0 \\pmod{1024}$. Since $qzxwvtnp$ must be divisible by 2017, this again yields the claimed value of $plmdqrfz$. "
    },
    "kernel_variant": {
      "question": "A positive integer N is called bi-unique if it can be represented as a sum of k consecutive positive integers  \n\n  N = a + (a+1) + … + (a+k−1)   (k > 1)                                           (★)\n\nfor exactly the two distinct lengths  \n\n  k₁ = 2017  and  k₂ = 2017² = 4 068 289,\n\nand for no other value of k greater than 1.  \nFor every such N write  \n\n  a₁ – the first term in its 2017-term decomposition,  \n  a₂ – the first term in its 4 068 289-term decomposition.  \n\nDetermine  \n\n  min { a₁ , a₂ }  \n\ntaken over all bi-unique integers N.\n\n\n\n───────────────────────────────────────────────",
      "solution": "Throughout we write a decomposition in the form (★).  \nThe sum of k consecutive integers equals k(2a+k–1)/2, whence  \n\n  2N = k(2a+k−1).                      (1)\n\nConversely, every factorisation 2N = km with  \n\n  • k > 1,  • k and m of opposite parity,  • m ≥ k+1        (2)\n\nproduces, and is produced by, a representation (★) with first term  \na = (m−k+1)/2.\n\nFix, once and for all, a bi-unique integer N.  \nPut  \n\n  m := 2a+k−1 > k, so that 2N = km.               (3)\n\n\n\n1.  The two prescribed decompositions  \nFor k₁ = 2017 and k₂ = 2017² write  \n\n  2N = 2017·X = 2017²·Y, where X = 2a₁+2016, Y = 2a₂+2017²−1.  (4)\n\nBoth 2017 and 2017² are odd; by (2) this forces X and Y to be even.  \nBecause 2017 is prime, (4) implies  \n\n  X = 2017 Y,  Y = 2^α s  with α ≥ 1, s odd.          (5)\n\nThus  \n\n  2N = 2017² · 2^α · s                       (6)\n\n\n\n2.  Eliminating the factor 2017 from s  \nSuppose, for contradiction, that 2017 divides s; write s = 2017 t with t odd.  \nEquation (6) then becomes  \n\n  2N = 2017³ · 2^α · t.                      (7)\n\nConsider the two factors  \n\n  u = 2^α  (even),  v = 2017³ t  (odd).  \n\nThey are of opposite parity; set  \n\n  k := min{u, v},  m := max{u, v}.  \n\nBoth k and m exceed 1, and because u ≠ v we have m ≥ k+1.  \nConsequently (k,m) satisfies (2), producing a decomposition of length k.  \nThis k is neither 2017 nor 2017² (it is even if k=u, or at least 2017³ if k=v).  \nHence a third decomposition exists – impossible for a bi-unique N.  \nTherefore  \n\n  2017 ∤ s.                         (8)\n\n\n\n3.  Proving that the odd factor s equals 1  \n\nWrite 2N again as in (6) with (8) in force.\n\nCase A.  s is composite.  \nLet p be its smallest odd prime divisor (p ≥ 3, p ≠ 2017).  \nChoose  \n\n  k = 2017 p   (odd divisor of 2N).            (9)\n\nIts partner equals  \n\n  m = 2N/k = 2017 · 2^α · (s/p),  an even integer.      (10)\n\nBecause s/p ≥ p and α ≥ 1,\n\n  m ≥ 2017·2·p > 2017 p = k.               (11)\n\nThus another legitimate decomposition of length k (≠2017,2017²) appears – contradiction.  \nHence s cannot be composite.\n\nCase B.  s is prime (by (8) necessarily s ≠ 2017).  \nWe first show that the forthcoming exclusion of even lengths demands  \n\n  2^α > s.                         (12)\n\nIndeed, let u be any odd divisor of 2017² s.  \nA potential even length has the form k = 2^ρ u (ρ ≥ 1); applying (1)–(2) as before yields the necessity  \n\n  2017² s/u < 2^α u + 1.              (13)\n\nPutting u = 1 gives 2^α > 2017² s ≥ s, proving (12).\n\nNow select  \n\n  k = 2017 s   (odd divisor of 2N).           (14)\n\nIts partner is m = 2017·2^α (even), and because of (12) we have m > k.  \nCondition (2) is fulfilled, yielding a third decomposition – contradiction.  \nTherefore Case B is impossible and we conclude  \n\n  s = 1.                         (15)\n\nConsequently  \n\n  2N = 2017² · 2^α  with α ≥ 1.             (16)\n\n\n\n4.  Barring even lengths – a lower bound for α  \n\nLet k be any even divisor of 2N; write k = 2^ρ u with u odd, ρ ≥ 1.  \nFrom (16),\n\n  m = 2N/k = 2017² · 2^{α-ρ}/u.            (17)\n\nOpposite parity forces α = ρ, so that  \n\n  (m, k) = (2017²/u, 2^α u).  \n\nCondition (2) requires m ≥ k+1, i.e.  \n\n  2017²/u ≥ 2^α u + 1.               (18)\n\nThe left‐hand side decreases and the right‐hand side increases with u;  \nthe most difficult case is u = 1, giving  \n\n  2017² ≥ 2^α + 1.                 (19)\n\nNumerically, 2²¹ = 2 097 152 < 2017² = 4 068 289 < 2²² = 4 194 304,  \nso (19) fails for α ≥ 22 and holds for α ≤ 21.  Therefore\n\n  α ≥ 22 is necessary and sufficient to eliminate every even length. (20)\n\n\n\n5.  Verifying that α ≥ 22 indeed yields bi-uniqueness  \n\nWith s = 1, the odd divisors of 2N are 1, 2017, 2017² only.  \nThus the only possible odd lengths are k = 2017 and k = 2017².  \nBy (20) no even length survives when α ≥ 22, so exactly the two required decompositions remain, and N is bi-unique.\n\n\n\n6.  Computing a₁ and a₂  \n\nFor any α ≥ 22 formulae (4)–(5) give  \n\n  Y = 2^α,  X = 2017 Y = 2017·2^α.  \n\nHence  \n\n  a₂ = (Y − 2017² + 1)/2 = (2^α − 4 068 289 + 1)/2,  \n\n  a₁ = (X − 2017 + 1)/2 = (2017·2^α − 2017 + 1)/2.     (21)\n\nBecause 2^α ≤ 2017·2^α − 2017 + 1 for α ≥ 1, we always have a₂ ≤ a₁,  \nso  \n\n  min{a₁, a₂} = a₂.                  (22)\n\nThe right‐hand side is strictly increasing in α, hence its minimum occurs at the minimal admissible exponent α = 22.  Substituting 2²² = 4 194 304,\n\n  a₂(min) = (4 194 304 − 4 068 289 + 1)/2 = 63 008.    (23)\n\nThe corresponding integer  \n\n  N₀ = 2017² · 2²¹                    (24)\n\nindeed possesses exactly the two decompositions (k = 2017 and k = 2017²) and no others.\n\nAnswer: 63 008.\n\n\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.846774",
        "was_fixed": false,
        "difficulty_analysis": "1.  Multiple interacting representations.  \nUnlike the original problem, which involved a single length of 2017, the enhanced variant imposes two interacting lengths 2017 and 2017².  Their coexistence forces a non-trivial simultaneous factorisation (equation (2)) that has to be analysed via unique factorisation.\n\n2.  Odd-versus-even length dichotomy.  \nProving that no even length can occur demands a delicate parity argument using the inequality e > k from Step 1; this is entirely absent from the original task.\n\n3.  Control of the full divisor structure.  \nThe solver must show that having exactly two admissible lengths forces N to possess precisely one odd prime squared in its factorisation (Step 2) and then check every possible divisor of 2N (Step 3).  This uses deeper number-theoretic tools (divisor classification, size estimates of powers of two) than the original solution.\n\n4.  Optimisation under an inequality constraint.  \nDetermining the minimal t that blocks all even lengths (inequality (5)) and still keeps both required starting integers positive adds an additional optimisation layer.\n\n5.  Size handling.  \nExplicit numbers on the order of four million (2017²) and four billion (a₁) must be managed accurately, increasing computational and conceptual load.\n\nAltogether these ingredients demand substantially more theory (parity arguments for all divisors, simultaneous Diophantine conditions) and longer chains of reasoning than the prime-and-power-of-two trick used in the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "A positive integer N is called bi-unique if it can be represented as a sum of k consecutive positive integers  \n\n  N = a + (a+1) + … + (a+k−1)   (k > 1)                                           (★)\n\nfor exactly the two distinct lengths  \n\n  k₁ = 2017  and  k₂ = 2017² = 4 068 289,\n\nand for no other value of k greater than 1.  \nFor every such N write  \n\n  a₁ – the first term in its 2017-term decomposition,  \n  a₂ – the first term in its 4 068 289-term decomposition.  \n\nDetermine  \n\n  min { a₁ , a₂ }  \n\ntaken over all bi-unique integers N.\n\n\n\n───────────────────────────────────────────────",
      "solution": "Throughout we write a decomposition in the form (★).  \nThe sum of k consecutive integers equals k(2a+k–1)/2, whence  \n\n  2N = k(2a+k−1).                      (1)\n\nConversely, every factorisation 2N = km with  \n\n  • k > 1,  • k and m of opposite parity,  • m ≥ k+1        (2)\n\nproduces, and is produced by, a representation (★) with first term  \na = (m−k+1)/2.\n\nFix, once and for all, a bi-unique integer N.  \nPut  \n\n  m := 2a+k−1 > k, so that 2N = km.               (3)\n\n\n\n1.  The two prescribed decompositions  \nFor k₁ = 2017 and k₂ = 2017² write  \n\n  2N = 2017·X = 2017²·Y, where X = 2a₁+2016, Y = 2a₂+2017²−1.  (4)\n\nBoth 2017 and 2017² are odd; by (2) this forces X and Y to be even.  \nBecause 2017 is prime, (4) implies  \n\n  X = 2017 Y,  Y = 2^α s  with α ≥ 1, s odd.          (5)\n\nThus  \n\n  2N = 2017² · 2^α · s                       (6)\n\n\n\n2.  Eliminating the factor 2017 from s  \nSuppose, for contradiction, that 2017 divides s; write s = 2017 t with t odd.  \nEquation (6) then becomes  \n\n  2N = 2017³ · 2^α · t.                      (7)\n\nConsider the two factors  \n\n  u = 2^α  (even),  v = 2017³ t  (odd).  \n\nThey are of opposite parity; set  \n\n  k := min{u, v},  m := max{u, v}.  \n\nBoth k and m exceed 1, and because u ≠ v we have m ≥ k+1.  \nConsequently (k,m) satisfies (2), producing a decomposition of length k.  \nThis k is neither 2017 nor 2017² (it is even if k=u, or at least 2017³ if k=v).  \nHence a third decomposition exists – impossible for a bi-unique N.  \nTherefore  \n\n  2017 ∤ s.                         (8)\n\n\n\n3.  Proving that the odd factor s equals 1  \n\nWrite 2N again as in (6) with (8) in force.\n\nCase A.  s is composite.  \nLet p be its smallest odd prime divisor (p ≥ 3, p ≠ 2017).  \nChoose  \n\n  k = 2017 p   (odd divisor of 2N).            (9)\n\nIts partner equals  \n\n  m = 2N/k = 2017 · 2^α · (s/p),  an even integer.      (10)\n\nBecause s/p ≥ p and α ≥ 1,\n\n  m ≥ 2017·2·p > 2017 p = k.               (11)\n\nThus another legitimate decomposition of length k (≠2017,2017²) appears – contradiction.  \nHence s cannot be composite.\n\nCase B.  s is prime (by (8) necessarily s ≠ 2017).  \nWe first show that the forthcoming exclusion of even lengths demands  \n\n  2^α > s.                         (12)\n\nIndeed, let u be any odd divisor of 2017² s.  \nA potential even length has the form k = 2^ρ u (ρ ≥ 1); applying (1)–(2) as before yields the necessity  \n\n  2017² s/u < 2^α u + 1.              (13)\n\nPutting u = 1 gives 2^α > 2017² s ≥ s, proving (12).\n\nNow select  \n\n  k = 2017 s   (odd divisor of 2N).           (14)\n\nIts partner is m = 2017·2^α (even), and because of (12) we have m > k.  \nCondition (2) is fulfilled, yielding a third decomposition – contradiction.  \nTherefore Case B is impossible and we conclude  \n\n  s = 1.                         (15)\n\nConsequently  \n\n  2N = 2017² · 2^α  with α ≥ 1.             (16)\n\n\n\n4.  Barring even lengths – a lower bound for α  \n\nLet k be any even divisor of 2N; write k = 2^ρ u with u odd, ρ ≥ 1.  \nFrom (16),\n\n  m = 2N/k = 2017² · 2^{α-ρ}/u.            (17)\n\nOpposite parity forces α = ρ, so that  \n\n  (m, k) = (2017²/u, 2^α u).  \n\nCondition (2) requires m ≥ k+1, i.e.  \n\n  2017²/u ≥ 2^α u + 1.               (18)\n\nThe left‐hand side decreases and the right‐hand side increases with u;  \nthe most difficult case is u = 1, giving  \n\n  2017² ≥ 2^α + 1.                 (19)\n\nNumerically, 2²¹ = 2 097 152 < 2017² = 4 068 289 < 2²² = 4 194 304,  \nso (19) fails for α ≥ 22 and holds for α ≤ 21.  Therefore\n\n  α ≥ 22 is necessary and sufficient to eliminate every even length. (20)\n\n\n\n5.  Verifying that α ≥ 22 indeed yields bi-uniqueness  \n\nWith s = 1, the odd divisors of 2N are 1, 2017, 2017² only.  \nThus the only possible odd lengths are k = 2017 and k = 2017².  \nBy (20) no even length survives when α ≥ 22, so exactly the two required decompositions remain, and N is bi-unique.\n\n\n\n6.  Computing a₁ and a₂  \n\nFor any α ≥ 22 formulae (4)–(5) give  \n\n  Y = 2^α,  X = 2017 Y = 2017·2^α.  \n\nHence  \n\n  a₂ = (Y − 2017² + 1)/2 = (2^α − 4 068 289 + 1)/2,  \n\n  a₁ = (X − 2017 + 1)/2 = (2017·2^α − 2017 + 1)/2.     (21)\n\nBecause 2^α ≤ 2017·2^α − 2017 + 1 for α ≥ 1, we always have a₂ ≤ a₁,  \nso  \n\n  min{a₁, a₂} = a₂.                  (22)\n\nThe right‐hand side is strictly increasing in α, hence its minimum occurs at the minimal admissible exponent α = 22.  Substituting 2²² = 4 194 304,\n\n  a₂(min) = (4 194 304 − 4 068 289 + 1)/2 = 63 008.    (23)\n\nThe corresponding integer  \n\n  N₀ = 2017² · 2²¹                    (24)\n\nindeed possesses exactly the two decompositions (k = 2017 and k = 2017²) and no others.\n\nAnswer: 63 008.\n\n\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.648520",
        "was_fixed": false,
        "difficulty_analysis": "1.  Multiple interacting representations.  \nUnlike the original problem, which involved a single length of 2017, the enhanced variant imposes two interacting lengths 2017 and 2017².  Their coexistence forces a non-trivial simultaneous factorisation (equation (2)) that has to be analysed via unique factorisation.\n\n2.  Odd-versus-even length dichotomy.  \nProving that no even length can occur demands a delicate parity argument using the inequality e > k from Step 1; this is entirely absent from the original task.\n\n3.  Control of the full divisor structure.  \nThe solver must show that having exactly two admissible lengths forces N to possess precisely one odd prime squared in its factorisation (Step 2) and then check every possible divisor of 2N (Step 3).  This uses deeper number-theoretic tools (divisor classification, size estimates of powers of two) than the original solution.\n\n4.  Optimisation under an inequality constraint.  \nDetermining the minimal t that blocks all even lengths (inequality (5)) and still keeps both required starting integers positive adds an additional optimisation layer.\n\n5.  Size handling.  \nExplicit numbers on the order of four million (2017²) and four billion (a₁) must be managed accurately, increasing computational and conceptual load.\n\nAltogether these ingredients demand substantially more theory (parity arguments for all divisors, simultaneous Diophantine conditions) and longer chains of reasoning than the prime-and-power-of-two trick used in the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}