{
  "index": "1971-B-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-6. Let \\( \\delta(x) \\) be the greatest odd divisor of the positive integer \\( x \\). Show that \\( \\sum_{n=1}^{x} \\delta(n) / n-2 x / 3 \\mid<1 \\), for all positive integers \\( x \\).",
  "solution": "B-6 Set\n\\[\nS(x)=\\sum_{n=1}^{x} \\frac{\\delta(n)}{n}\n\\]\n\nNote that \\( \\delta(2 m+1)=2 m+1, \\delta(2 m)=\\delta(m) \\) and that \\( S(2 x+1)=S(2 x)+1 \\). Dividing the summation for \\( S(2 x) \\) into even and odd values of the index produces the following relation:\n\\[\nS(2 x)=\\sum_{m=1}^{x} \\frac{\\delta(2 m)}{2 m}+\\sum_{m=1}^{x} \\frac{\\delta(2 m-1)}{2 m-1}=\\frac{1}{2} S(x)+x .\n\\]\n\nIf we denote \\( S(x)-\\frac{2 x}{3} \\) by \\( F(x) \\), the above relations translate into\n\\[\nF(2 x)=\\frac{1}{2} F(x), \\text { and } F(2 x+1)=F(2 x)+\\frac{1}{3}\n\\]\n\nNow induction can be used to show that \\( 0<F(x)<\\frac{2}{3} \\), for all positive integers \\( x \\). This result is sharper than that requested.",
  "vars": [
    "x",
    "n",
    "m"
  ],
  "params": [
    "S",
    "F",
    "\\\\delta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "integerx",
        "n": "indexn",
        "m": "indexm",
        "S": "series",
        "F": "function",
        "\\delta": "oddfactor"
      },
      "question": "B-6. Let \\( oddfactor(integerx) \\) be the greatest odd divisor of the positive integer \\( integerx \\). Show that \\( \\sum_{indexn=1}^{integerx} oddfactor(indexn) / indexn-2 integerx / 3 \\mid<1 \\), for all positive integers \\( integerx \\).",
      "solution": "B-6 Set\n\\[\nseries(integerx)=\\sum_{indexn=1}^{integerx} \\frac{oddfactor(indexn)}{indexn}\n\\]\n\nNote that \\( oddfactor(2 indexm+1)=2 indexm+1, oddfactor(2 indexm)=oddfactor(indexm) \\) and that \\( series(2 integerx+1)=series(2 integerx)+1 \\). Dividing the summation for \\( series(2 integerx) \\) into even and odd values of the index produces the following relation:\n\\[\nseries(2 integerx)=\\sum_{indexm=1}^{integerx} \\frac{oddfactor(2 indexm)}{2 indexm}+\\sum_{indexm=1}^{integerx} \\frac{oddfactor(2 indexm-1)}{2 indexm-1}=\\frac{1}{2} series(integerx)+integerx .\n\\]\n\nIf we denote \\( series(integerx)-\\frac{2 integerx}{3} \\) by \\( function(integerx) \\), the above relations translate into\n\\[\nfunction(2 integerx)=\\frac{1}{2} function(integerx), \\text { and } function(2 integerx+1)=function(2 integerx)+\\frac{1}{3}\n\\]\n\nNow induction can be used to show that \\( 0<function(integerx)<\\frac{2}{3} \\), for all positive integers \\( integerx \\). This result is sharper than that requested."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "cardboard",
        "n": "chandelier",
        "m": "toothpaste",
        "S": "coffeecup",
        "F": "honeycomb",
        "\\\\delta": "riverbank"
      },
      "question": "B-6. Let \\( riverbank(cardboard) \\) be the greatest odd divisor of the positive integer \\( cardboard \\). Show that \\( \\sum_{chandelier=1}^{cardboard} riverbank(chandelier) / chandelier-2 cardboard / 3 \\mid<1 \\), for all positive integers \\( cardboard \\).",
      "solution": "B-6 Set\n\\[\ncoffeecup(cardboard)=\\sum_{chandelier=1}^{cardboard} \\frac{riverbank(chandelier)}{chandelier}\n\\]\n\nNote that \\( riverbank(2 toothpaste+1)=2 toothpaste+1, riverbank(2 toothpaste)=riverbank(toothpaste) \\) and that \\( coffeecup(2 cardboard+1)=coffeecup(2 cardboard)+1 \\). Dividing the summation for \\( coffeecup(2 cardboard) \\) into even and odd values of the index produces the following relation:\n\\[\ncoffeecup(2 cardboard)=\\sum_{toothpaste=1}^{cardboard} \\frac{riverbank(2 toothpaste)}{2 toothpaste}+\\sum_{toothpaste=1}^{cardboard} \\frac{riverbank(2 toothpaste-1)}{2 toothpaste-1}=\\frac{1}{2} coffeecup(cardboard)+cardboard .\n\\]\n\nIf we denote \\( coffeecup(cardboard)-\\frac{2 cardboard}{3} \\) by \\( honeycomb(cardboard) \\), the above relations translate into\n\\[\nhoneycomb(2 cardboard)=\\frac{1}{2} honeycomb(cardboard), \\text { and } honeycomb(2 cardboard+1)=honeycomb(2 cardboard)+\\frac{1}{3}\n\\]\n\nNow induction can be used to show that \\( 0<honeycomb(cardboard)<\\frac{2}{3} \\), for all positive integers \\( cardboard \\). This result is sharper than that requested."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "n": "stagnantindex",
        "m": "steadyparam",
        "S": "difference",
        "F": "staticval",
        "\\delta": "leastevenfactor"
      },
      "question": "B-6. Let \\( leastevenfactor(constantvalue) \\) be the greatest odd divisor of the positive integer \\( constantvalue \\). Show that \\( \\sum_{stagnantindex=1}^{constantvalue} leastevenfactor(stagnantindex) / stagnantindex-2 constantvalue / 3 \\mid<1 \\), for all positive integers \\( constantvalue \\).",
      "solution": "B-6 Set\n\\[\ndifference(constantvalue)=\\sum_{stagnantindex=1}^{constantvalue} \\frac{leastevenfactor(stagnantindex)}{stagnantindex}\n\\]\n\nNote that \\( leastevenfactor(2 steadyparam+1)=2 steadyparam+1, leastevenfactor(2 steadyparam)=leastevenfactor(steadyparam) \\) and that \\( difference(2 constantvalue+1)=difference(2 constantvalue)+1 \\). Dividing the summation for \\( difference(2 constantvalue) \\) into even and odd values of the index produces the following relation:\n\\[\ndifference(2 constantvalue)=\\sum_{steadyparam=1}^{constantvalue} \\frac{leastevenfactor(2 steadyparam)}{2 steadyparam}+\\sum_{steadyparam=1}^{constantvalue} \\frac{leastevenfactor(2 steadyparam-1)}{2 steadyparam-1}=\\frac{1}{2} difference(constantvalue)+constantvalue .\n\\]\n\nIf we denote \\( difference(constantvalue)-\\frac{2 constantvalue}{3} \\) by \\( staticval(constantvalue) \\), the above relations translate into\n\\[\nstaticval(2 constantvalue)=\\frac{1}{2} staticval(constantvalue), \\text { and } staticval(2 constantvalue+1)=staticval(2 constantvalue)+\\frac{1}{3}\n\\]\n\nNow induction can be used to show that \\( 0<staticval(constantvalue)<\\frac{2}{3} \\), for all positive integers \\( constantvalue \\). This result is sharper than that requested."
    },
    "garbled_string": {
      "map": {
        "x": "qmdzofpl",
        "n": "ycvtrmna",
        "m": "hlgxepso",
        "S": "abkyrteq",
        "F": "mqslpfda",
        "\\delta": "cswpejrt"
      },
      "question": "B-6. Let \\( cswpejrt(qmdzofpl) \\) be the greatest odd divisor of the positive integer \\( qmdzofpl \\). Show that \\( \\sum_{ycvtrmna=1}^{qmdzofpl} cswpejrt(ycvtrmna) / ycvtrmna-2 qmdzofpl / 3 \\mid<1 \\), for all positive integers \\( qmdzofpl \\).",
      "solution": "B-6 Set\n\\[\nabkyrteq(qmdzofpl)=\\sum_{ycvtrmna=1}^{qmdzofpl} \\frac{cswpejrt(ycvtrmna)}{ycvtrmna}\n\\]\n\nNote that \\( cswpejrt(2 hlgxepso+1)=2 hlgxepso+1, cswpejrt(2 hlgxepso)=cswpejrt(hlgxepso) \\) and that \\( abkyrteq(2 qmdzofpl+1)=abkyrteq(2 qmdzofpl)+1 \\). Dividing the summation for \\( abkyrteq(2 qmdzofpl) \\) into even and odd values of the index produces the following relation:\n\\[\nabkyrteq(2 qmdzofpl)=\\sum_{hlgxepso=1}^{qmdzofpl} \\frac{cswpejrt(2 hlgxepso)}{2 hlgxepso}+\\sum_{hlgxepso=1}^{qmdzofpl} \\frac{cswpejrt(2 hlgxepso-1)}{2 hlgxepso-1}=\\frac{1}{2} abkyrteq(qmdzofpl)+qmdzofpl .\n\\]\n\nIf we denote \\( abkyrteq(qmdzofpl)-\\frac{2 qmdzofpl}{3} \\) by \\( mqslpfda(qmdzofpl) \\), the above relations translate into\n\\[\nmqslpfda(2 qmdzofpl)=\\frac{1}{2} mqslpfda(qmdzofpl), \\text { and } mqslpfda(2 qmdzofpl+1)=mqslpfda(2 qmdzofpl)+\\frac{1}{3}\n\\]\n\nNow induction can be used to show that \\( 0<mqslpfda(qmdzofpl)<\\frac{2}{3} \\), for all positive integers \\( qmdzofpl \\). This result is sharper than that requested."
    },
    "kernel_variant": {
      "question": "Let $p$ be a fixed prime, $p\\ge 2$.\n\nFor every positive integer $k$ write $k=p^{v_{p}(k)}\\delta_{p}(k)$ with $\\gcd\\!\\bigl(\\delta_{p}(k),p\\bigr)=1$ and put\n\\[\n\\delta_{p}(k):=\\frac{k}{p^{v_{p}(k)}},\\qquad\nS_{p}(0):=0,\\qquad\nS_{p}(N):=\\sum_{n=1}^{N}\\frac{\\delta_{p}(n)}{n}\\quad(N\\ge 1).\n\\]\n\n1. (Sharp one–sided estimate) Show that for every $N\\ge 1$\n\\[\n0<S_{p}(N)-\\frac{pN}{p+1}<\\frac{p}{p+1}.\n\\]\n\n2. For $N\\ge 0$ define\n\\[\nF_{p}(N):=S_{p}(N)-\\frac{pN}{p+1},\\qquad \n\\mathcal F_{p}:=\\bigl\\{F_{p}(N):N\\in\\mathbb N\\bigr\\}.\n\\]\nProve that $\\mathcal F_{p}$ is dense in the open interval $\\bigl(0,\\dfrac{p}{p+1}\\bigr)$.  Equivalently, for every real number\n\\[\n0<\\alpha<\\frac{p}{p+1}\n\\]\nthere exists a strictly increasing sequence of positive integers\n\\[\nN_{1}<N_{2}<\\dots,\\qquad\\text{such that } F_{p}(N_{k})\\longrightarrow\\alpha .\n\\]",
      "solution": "Throughout write\n\\[\nS(N):=S_{p}(N),\\qquad \nF(N):=F_{p}(N)=S(N)-\\frac{pN}{p+1}\\quad(N\\ge 0). \\tag{1}\n\\]\n\n\\textbf{Step 1.  A recurrence for $S$.}\n\nPartition $\\{1,\\dots ,pX\\}$ into residue classes modulo $p$:\n\\[\nS(pX)=\\sum_{m=1}^{X}\\frac{\\delta_{p}(pm)}{pm}\n+\\sum_{r=1}^{p-1}\\sum_{m=0}^{X-1}\\frac{\\delta_{p}(pm+r)}{pm+r}.\n\\]\nBecause $\\delta_{p}(pm)=\\delta_{p}(m)$ and $\\delta_{p}(pm+r)=pm+r$ for $1\\le r\\le p-1$,\n\\[\nS(pX)=\\frac{1}{p}\\,S(X)+(p-1)X. \\tag{2}\n\\]\n\n\\textbf{Step 2.  Recurrences for $F$.}\n\nSubstituting $S(X)=F(X)+\\dfrac{pX}{p+1}$ into (2) yields\n\\[\nF(pX)=\\frac{1}{p}\\,F(X). \\tag{3}\n\\]\nMoreover, $S(pX+t)=S(pX)+t$ for $1\\le t\\le p-1$, hence\n\\[\nF(pX+t)=F(pX)+\\frac{t}{p+1}. \\tag{4}\n\\]\n\n\\textbf{Step 3.  The sharp estimate.}\n\nWe prove by strong induction on $N$ that\n\\[\n0<F(N)<\\frac{p}{p+1}\\qquad(N\\ge 1). \\tag{5}\n\\]\n\n\\emph{Basis $N=1$.}  Using (1) we have $F(1)=1-\\dfrac{p}{p+1}=\\dfrac{1}{p+1}$.\n\n\\emph{Induction step.}  Write $N=pX+t$, $0\\le t\\le p-1$.\n\n$\\bullet$ If $t=0$, then (3) gives $F(N)=p^{-1}F(X)$; the induction hypothesis for $X$ implies $0<F(N)<\\dfrac{p}{p+1}$.\n\n$\\bullet$ If $1\\le t\\le p-1$, then (4) yields\n$F(N)=p^{-1}F(X)+\\dfrac{t}{p+1}$.  Using $0<p^{-1}F(X)<(p+1)^{-1}$ we obtain\n\\[\n\\frac{t}{p+1}<F(N)<\\frac{t}{p+1}+\\frac{1}{p+1}\\le\\frac{p}{p+1}.\n\\]\nThus (5) holds, completing Part 1.  \\qed\n\n\\textbf{Step 4.  A closed formula and density of $\\Phi(\\mathbb N)$.}\n\nLet\n\\[\nN=\\varepsilon_{d}p^{d}+\\varepsilon_{d-1}p^{d-1}+\\dots+\\varepsilon_{1}p+\\varepsilon_{0},\n\\qquad\n\\varepsilon_{d}\\neq 0,\\; 0\\le\\varepsilon_{j}\\le p-1. \\tag{6}\n\\]\nIterating (3)–(4) from the most to the least significant digit gives\n\\[\nF(N)=\\frac{\\varepsilon_{0}+\\varepsilon_{1}p^{-1}+\\dots+\\varepsilon_{d}p^{-d}}{p+1}. \\tag{7}\n\\]\nDefine\n\\[\n\\Phi(N):=(p+1)F(N)\\qquad(N\\ge 0). \\tag{8}\n\\]\nThen\n\\[\n\\Phi(N)=\\varepsilon_{0}+\\varepsilon_{1}p^{-1}+\\dots+\\varepsilon_{d}p^{-d}. \\tag{9}\n\\]\n\nFor $k\\ge 0$ set\n\\[\n\\mathcal A_{k}:=\\Bigl\\{\\frac{a}{p^{k}}\\;:\\;1\\le a\\le p^{k+1}-1\\Bigr\\}. \\tag{10}\n\\]\nIf $N$ has highest digit index $d=k$, multiplying (9) by $p^{k}$ produces an integer $a$ with $1\\le a\\le p^{k+1}-1$; hence\n\\[\n\\Phi(\\mathbb N)=\\bigcup_{k\\ge 0}\\mathcal A_{k}. \\tag{11}\n\\]\nConversely, every fraction $a/p^{k}$ in $\\mathcal A_{k}$ arises from some $N$, so (11) is an equality.\n\n\\emph{Density of $\\Phi(\\mathbb N)$ in $(0,p)$.}  \nGiven $x\\in(0,p)$ and $\\varepsilon>0$, choose $k$ with $p^{-k}<\\varepsilon$ and put $a=\\lfloor x p^{k}\\rfloor$.  Then $a/p^{k}\\in\\mathcal A_{k}$ and\n\\[\n\\bigl|x-\\frac{a}{p^{k}}\\bigr|<\\frac{1}{p^{k}}<\\varepsilon ,\n\\]\nso $\\Phi(\\mathbb N)$ is dense in $(0,p)$, and therefore\n\\[\n\\mathcal F_{p}=F(\\mathbb N)=\\frac{1}{p+1}\\Phi(\\mathbb N)\n\\]\nis dense in $\\bigl(0,\\dfrac{p}{p+1}\\bigr)$.\n\n\\textbf{Step 5.  A concrete strictly increasing sequence approaching any prescribed value.}\n\nFix $\\alpha\\in\\bigl(0,\\dfrac{p}{p+1}\\bigr)$ and set $\\beta:=(p+1)\\alpha\\in(0,p)$.\nTake the non–terminating base–$p$ expansion\n\\[\n\\beta=b_{0}+b_{1}p^{-1}+b_{2}p^{-2}+\\dots,\\qquad 0\\le b_{j}\\le p-1, \\tag{12}\n\\]\nwhich has infinitely many indices with $b_{j}<p-1$.  List them as\n\\[\ns_{1}<s_{2}<\\dots,\\qquad b_{s_{k}}<p-1\\ (k\\ge 1). \\tag{13}\n\\]\nDefine\n\\[\n\\beta_{k}^{\\ast}:=\\sum_{j=0}^{s_{k}}b_{j}p^{-j}+p^{-(s_{k}+1)},\\qquad\nN_{k}:=\\sum_{j=0}^{s_{k}}b_{j}p^{\\,j}+p^{\\,s_{k}+1}. \\tag{14}\n\\]\n(The base–$p$ digits of $N_{k}$ are $b_{0},b_{1},\\dots,b_{s_{k}},1,0,0,\\dots$.)\n\nUsing (7) with $d=s_{k}+1$ we have\n\\[\nF(N_{k})=\\frac{\\beta_{k}^{\\ast}}{p+1}. \\tag{15}\n\\]\n\n\\emph{Approximation.}  \nBecause $0\\le b_{j}\\le p-1$ for all $j>s_{k}$,\n\\[\n\\begin{aligned}\n\\lvert\\beta_{k}^{\\ast}-\\beta\\rvert\n&=p^{-(s_{k}+1)}-\\sum_{j=s_{k}+1}^{\\infty}b_{j}p^{-j}\\\\\n&\\le p^{-(s_{k}+1)}+\\sum_{j=s_{k}+1}^{\\infty}(p-1)p^{-j}  \\\\\n&=(p+1)p^{-(s_{k}+1)}. \n\\end{aligned} \\tag{16}\n\\]\nDividing by $p+1$ and using $s_{k}\\ge k$ gives\n\\[\n\\lvert F(N_{k})-\\alpha\\rvert\n\\le p^{-(s_{k}+1)}\n\\le p^{-(k+1)}\\xrightarrow[k\\to\\infty]{}0. \\tag{17}\n\\]\n\n\\emph{Monotonicity of $\\bigl(N_{k}\\bigr)$.}  \nSince $s_{k+1}>s_{k}$, the $(s_{k+1}+1)$-st digit of $N_{k+1}$ equals $1$ while that digit of $N_{k}$ is $0$; higher digits coincide.  Therefore\n\\[\nN_{k+1}-N_{k}\\ge p^{\\,s_{k+1}+1}>0,\\qquad\\text{so } N_{1}<N_{2}<\\dots . \\tag{18}\n\\]\n\nCombining (17) and (18) we obtain a strictly increasing sequence of positive integers $N_{k}$ satisfying $F(N_{k})\\to\\alpha$, completing Part 2. \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.601908",
        "was_fixed": false,
        "difficulty_analysis": "1. Generalisation to an arbitrary prime p introduces an extra parameter and forces the solver to look for relations that survive under base-p rather than base-2 arithmetic; the original problem dealt only with p=2.\n\n2. The proof now requires two non-trivial functional equations, (3) and (4), derived from a careful residue-class decomposition mod p.  In the original task a single binary recursion sufficed.\n\n3. Part 2 is entirely new.  Establishing density obliges the contestant to recognise that (3)–(4) encode a base-p dynamical system, then to synthesise a constructive, digit-by-digit argument reminiscent of Cantor-set or p-adic techniques.  No such conceptual layer is present in the original exercise.\n\n4. Bounding both below and above (with a sharp, parameter-dependent constant) is tougher than proving a one-sided estimate of size <1.  Showing that the bound p/(p+1) is optimal (it is attained as a limit) further deepens the argument.\n\n5. Altogether the enhanced variant intertwines modular decomposition, functional recursions, induction, real-analysis limits, and number-representation theory—considerably broader and more sophisticated than the single-recursion estimate required in the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $p$ be a fixed prime, $p\\ge 2$.\n\nFor every positive integer $k$ write $k=p^{v_{p}(k)}\\delta_{p}(k)$ with $\\gcd\\!\\bigl(\\delta_{p}(k),p\\bigr)=1$ and put\n\\[\n\\delta_{p}(k):=\\frac{k}{p^{v_{p}(k)}},\\qquad\nS_{p}(0):=0,\\qquad\nS_{p}(N):=\\sum_{n=1}^{N}\\frac{\\delta_{p}(n)}{n}\\quad(N\\ge 1).\n\\]\n\n1. (Sharp one–sided estimate) Prove that for every $N\\ge 1$\n\\[\n0<S_{p}(N)-\\frac{pN}{p+1}<\\frac{p}{p+1}.\n\\]\n\n2. For $N\\ge 0$ put\n\\[\nF_{p}(N):=S_{p}(N)-\\frac{pN}{p+1},\\qquad \n\\mathcal F_{p}:=\\bigl\\{F_{p}(N):N\\in\\mathbb N\\bigr\\}.\n\\]\nShow that $\\mathcal F_{p}$ is dense in the open interval $\\bigl(0,\\dfrac{p}{p+1}\\bigr)$.  Equivalently, for every real number\n\\[\n0<\\alpha<\\frac{p}{p+1}\n\\]\nthere exists an increasing sequence of positive integers\n\\[\nN_{1}<N_{2}<\\dots,\\qquad\\text{such that } F_{p}(N_{k})\\longrightarrow\\alpha .\n\\]",
      "solution": "Throughout write\n\\[\nS(N):=S_{p}(N),\\qquad \nF(N):=F_{p}(N)=S(N)-\\frac{pN}{p+1}\\quad(N\\ge 0). \\tag{1}\n\\]\n\n\\textbf{Step 1.  A recurrence for $S$.}\n\nPartition $\\{1,\\dots ,pX\\}$ into residue classes modulo $p$:\n\\[\nS(pX)=\\sum_{m=1}^{X}\\frac{\\delta_{p}(pm)}{pm}\n+\\sum_{r=1}^{p-1}\\sum_{m=0}^{X-1}\\frac{\\delta_{p}(pm+r)}{pm+r}.\n\\]\nBecause $\\delta_{p}(pm)=\\delta_{p}(m)$ and $\\delta_{p}(pm+r)=pm+r$ for $1\\le r\\le p-1$,\n\\[\nS(pX)=\\frac{1}{p}\\,S(X)+(p-1)X. \\tag{2}\n\\]\n\n\\textbf{Step 2.  Recurrences for $F$.}\n\nSubstituting $S(X)=F(X)+\\dfrac{pX}{p+1}$ into (2) yields\n\\[\nF(pX)=\\frac{1}{p}\\,F(X). \\tag{3}\n\\]\nMoreover, $S(pX+t)=S(pX)+t$ for $1\\le t\\le p-1$, hence\n\\[\nF(pX+t)=F(pX)+\\frac{t}{p+1}. \\tag{4}\n\\]\n\n\\textbf{Step 3.  The sharp estimate.}\n\nWe prove by strong induction on $N$ that\n\\[\n0<F(N)<\\frac{p}{p+1}\\qquad(N\\ge 1). \\tag{5}\n\\]\n\n\\emph{Basis $N=1$.}  Using (1) we have $F(1)=1-\\dfrac{p}{p+1}=\\dfrac{1}{p+1}$.\n\n\\emph{Induction step.}  Write $N=pX+t$, $0\\le t\\le p-1$.\n\n$\\bullet$ If $t=0$, then (3) gives $F(N)=p^{-1}F(X)$; the induction hypothesis for $X$ implies $0<F(N)<\\dfrac{p}{p+1}$.\n\n$\\bullet$ If $1\\le t\\le p-1$, then (4) yields\n$F(N)=p^{-1}F(X)+\\dfrac{t}{p+1}$.  Using $0<p^{-1}F(X)<(p+1)^{-1}$ we obtain\n\\[\n\\frac{t}{p+1}<F(N)<\\frac{t}{p+1}+\\frac{1}{p+1}\\le\\frac{p}{p+1}.\n\\]\nThus (5) holds, completing Part 1.  \\qed\n\n\\textbf{Step 4.  A closed formula and density of $\\Phi(\\mathbb N)$.}\n\nLet\n\\[\nN=\\varepsilon_{d}p^{d}+\\varepsilon_{d-1}p^{d-1}+\\dots+\\varepsilon_{1}p+\\varepsilon_{0},\n\\qquad\n\\varepsilon_{d}\\neq 0,\\; 0\\le\\varepsilon_{j}\\le p-1. \\tag{6}\n\\]\nIterating (3)–(4) from the most to the least significant digit gives\n\\[\nF(N)=\\frac{\\varepsilon_{0}+\\varepsilon_{1}p^{-1}+\\dots+\\varepsilon_{d}p^{-d}}{p+1}. \\tag{7}\n\\]\nDefine\n\\[\n\\Phi(N):=(p+1)F(N)\\qquad(N\\ge 0). \\tag{8}\n\\]\nThen\n\\[\n\\Phi(N)=\\varepsilon_{0}+\\varepsilon_{1}p^{-1}+\\dots+\\varepsilon_{d}p^{-d}. \\tag{9}\n\\]\n\nFor $k\\ge 0$ set\n\\[\n\\mathcal A_{k}:=\\Bigl\\{\\frac{a}{p^{k}}\\;:\\;1\\le a\\le p^{k+1}-1\\Bigr\\}. \\tag{10}\n\\]\nIf $N$ has highest digit index $d=k$, multiplying (9) by $p^{k}$ produces an integer $a$ with $1\\le a\\le p^{k+1}-1$; hence\n\\[\n\\Phi(\\mathbb N)=\\bigcup_{k\\ge 0}\\mathcal A_{k}. \\tag{11}\n\\]\nConversely, any fraction $a/p^{k}$ in $\\mathcal A_{k}$ can be produced in the form (9), so inclusion in (11) is two–sided.\n\n\\emph{Density of $\\Phi(\\mathbb N)$ in $(0,p)$.}  \nGiven $x\\in(0,p)$ and $\\varepsilon>0$, choose $k$ with $p^{-k}<\\varepsilon$ and put $a=\\lfloor x p^{k}\\rfloor$.  Then $a/p^{k}\\in\\mathcal A_{k}$ and\n\\[\n\\bigl|x-\\frac{a}{p^{k}}\\bigr|<\\frac{1}{p^{k}}<\\varepsilon .\n\\]\nHence $\\Phi(\\mathbb N)$ is dense in $(0,p)$, and therefore\n\\[\n\\mathcal F_{p}=F(\\mathbb N)=\\frac{1}{p+1}\\Phi(\\mathbb N)\n\\]\nis dense in $\\bigl(0,\\dfrac{p}{p+1}\\bigr)$, completing Part 2 except for the explicit increasing sequence.\n\n\\textbf{Step 5.  A concrete strictly increasing sequence converging to a prescribed value.}\n\nFix $\\alpha\\in\\bigl(0,\\dfrac{p}{p+1}\\bigr)$ and set $\\beta:=(p+1)\\alpha\\in(0,p)$.\nTake the non–terminating base–$p$ expansion\n\\[\n\\beta=b_{0}+b_{1}p^{-1}+b_{2}p^{-2}+\\dots,\\qquad 0\\le b_{j}\\le p-1, \\tag{12}\n\\]\nwhich has infinitely many indices with $b_{j}<p-1$.  Let $s_{1}<s_{2}<\\dots$ list those indices:\n\\[\nb_{s_{k}}<p-1\\quad(k\\ge 1). \\tag{13}\n\\]\nDefine for each $k$\n\\[\n\\beta_{k}^{\\ast}:=\\sum_{j=0}^{s_{k}}b_{j}p^{-j}+p^{-(s_{k}+1)},\\qquad\nN_{k}:=\\sum_{j=0}^{s_{k}}b_{j}p^{\\,j}+p^{\\,s_{k}+1}. \\tag{14}\n\\]\n(The base–$p$ digits of $N_{k}$ are $b_{0},b_{1},\\dots,b_{s_{k}},1,0,0,\\dots$.)\n\nUsing (7) with $d=s_{k}+1$ we have\n\\[\nF(N_{k})=\\frac{\\beta_{k}^{\\ast}}{p+1}. \\tag{15}\n\\]\n\n\\emph{Approximation.}  \nSince $0\\le b_{j}\\le p-1$ for all $j>s_{k}$,\n\\[\n\\lvert\\beta_{k}^{\\ast}-\\beta\\rvert\n=p^{-(s_{k}+1)}-\\sum_{j=s_{k}+1}^{\\infty}b_{j}p^{-j}\n\\le p^{-(s_{k}+1)}+\\sum_{j=s_{k}+1}^{\\infty}(p-1)p^{-j}\n\\le p^{-s_{k}}. \\tag{16}\n\\]\nConsequently\n\\[\n\\lvert F(N_{k})-\\alpha\\rvert\n=\\frac{\\lvert\\beta_{k}^{\\ast}-\\beta\\rvert}{p+1}\n\\le\\frac{p^{-s_{k}}}{p+1}<p^{-s_{k}}\\le p^{-k}\\xrightarrow[k\\to\\infty]{}0. \\tag{17}\n\\]\n\n\\emph{Monotonicity of $\\bigl(N_{k}\\bigr)$.}  \nBecause $s_{k+1}>s_{k}$, the $(s_{k+1}+1)$-st digit of $N_{k+1}$ equals $1$ while that digit of $N_{k}$ is $0$; higher digits coincide.  Therefore\n\\[\nN_{k+1}-N_{k}\\ge p^{\\,s_{k+1}+1}>0,\\qquad\\text{so } N_{1}<N_{2}<\\dots . \\tag{18}\n\\]\n\nCombining (17) and (18) we obtain a strictly increasing sequence of positive integers $N_{k}$ such that $F(N_{k})\\to\\alpha$, as required. \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.481709",
        "was_fixed": false,
        "difficulty_analysis": "1. Generalisation to an arbitrary prime p introduces an extra parameter and forces the solver to look for relations that survive under base-p rather than base-2 arithmetic; the original problem dealt only with p=2.\n\n2. The proof now requires two non-trivial functional equations, (3) and (4), derived from a careful residue-class decomposition mod p.  In the original task a single binary recursion sufficed.\n\n3. Part 2 is entirely new.  Establishing density obliges the contestant to recognise that (3)–(4) encode a base-p dynamical system, then to synthesise a constructive, digit-by-digit argument reminiscent of Cantor-set or p-adic techniques.  No such conceptual layer is present in the original exercise.\n\n4. Bounding both below and above (with a sharp, parameter-dependent constant) is tougher than proving a one-sided estimate of size <1.  Showing that the bound p/(p+1) is optimal (it is attained as a limit) further deepens the argument.\n\n5. Altogether the enhanced variant intertwines modular decomposition, functional recursions, induction, real-analysis limits, and number-representation theory—considerably broader and more sophisticated than the single-recursion estimate required in the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}