{
  "index": "2009-A-5",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Is there a finite abelian group $G$ such that the product of the\norders of all its elements is $2^{2009}$?",
  "solution": "No, there is no such group.\nBy the structure theorem for finitely generated abelian groups,\n$G$ can be written as a product of cyclic groups.\nIf any of these factors has odd order, then $G$ has an element of odd order,\nso the product of the orders of all of its elements cannot be a power of 2.\n\nWe may thus consider only abelian $2$-groups hereafter.\nFor such a group $G$, the product of the orders of all of its elements\nhas the form $2^{k(G)}$ for some nonnegative integer $G$, and we must show\nthat it is impossible to achieve $k(G) = 2009$.\nAgain by the structure theorem, we may write\n\\[\nG \\cong \\prod_{i=1}^\\infty (\\ZZ/2^i \\ZZ)^{e_i}\n\\]\nfor some nonnegative integers $e_1,e_2,\\dots$, all but finitely many of\nwhich are $0$.\n\nFor any nonnegative integer $m$, the elements of $G$ of order at most $2^m$\nform a subgroup isomorphic to\n\\[\n\\prod_{i=1}^\\infty (\\ZZ/2^{\\min\\{i,m\\}} \\ZZ)^{e_i},\n\\]\nwhich has $2^{s_m}$ elements for $s_m = \\sum_{i=1}^\\infty \\min\\{i,m\\} e_i$.\nHence\n\\[\nk(G) = \\sum_{i=1}^\\infty i(2^{s_i} - 2^{s_{i-1}}).\n\\]\nSince $s_1 \\leq s_2 \\leq \\cdots$, $k(G)+1$ is always divisible by $2^{s_1}$.\nIn particular, $k(G) = 2009$ forces $s_1 \\leq 1$.\n\nHowever, the only cases where $s_1 \\leq 1$ are where all of the $e_i$ are $0$,\nin which case $k(G) = 0$, or where $e_i = 1$ for some $i$ and $e_j = 0$\nfor $j \\neq i$, in which case $k(G) = (i-1)2^i + 1$.\nThe right side is a strictly increasing function\nof $i$ which equals $1793$ for $i=8$ and $4097$ for $i=9$, so it can never equal\n2009. This proves the claim.\n\n\\textbf{Remark.} One can also arrive at the key congruence by dividing $G$\ninto equivalence classes, by declaring two elements to be equivalent if they generate the\nsame cyclic subgroup of $G$.\nFor $h>0$, an element of order $2^h$ belongs to an equivalence class of size $2^{h-1}$,\nso the products of the orders of the elements of this equivalence class is $2^j$\nfor $j = h 2^{h-1}$. This quantity is divisible by 4 as long as $h > 1$;\nthus to have $k(G) \\equiv 1 \\pmod{4}$, the number of elements of $G$ of order 2 must be\ncongruent to 1 modulo 4. However, there are exactly $2^e-1$ such elements,\nfor $e$ the number of cyclic factors of $G$. Hence $e = 1$, and one concludes as in the\ngiven solution.",
  "vars": [
    "G",
    "k",
    "m",
    "i",
    "j",
    "h",
    "s_m",
    "s_i",
    "s_i-1",
    "s_1"
  ],
  "params": [
    "e",
    "e_i",
    "e_j"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "G": "totalgrp",
        "k": "orderpower",
        "m": "boundind",
        "i": "indexprim",
        "j": "indexsec",
        "h": "heightvar",
        "s_m": "sumbound",
        "s_i": "sumindex",
        "s_i-1": "sumprev",
        "s_1": "sumfirst",
        "e": "expbase",
        "e_i": "expindex",
        "e_j": "expalt"
      },
      "question": "Is there a finite abelian group $totalgrp$ such that the product of the\norders of all its elements is $2^{2009}$?",
      "solution": "No, there is no such group.\nBy the structure theorem for finitely generated abelian groups,\n$totalgrp$ can be written as a product of cyclic groups.\nIf any of these factors has odd order, then $totalgrp$ has an element of odd order,\nso the product of the orders of all of its elements cannot be a power of 2.\n\nWe may thus consider only abelian $2$-groups hereafter.\nFor such a group $totalgrp$, the product of the orders of all of its elements\nhas the form $2^{orderpower(totalgrp)}$ for some nonnegative integer $totalgrp$, and we must show\nthat it is impossible to achieve $orderpower(totalgrp) = 2009$.\nAgain by the structure theorem, we may write\n\\[\ntotalgrp \\cong \\prod_{indexprim=1}^\\infty (\\mathbb{Z}/2^{indexprim} \\mathbb{Z})^{expindex}\n\\]\nfor some nonnegative integers $expbase_1,expbase_2,\\dots$, all but finitely many of\nwhich are $0$.\n\nFor any nonnegative integer $boundind$, the elements of $totalgrp$ of order at most $2^{boundind}$\nform a subgroup isomorphic to\n\\[\n\\prod_{indexprim=1}^\\infty (\\mathbb{Z}/2^{\\min\\{indexprim,boundind\\}} \\mathbb{Z})^{expindex},\n\\]\nwhich has $2^{sumbound}$ elements for $sumbound = \\sum_{indexprim=1}^\\infty \\min\\{indexprim,boundind\\} expindex$.\nHence\n\\[\norderpower(totalgrp) = \\sum_{indexprim=1}^\\infty indexprim(2^{sumindex} - 2^{sumprev}).\n\\]\nSince $sumfirst \\leq sumindex \\leq \\cdots$, $orderpower(totalgrp)+1$ is always divisible by $2^{sumfirst}$.\nIn particular, $orderpower(totalgrp) = 2009$ forces $sumfirst \\leq 1$.\n\nHowever, the only cases where $sumfirst \\leq 1$ are where all of the $expindex$ are $0$,\nin which case $orderpower(totalgrp) = 0$, or where $expindex = 1$ for some $indexprim$ and $expalt = 0$\nfor $indexsec \\neq indexprim$, in which case $orderpower(totalgrp) = (indexprim-1)2^{indexprim} + 1$.\nThe right side is a strictly increasing function\nof $indexprim$ which equals $1793$ for $indexprim=8$ and $4097$ for $indexprim=9$, so it can never equal\n2009. This proves the claim.\n\n\\textbf{Remark.} One can also arrive at the key congruence by dividing $totalgrp$\ninto equivalence classes, by declaring two elements to be equivalent if they generate the\nsame cyclic subgroup of $totalgrp$.\nFor $heightvar>0$, an element of order $2^{heightvar}$ belongs to an equivalence class of size $2^{heightvar-1}$,\nso the products of the orders of the elements of this equivalence class is $2^{indexsec}$\nfor $indexsec = heightvar 2^{heightvar-1}$. This quantity is divisible by 4 as long as $heightvar > 1$;\nthus to have $orderpower(totalgrp) \\equiv 1 \\pmod{4}$, the number of elements of $totalgrp$ of order 2 must be\ncongruent to 1 modulo 4. However, there are exactly $2^{expbase}-1$ such elements,\nfor $expbase$ the number of cyclic factors of $totalgrp$. Hence $expbase = 1$, and one concludes as in the\ngiven solution."
    },
    "descriptive_long_confusing": {
      "map": {
        "G": "dawncircle",
        "k": "mistheight",
        "m": "windyroad",
        "i": "riverstone",
        "j": "willowseed",
        "h": "cloudgate",
        "s_m": "embertrail",
        "s_i": "shadowpath",
        "s_i-1": "quietvalley",
        "s_1": "silentbrook",
        "e": "galefactor",
        "e_i": "echochamber",
        "e_j": "hollowcrest"
      },
      "question": "Is there a finite abelian group $dawncircle$ such that the product of the\norders of all its elements is $2^{2009}$?",
      "solution": "No, there is no such group.\nBy the structure theorem for finitely generated abelian groups,\n$dawncircle$ can be written as a product of cyclic groups.\nIf any of these factors has odd order, then $dawncircle$ has an element of odd order,\nso the product of the orders of all of its elements cannot be a power of 2.\n\nWe may thus consider only abelian $2$-groups hereafter.\nFor such a group $dawncircle$, the product of the orders of all of its elements\nhas the form $2^{mistheight(dawncircle)}$ for some nonnegative integer $dawncircle$, and we must show\nthat it is impossible to achieve $mistheight(dawncircle) = 2009$.\nAgain by the structure theorem, we may write\n\\[\ndawncircle \\cong \\prod_{riverstone=1}^\\infty (\\ZZ/2^{riverstone} \\ZZ)^{echochamber}\n\\]\nfor some nonnegative integers $galefactor_1,galefactor_2,\\dots$, all but finitely many of\nwhich are $0$.\n\nFor any nonnegative integer $windyroad$, the elements of $dawncircle$ of order at most $2^{windyroad}$\nform a subgroup isomorphic to\n\\[\n\\prod_{riverstone=1}^\\infty (\\ZZ/2^{\\min\\{riverstone,windyroad\\}} \\ZZ)^{echochamber},\n\\]\nwhich has $2^{embertrail}$ elements for $embertrail = \\sum_{riverstone=1}^\\infty \\min\\{riverstone,windyroad\\} echochamber$.\nHence\n\\[\nmistheight(dawncircle) = \\sum_{riverstone=1}^\\infty riverstone(2^{shadowpath} - 2^{quietvalley}).\n\\]\nSince $silentbrook \\leq s_2 \\leq \\cdots$, $mistheight(dawncircle)+1$ is always divisible by $2^{silentbrook}$.\nIn particular, $mistheight(dawncircle) = 2009$ forces $silentbrook \\leq 1$.\n\nHowever, the only cases where $silentbrook \\leq 1$ are where all of the $echochamber$ are $0$,\nin which case $mistheight(dawncircle) = 0$, or where $echochamber = 1$ for some $riverstone$ and $hollowcrest = 0$\nfor $willowseed \\neq riverstone$, in which case $mistheight(dawncircle) = (riverstone-1)2^{riverstone} + 1$.\nThe right side is a strictly increasing function\nof $riverstone$ which equals $1793$ for $riverstone=8$ and $4097$ for $riverstone=9$, so it can never equal\n2009. This proves the claim.\n\n\\textbf{Remark.} One can also arrive at the key congruence by dividing $dawncircle$\ninto equivalence classes, by declaring two elements to be equivalent if they generate the\nsame cyclic subgroup of $dawncircle$.\nFor $cloudgate>0$, an element of order $2^{cloudgate}$ belongs to an equivalence class of size $2^{cloudgate-1}$,\nso the products of the orders of the elements of this equivalence class is $2^{willowseed}$\nfor $willowseed = cloudgate 2^{cloudgate-1}$. This quantity is divisible by 4 as long as $cloudgate > 1$;\nthus to have $mistheight(dawncircle) \\equiv 1 \\pmod{4}$, the number of elements of $dawncircle$ of order 2 must be\ncongruent to 1 modulo 4. However, there are exactly $2^{galefactor}-1$ such elements,\nfor $galefactor$ the number of cyclic factors of $dawncircle$. Hence $galefactor = 1$, and one concludes as in the\ngiven solution."
    },
    "descriptive_long_misleading": {
      "map": {
        "G": "disordering",
        "k": "voidvalue",
        "m": "negativezone",
        "i": "wholeindex",
        "j": "singularone",
        "h": "shallowness",
        "s_m": "gapmeasure",
        "s_i": "splitvalue",
        "s_i-1": "splitminus",
        "s_1": "splitone",
        "e": "shrinker",
        "e_i": "shrinkidx",
        "e_j": "shrinkjidx"
      },
      "question": "Is there a finite abelian group $disordering$ such that the product of the\norders of all its elements is $2^{2009}$?",
      "solution": "No, there is no such group.\nBy the structure theorem for finitely generated abelian groups,\n$disordering$ can be written as a product of cyclic groups.\nIf any of these factors has odd order, then $disordering$ has an element of odd order,\nso the product of the orders of all of its elements cannot be a power of 2.\n\nWe may thus consider only abelian $2$-groups hereafter.\nFor such a group $disordering$, the product of the orders of all of its elements\nhas the form $2^{voidvalue(disordering)}$ for some nonnegative integer $disordering$, and we must show\nthat it is impossible to achieve $voidvalue(disordering) = 2009$.\nAgain by the structure theorem, we may write\n\\[\ndisordering \\cong \\prod_{wholeindex=1}^\\infty (\\ZZ/2^{wholeindex} \\ZZ)^{shrinkidx}\n\\]\nfor some nonnegative integers $shrinkidx_{1},shrinkidx_{2},\\dots$, all but finitely many of\nwhich are $0$.\n\nFor any nonnegative integer $negativezone$, the elements of $disordering$ of order at most $2^{negativezone}$\nform a subgroup isomorphic to\n\\[\n\\prod_{wholeindex=1}^\\infty (\\ZZ/2^{\\min\\{wholeindex,negativezone\\}} \\ZZ)^{shrinkidx},\n\\]\nwhich has $2^{gapmeasure}$ elements for $gapmeasure = \\sum_{wholeindex=1}^\\infty \\min\\{wholeindex,negativezone\\} shrinkidx$.\nHence\n\\[\nvoidvalue(disordering) = \\sum_{wholeindex=1}^\\infty wholeindex(2^{splitvalue} - 2^{splitminus}).\n\\]\nSince $splitone \\leq splitvalue_{2} \\leq \\cdots$, $voidvalue(disordering)+1$ is always divisible by $2^{splitone}$.\nIn particular, $voidvalue(disordering) = 2009$ forces $splitone \\leq 1$.\n\nHowever, the only cases where $splitone \\leq 1$ are where all of the $shrinkidx$ are $0$,\nin which case $voidvalue(disordering) = 0$, or where $shrinkidx = 1$ for some $wholeindex$ and $shrinkjidx = 0$\nfor $singularone \\neq wholeindex$, in which case $voidvalue(disordering) = (wholeindex-1)2^{wholeindex} + 1$.\nThe right side is a strictly increasing function\nof $wholeindex$ which equals $1793$ for $wholeindex=8$ and $4097$ for $wholeindex=9$, so it can never equal\n2009. This proves the claim.\n\n\\textbf{Remark.} One can also arrive at the key congruence by dividing $disordering$\ninto equivalence classes, by declaring two elements to be equivalent if they generate the\nsame cyclic subgroup of $disordering$.\nFor $shallowness>0$, an element of order $2^{shallowness}$ belongs to an equivalence class of size $2^{shallowness-1}$,\nso the products of the orders of the elements of this equivalence class is $2^{singularone}$\nfor $singularone = shallowness 2^{shallowness-1}$. This quantity is divisible by 4 as long as $shallowness > 1$;\nthus to have $voidvalue(disordering) \\equiv 1 \\pmod{4}$, the number of elements of $disordering$ of order 2 must be\ncongruent to 1 modulo 4. However, there are exactly $2^{shrinker}-1$ such elements,\nfor $shrinker$ the number of cyclic factors of $disordering$. Hence $shrinker = 1$, and one concludes as in the\ngiven solution."
    },
    "garbled_string": {
      "map": {
        "G": "qzxwvtnp",
        "k": "hjgrksla",
        "m": "ptrnkeus",
        "i": "lgvadfme",
        "j": "rsubkoca",
        "h": "dymqslre",
        "s_m": "uehnzpal",
        "s_i": "kdpovhxr",
        "s_i-1": "gtrnslca",
        "s_1": "mznvcxqe",
        "e": "xqldbeuw",
        "e_i": "bvumkzsa",
        "e_j": "pczwrlan"
      },
      "question": "Is there a finite abelian group $qzxwvtnp$ such that the product of the\norders of all its elements is $2^{2009}$?",
      "solution": "No, there is no such group.\nBy the structure theorem for finitely generated abelian groups,\n$qzxwvtnp$ can be written as a product of cyclic groups.\nIf any of these factors has odd order, then $qzxwvtnp$ has an element of odd order,\nso the product of the orders of all of its elements cannot be a power of 2.\n\nWe may thus consider only abelian $2$-groups hereafter.\nFor such a group $qzxwvtnp$, the product of the orders of all of its elements\nhas the form $2^{hjgrksla(qzxwvtnp)}$ for some nonnegative integer $qzxwvtnp$, and we must show\nthat it is impossible to achieve $hjgrksla(qzxwvtnp) = 2009$.\nAgain by the structure theorem, we may write\n\\[\nqzxwvtnp \\cong \\prod_{lgvadfme=1}^\\infty (\\ZZ/2^{lgvadfme} \\ZZ)^{bvumkzsa}\n\\]\nfor some nonnegative integers $bvumkzsa_1,bvumkzsa_2,\\dots$, all but finitely many of\nwhich are $0$.\n\nFor any nonnegative integer $ptrnkeus$, the elements of $qzxwvtnp$ of order at most $2^{ptrnkeus}$\nform a subgroup isomorphic to\n\\[\n\\prod_{lgvadfme=1}^\\infty (\\ZZ/2^{\\min\\{lgvadfme,ptrnkeus\\}} \\ZZ)^{bvumkzsa},\n\\]\nwhich has $2^{uehnzpal}$ elements for $uehnzpal = \\sum_{lgvadfme=1}^\\infty \\min\\{lgvadfme,ptrnkeus\\} bvumkzsa$.\nHence\n\\[\nhjgrksla(qzxwvtnp) = \\sum_{lgvadfme=1}^\\infty lgvadfme(2^{kdpovhxr} - 2^{gtrnslca}).\n\\]\nSince $mznvcxqe \\leq kdpovhxr_2 \\leq \\cdots$, $hjgrksla(qzxwvtnp)+1$ is always divisible by $2^{mznvcxqe}$.\nIn particular, $hjgrksla(qzxwvtnp) = 2009$ forces $mznvcxqe \\leq 1$.\n\nHowever, the only cases where $mznvcxqe \\leq 1$ are where all of the $bvumkzsa$ are $0$,\nin which case $hjgrksla(qzxwvtnp) = 0$, or where $bvumkzsa = 1$ for some $lgvadfme$ and $pczwrlan = 0$\nfor $rsubkoca \\neq lgvadfme$, in which case $hjgrksla(qzxwvtnp) = (lgvadfme-1)2^{lgvadfme} + 1$.\nThe right side is a strictly increasing function\nof $lgvadfme$ which equals $1793$ for $lgvadfme=8$ and $4097$ for $lgvadfme=9$, so it can never equal\n2009. This proves the claim.\n\n\\textbf{Remark.} One can also arrive at the key congruence by dividing $qzxwvtnp$\ninto equivalence classes, by declaring two elements to be equivalent if they generate the\nsame cyclic subgroup of $qzxwvtnp$.\nFor $dymqslre>0$, an element of order $2^{dymqslre}$ belongs to an equivalence class of size $2^{dymqslre-1}$,\nso the products of the orders of the elements of this equivalence class is $2^{rsubkoca}$\nfor $rsubkoca = dymqslre 2^{dymqslre-1}$. This quantity is divisible by 4 as long as $dymqslre > 1$;\nthus to have $hjgrksla(qzxwvtnp) \\equiv 1 \\pmod{4}$, the number of elements of $qzxwvtnp$ of order 2 must be\ncongruent to 1 modulo 4. However, there are exactly $2^{xqldbeuw}-1$ such elements,\nfor $xqldbeuw$ the number of cyclic factors of $qzxwvtnp$. Hence $xqldbeuw = 1$, and one concludes as in the\ngiven solution."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nP(G)=\\prod_{g\\in G}\\operatorname{ord}(g)\n\\]\nbe the product of the orders of all elements of a finite abelian group $G$, and put  \n\\[\n\\mathcal{K}_{2}=\\Bigl\\{k\\ge 0:\\,\\exists \\text{ finite abelian }G\\text{ with }P(G)=2^{\\,k}\\Bigr\\}.\n\\]\n\nThroughout, finite abelian $2$–groups are written additively.  \nIf  \n\\[\nG\\cong\\bigoplus_{j=1}^{t}\\mathbf{Z}/2^{\\lambda_{j}}\\mathbf{Z}\n\\quad(\\lambda_{1}\\ge\\lambda_{2}\\ge\\dots\\ge\\lambda_{t}\\ge 1),\n\\]\ndefine multiplicities  \n\\[\ne_{m}(G)=\\bigl|\\{\\,j\\mid\\lambda_{j}\\ge m\\,\\}\\bigr|\\qquad(m\\ge 1),\n\\]\nand partial sums  \n\\[\nS_{m}(G)=\\sum_{q=1}^{m}e_{q}(G)\\qquad(1\\le m\\le\\lambda_{1}).\\tag{$\\dagger$}\n\\]\nThus $0<S_{1}(G)<\\dots<S_{\\lambda_{1}}(G)$.\n\nA strictly increasing sequence  \n\\[\n0<a_{1}<a_{2}<\\dots<a_{r}\\tag{$\\ast$}\n\\]\nis called $2$–admissible if its consecutive gaps  \n\\[\nd_{i}=a_{i}-a_{i-1}\\qquad(a_{0}:=0)\\tag{$\\ast\\ast$}\n\\]\nsatisfy $d_{1}\\ge d_{2}\\ge\\dots\\ge d_{r}>0$.\n\n1.  Show that for every $k>0$\n\\[\nk\\in\\mathcal{K}_{2}\\Longleftrightarrow\n\\exists \\text{ $2$–admissible }(a_{1},\\dots ,a_{r})\\text{ with }\nk=r\\cdot 2^{a_{r}}-\\bigl(1+2^{a_{1}}+\\dots+2^{a_{r-1}}\\bigr).\\tag{$\\star$}\n\\]\n(The exponent $k=0$ is realised by the trivial group.)  \nMoreover prove that the assignments  \n\\[\nG\\longmapsto\\bigl(S_{1}(G),\\dots,S_{\\lambda_{1}}(G)\\bigr),\\qquad\n(a_{1},\\dots ,a_{r})\\stackrel{(\\star)}{\\longmapsto}k\n\\]\ngive bijections\n\\[\n\\bigl\\{\\text{isomorphism classes of non-trivial }2\\text{–groups}\\bigr\\}\n\\longleftrightarrow\n\\bigl\\{\\text{$2$–admissible sequences}\\bigr\\}\n\\longleftrightarrow\n\\mathcal{K}_{2}\\setminus\\{0\\}.\n\\]\n\n2.  For every admissible $k$ determine the number of pairwise non-isomorphic finite abelian groups $G$ with $P(G)=2^{\\,k}$, and decide whether this number can ever exceed $1$.\n\n3.  Decide whether $7001\\in\\mathcal{K}_{2}$; if the answer is affirmative, state how many non-isomorphic groups realise this value and describe them up to isomorphism.",
      "solution": "Throughout $v_{2}(\\,\\cdot\\,)$ denotes the $2$-adic valuation; all groups are written additively.\n\n\\textbf{A.  Invariant factors, multiplicities and a fundamental formula.}  \nBy the structure theorem every finite abelian $2$–group has a decomposition  \n\\[\nG\\cong\\bigoplus_{j=1}^{t}\\mathbf{Z}/2^{\\lambda_{j}}\\mathbf{Z}\n\\quad(\\lambda_{1}\\ge\\dots\\ge\\lambda_{t}\\ge 1).\\tag{1}\n\\]\nPut $h=\\lambda_{1}$ and define  \n\\[\ne_{m}=\\#\\{\\,j\\mid\\lambda_{j}\\ge m\\,\\}\\quad(1\\le m\\le h),\\qquad\nS_{m}=\\sum_{q=1}^{m}e_{q},\\qquad S_{0}=0.\\tag{2}\n\\]\nThen $0<S_{1}<\\dots<S_{h}$ and $S_{h}=t$.  \nThe subgroup $G[2^{m}]$ of elements annihilated by $2^{m}$ has order $2^{S_{m}}$; hence there are $2^{S_{m}}-2^{S_{m-1}}$ elements of exact order $2^{m}$.  \nTaking the product of element orders gives\n\\[\nk(G)=\\sum_{m=1}^{h}m\\bigl(2^{S_{m}}-2^{S_{m-1}}\\bigr)\n      =h\\cdot 2^{S_{h}}-\\sum_{m=0}^{h-1}2^{S_{m}}.\\tag{3}\n\\]\n\n\\textbf{B.  From a $2$–group to a $2$–admissible sequence.}  \nSet $a_{i}=S_{i}\\;(1\\le i\\le h)$.  \nBecause $e_{i}=S_{i}-S_{i-1}>0$ and $e_{i+1}\\le e_{i}$, the gaps $d_{i}=a_{i}-a_{i-1}=e_{i}$ are weakly decreasing and positive, so  \n\\[\n(a_{1},\\dots ,a_{h}) \\text{ is $2$–admissible.}\n\\]\nEquation (3) is precisely formula $(\\star)$ with $r=h$.  Thus every non-trivial $2$–group yields a $2$–admissible sequence and an exponent $k$ with $2^{k}=P(G)$.\n\n\\textbf{C.  From a $2$–admissible sequence to a $2$–group.}  \nConversely, let $(a_{1},\\dots ,a_{r})$ be $2$–admissible and put\n\\[\ne_{m}=a_{m}-a_{m-1}\\quad(1\\le m\\le r).\n\\]\nThe sequence $(e_{1},\\dots ,e_{r})$ is strictly positive and weakly decreasing, hence there exists a unique list  \n\\[\n\\lambda_{1}\\ge\\dots\\ge\\lambda_{t}\\ge 1\n\\quad(t=e_{1})\n\\]\ngiven by\n\\[\n\\lambda_{j}=\\max\\{\\,m\\mid e_{m}\\ge j\\,\\}\\qquad(1\\le j\\le t).\\tag{4}\n\\]\nWith these $\\lambda_{j}$ formula (1) defines a $2$–group $G$ for which $S_{m}=a_{m}$ and, by (3), $k(G)=k$ from $(\\star)$.  \nHence\n\\[\n\\{\\text{non-trivial }2\\text{–groups}\\}\\twoheadrightarrow\n\\{\\text{$2$–admissible sequences}\\}\\twoheadrightarrow\n\\mathcal{K}_{2}\\setminus\\{0\\}\\tag{5}\n\\]\nare both surjections.\n\n\\textbf{D.  Injectivity of the map $(\\ast)\\mapsto k$.}  \nLet  \n\\[\nA=(a_{1},\\dots ,a_{r}),\\qquad\nB=(b_{1},\\dots ,b_{s})\n\\]\nbe $2$–admissible and assume $k(A)=k(B)=k$.  We prove $A=B$.\n\n\\emph{Step 1: the first entry is determined by $k$.}  \nBecause\n\\[\nk+1=r\\cdot 2^{a_{r}}-\\sum_{i=1}^{r-1}2^{a_{i}}\n          =s\\cdot 2^{b_{s}}-\\sum_{j=1}^{s-1}2^{b_{j}},\n\\]\nevery term on the right is divisible by $2^{a_{1}}$ (respectively $2^{b_{1}}$).  \nHence\n\\[\nv_{2}(k+1)=a_{1}=b_{1}.\\tag{6}\n\\]\n\n\\emph{Step 2: reduction to shorter sequences.}  \nPut\n\\[\nk^{\\prime}:=\\frac{k+1-2^{a_{1}}}{2^{a_{1}}}\n           =r\\cdot 2^{a_{r}-a_{1}}-\\sum_{i=2}^{r-1}2^{a_{i}-a_{1}}\n           =s\\cdot 2^{b_{s}-a_{1}}-\\sum_{j=2}^{s-1}2^{b_{j}-a_{1}}.\\tag{7}\n\\]\nSet\n\\[\nA^{\\prime}=(a_{2}-a_{1},\\dots ,a_{r}-a_{1}),\\qquad\nB^{\\prime}=(b_{2}-a_{1},\\dots ,b_{s}-a_{1}).\n\\]\nBecause the original gaps $d_{i}$ are weakly decreasing, the new sequences $A^{\\prime}$ and $B^{\\prime}$ are again $2$–admissible (possibly empty).  \nMoreover $k^{\\prime}+1$ is obtained from $A^{\\prime}$ and $B^{\\prime}$ via $(\\star)$ exactly as $k+1$ was obtained from $A$ and $B$.\n\n\\emph{Step 3: induction.}  \nIterating the construction of Step 2 we build successive quotients\n\\[\nk_{0}+1=k+1,\\quad\nk_{1}+1=k^{\\prime}+1,\\quad\nk_{2}+1,\\dots\n\\]\nand successively shorter $2$–admissible sequences; on each level the first entry is determined by the $2$-adic valuation, by (6).  \nThe process terminates when the sequence becomes empty, showing that \\emph{each entry of $A$ and of $B$ is uniquely determined by $k$}.  Hence $A=B$.\n\n\\emph{Alternative direct check for the last entry.}  \nIf one prefers to stay on one level, it suffices to distinguish three cases.\n\n\\smallskip\n\\noindent\n(i) $r\\neq s$.  \nAssume $r<s$ and $a_{i}=b_{i}$ for $i\\le r$.  Then\n\\[\nk+1=r\\cdot 2^{a_{r}}-\\!\\!\\sum_{i=1}^{r-1}\\!\\!2^{a_{i}}\n     <s\\cdot 2^{b_{s}}-\\!\\!\\sum_{j=1}^{s-1}\\!\\!2^{b_{j}}=k+1,\n\\]\na contradiction.\n\n\\smallskip\n\\noindent\n(ii) $r=s$ and $t<r$ is the first index with $a_{t}\\neq b_{t}$.  \nExactly as in the original solution one reduces modulo $2^{a_{t}+1}$ to obtain a contradiction.\n\n\\smallskip\n\\noindent\n(iii) $r=s$ and the sequences differ \\emph{only} in the last entry.  \nIf $a_{r}<b_{r}$, then\n\\[\nk+1=r\\cdot 2^{a_{r}}-\\sum_{i=1}^{r-1}2^{a_{i}}\n     <r\\cdot 2^{b_{r}}-\\sum_{i=1}^{r-1}2^{a_{i}}=k+1,\n\\]\nagain impossible.  Thus $a_{r}=b_{r}$.  \n\n\\smallskip\nEither way the map $(\\ast)\\mapsto k$ is injective.  Combined with (5) we have the bijections\n\\[\n\\bigl\\{\\text{non-trivial }2\\text{–groups}\\bigr\\}\n\\longleftrightarrow\n\\bigl\\{\\text{$2$–admissible sequences}\\bigr\\}\n\\longleftrightarrow\n\\mathcal{K}_{2}\\setminus\\{0\\}.\\tag{8}\n\\]\n\n\\textbf{E.  A convenient $2$-adic test.}  \nPutting $m=0$ in (3) yields\n\\[\nk+1=h\\cdot 2^{S_{h}}-\\sum_{m=1}^{h-1}2^{S_{m}}.\\tag{9}\n\\]\nAll summands are divisible by $2^{S_{1}}$, while the quotient is odd, so\n\\[\nv_{2}(k+1)=S_{1}=e_{1}=t,\\tag{10}\n\\]\nthe number of invariant factors of every group realising $k$.\n\n\\textbf{F.  Counting isomorphism classes for a fixed $k$.}  \nBecause of (8) each $k\\in\\mathcal{K}_{2}\\setminus\\{0\\}$ originates from a \\emph{unique} $2$–admissible sequence and therefore from a \\emph{unique} structure sequence $(\\lambda_{1},\\dots ,\\lambda_{t})$.  \nConsequently\n\\[\n\\forall\\,k\\in\\mathcal{K}_{2}\\qquad\n\\bigl|\\{\\,G\\mid P(G)=2^{\\,k}\\}/\\!\\!\\cong\\bigr|=\n\\begin{cases}\n1,&k>0,\\\\\n1,&k=0 \\text{ (the trivial group).}\n\\end{cases}\\tag{11}\n\\]\nThe required number never exceeds $1$.\n\n\\textbf{G.  The cyclic case.}  \nFor $t=1$ we have $\\lambda=(h)$ with $h\\ge 1$ and $S_{m}=\\min\\{m,h\\}$.  Substituting into (3) gives\n\\[\nk=(h-1)\\cdot 2^{h}+1\\qquad(h\\ge 1).\\tag{12}\n\\]\n\n\\textbf{H.  The test value $k=7001$.}  \nCompute $k+1=7002=2\\cdot 3501$, so $v_{2}(k+1)=1$.  \nBy (10) any realising group would have $t=1$ and hence be cyclic; thus (12) must satisfy\n\\[\n(h-1)\\cdot 2^{h}=7000.\\tag{13}\n\\]\nSince $v_{2}(7000)=3$, necessarily $h\\le 3$.  Direct inspection\n\\[\nh=1\\Longrightarrow 0,\\qquad\nh=2\\Longrightarrow 4,\\qquad\nh=3\\Longrightarrow 16,\n\\]\nshows that (13) has no solution.  Therefore\n\\[\n7001\\notin\\mathcal{K}_{2}.\\tag{14}\n\\]\n\n\\textbf{I.  Final answers.}  \n\n(1)  For $k>0$, $k\\in\\mathcal{K}_{2}$ iff there is a $2$–admissible sequence satisfying $(\\star)$; the correspondences in (8) are bijective.\n\n(2)  For every $k\\in\\mathcal{K}_{2}$ exactly \\emph{one} finite abelian group (up to isomorphism) satisfies $P(G)=2^{\\,k}$; the number never exceeds $1$.\n\n(3)  One has $7001\\notin\\mathcal{K}_{2}$; hence no finite abelian group realises the value $2^{7001}$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.813749",
        "was_fixed": false,
        "difficulty_analysis": "The original task asked for the mere existence of a group whose order-product is a fixed power of 2.  The enhanced variant demands\n\n1. A complete characterisation of *all* exponents k that can occur (Problem (1)),  \n2. An exact enumeration of the isomorphism classes for every admissible k (Problem (2)), and  \n3. An explicit decision for a concrete large k, namely 7001 (Problem (3)).\n\nSolving this requires:\n\n• Encoding abelian 2–groups via integer partitions and analysing the cumulative minima S_m;  \n• Deriving and manipulating a non-obvious closed formula (3) through telescoping sums;  \n• Proving the *bijectivity* of the map λ ↔ k(λ), which forces a delicate 2-adic analysis rather than the single congruence used in the original solution;  \n• Interpreting binary strings as concatenations of structured blocks imposed by the partition, an additional combinatorial layer absent from the kernel problem;  \n• Producing a proof that the enumeration function is identically 1, again relying on the injectivity part of the bijection.\n\nThese ingredients combine group theory, valuation theory, partition combinatorics, and binary arithmetic, far exceeding the single congruence argument needed before."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nP(G)=\\prod_{g\\in G}\\operatorname{ord}(g)\n\\]\nbe the product of the orders of all elements of a finite abelian group $G$, and put  \n\\[\n\\mathcal{K}_{2}\\;=\\;\\Bigl\\{k\\ge 0\\;:\\;\\exists\\;\\text{finite abelian }G\\text{ with }P(G)=2^{\\,k}\\Bigr\\}.\n\\]\n\nThroughout, finite abelian $2$–groups are written additively.  \nFor such a group  \n\\[\nG\\cong\\bigoplus_{j=1}^{t}\\mathbf{Z}/2^{\\lambda_{j}}\\mathbf{Z}\n\\qquad(\\lambda_{1}\\ge\\lambda_{2}\\ge\\dots\\ge\\lambda_{t}\\ge 1),\n\\]\ndefine the multiplicities  \n\\[\ne_{m}(G):=\\bigl|\\{\\,j\\mid\\lambda_{j}\\ge m\\,\\}\\bigr| \\qquad(m\\ge 1).\n\\]\nThe finite sequence $e_{1}(G),e_{2}(G),\\dots ,e_{\\lambda_{1}}(G)$ is strictly positive and weakly decreasing; conversely every such sequence occurs for a unique isomorphism class of $2$–groups.\n\nPut  \n\\[\nS_{m}(G):=\\sum_{q=1}^{m}e_{q}(G)\\qquad(1\\le m\\le\\lambda_{1}).\\tag{$\\dagger$}\n\\]\nHence $0<S_{1}<\\dots<S_{\\lambda_{1}}$.\n\nA strictly increasing finite sequence  \n\\[\n0<a_{1}<a_{2}<\\dots<a_{r}\\tag{$\\ast$}\n\\]\nis called $2$–admissible if its consecutive gaps  \n\\[\nd_{i}:=a_{i}-a_{i-1}\\qquad(a_{0}:=0)\\tag{$\\ast\\ast$}\n\\]\nsatisfy $d_{1}\\ge d_{2}\\ge\\dots\\ge d_{r}>0$.\n\n1.\\; Show that for every positive integer $k$ one has $k\\in\\mathcal{K}_{2}$ if and only if there exists  \n   a $2$–admissible sequence $(\\ast)$ such that  \n   \\[\n   k=r\\cdot 2^{a_{r}}-\\bigl(1+2^{a_{1}}+2^{a_{2}}+\\dots+2^{a_{r-1}}\\bigr).\\tag{$\\star$}\n   \\]\n   (The exponent $k=0$ is realised by the trivial group.)  \n   Moreover prove that the assignments  \n   \\[\n   G\\longmapsto\\bigl(S_{1}(G),\\dots,S_{\\lambda_{1}}(G)\\bigr)\\quad\\text{and}\\quad\n   (a_{1},\\dots,a_{r})\\longmapsto k\\text{ via }(\\star)\n   \\]\n   give bijections  \n   \\[\n   \\{\\text{isomorphism classes of non-trivial }2\\text{–groups}\\}\n   \\;\\longleftrightarrow\\;\n   \\{\\text{$2$–admissible sequences}\\}\n   \\;\\longleftrightarrow\\;\n   \\mathcal{K}_{2}\\setminus\\{0\\}.\n   \\]\n\n2.\\; For every admissible exponent $k$ determine the exact number of pairwise non-isomorphic finite abelian groups $G$ with $P(G)=2^{\\,k}$, and decide whether this number can ever exceed $1$.\n\n3.\\; Decide whether $7001\\in\\mathcal{K}_{2}$; if the answer is affirmative, state how many non-isomorphic groups realise this value and describe them up to isomorphism.",
      "solution": "Throughout $v_{2}(\\,\\cdot\\,)$ denotes the $2$-adic valuation and all groups are written additively.\n\n\\textbf{A.  Invariant factors, multiplicities and the fundamental formula.}  \nBy the structure theorem every finite abelian $2$–group is  \n\\[\nG\\cong\\bigoplus_{j=1}^{t}\\mathbf{Z}/2^{\\lambda_{j}}\\mathbf{Z}\n\\qquad(\\lambda_{1}\\ge\\dots\\ge\\lambda_{t}\\ge 1).\\tag{1}\n\\]\nPut $h:=\\lambda_{1}$ and define  \n\\[\ne_{m}:=\\#\\{\\,j\\mid\\lambda_{j}\\ge m\\,\\}\\quad(1\\le m\\le h),\\qquad\nS_{m}:=\\sum_{q=1}^{m}e_{q},\\qquad S_{0}:=0.\\tag{2}\n\\]\nThen $0<S_{1}<\\dots<S_{h}$ and $S_{h}=t$.  \nLet $G[2^{m}]$ be the subgroup of elements annihilated by $2^{m}$.  One has $\\lvert G[2^{m}]\\rvert=2^{S_{m}}$; hence the number of elements of exact order $2^{m}$ equals $2^{S_{m}}-2^{S_{m-1}}$.  \nMultiplying the orders of all elements gives  \n\\[\nk(G)=\\sum_{m=1}^{h}m\\bigl(2^{S_{m}}-2^{S_{m-1}}\\bigr)\n      =h\\cdot 2^{S_{h}}-\\sum_{m=0}^{h-1}2^{S_{m}}.\\tag{3}\n\\]\n\n\\textbf{B.  From a $2$–group to a $2$–admissible sequence.}  \nPut $a_{i}:=S_{i}\\;(1\\le i\\le h)$.  \nBecause $e_{i}=S_{i}-S_{i-1}>0$ and $e_{i+1}\\le e_{i}$, the gaps $d_{i}=a_{i}-a_{i-1}=e_{i}$ are weakly decreasing and positive.  Thus $\\bigl(a_{1},\\dots,a_{h}\\bigr)$ is $2$-admissible.  \nEquation (3) rewrites exactly as $(\\star)$ with $r=h$.\n\n\\textbf{C.  From a $2$–admissible sequence to a $2$–group.}  \nConversely, let $\\bigl(a_{1},\\dots,a_{r}\\bigr)$ be $2$-admissible and set $e_{m}:=a_{m}-a_{m-1}$ $(1\\le m\\le r)$.  \nThe weakly decreasing positive sequence $(e_{1},\\dots,e_{r})$ satisfies (2) with $h:=r$.  \nDefine  \n\\[\n\\lambda_{j}:=\\max\\{\\,m\\mid e_{m}\\ge j\\,\\}\\quad(1\\le j\\le e_{1}).\\tag{4}\n\\]\nThen $(\\lambda_{1},\\dots,\\lambda_{t})$ as in (1) yields a $2$–group $G$ with $S_{m}=a_{m}$, and (3) gives $k(G)=k$ from $(\\star)$.  \nHence we obtain a surjection  \n\\[\n\\{\\text{isomorphism classes of non-trivial }2\\text{–groups}\\}\\twoheadrightarrow\n\\mathcal{K}_{2}\\setminus\\{0\\}.\\tag{5}\n\\]\n\n\\textbf{D.  Uniqueness of the representation – injectivity of $(\\star)$.}  \nWe prove that the map $(\\ast)\\mapsto k$ defined by $(\\star)$ is injective.  \n\nAssume two $2$–admissible sequences  \n\\[\nA=\\bigl(a_{1},\\dots,a_{r}\\bigr),\\qquad\nB=\\bigl(b_{1},\\dots,b_{s}\\bigr)\n\\]\nsatisfy $(\\star)$ with the same $k$.  \nLet $t$ be the smallest index with $a_{t}\\neq b_{t}$ and assume $a_{t}<b_{t}$; if no such $t$ exists the sequences coincide and we are done.\n\nWrite $k+1$ in the two $(\\star)$–forms:\n\\[\nk+1\n=r\\cdot 2^{a_{r}}-\\sum_{i=1}^{r-1}2^{a_{i}}\n=s\\cdot 2^{b_{s}}-\\sum_{j=1}^{s-1}2^{b_{j}}.\\tag{6}\n\\]\nReduce (6) modulo $2^{a_{t}+1}$.\n\n\\emph{For the $A$–expression:}  \nAll terms $2^{a_{i}}$ with $i\\ge t$ except $i=t$ are divisible by $2^{a_{t}+1}$, and $r\\cdot 2^{a_{r}}$ is as well because $a_{r}>a_{t}$.  Hence\n\\[\nk+1\\equiv-2^{a_{t}}-\\sum_{i=1}^{t-1}2^{a_{i}}\\pmod{2^{a_{t}+1}}.\\tag{7}\n\\]\n\n\\emph{For the $B$–expression:}  \nSince $b_{t}\\ge a_{t}+1$, every term on the right-hand side of the $B$–expression is divisible by $2^{a_{t}+1}$ except possibly those with $j<t$.  Consequently\n\\[\nk+1\\equiv-\\sum_{i=1}^{t-1}2^{a_{i}}\\pmod{2^{a_{t}+1}}.\\tag{8}\n\\]\n\nSubtracting (8) from (7) gives  \n\\[\n0\\equiv -2^{a_{t}}\\pmod{2^{a_{t}+1}},\n\\]\na contradiction.  Therefore no two distinct $2$–admissible sequences can produce the same $k$, and the map $(\\ast)\\mapsto k$ is injective.  Together with the surjectivity proved in B–C we obtain the desired bijections:\n\\[\n\\{\\text{isomorphism classes of non-trivial }2\\text{–groups}\\}\n\\;\\longleftrightarrow\\;\n\\{\\text{$2$–admissible sequences}\\}\n\\;\\longleftrightarrow\\;\n\\mathcal{K}_{2}\\setminus\\{0\\}.\\tag{9}\n\\]\n\n\\textbf{E.  A useful $2$–adic test.}  \nSetting $m=0$ in (3) we have\n\\[\nk+1=h\\cdot 2^{S_{h}}-\\sum_{m=1}^{h-1}2^{S_{m}}.\\tag{10}\n\\]\nAll terms on the right-hand side are divisible by $2^{S_{1}}$, but the quotient is odd, so\n\\[\nv_{2}(k+1)=S_{1}=e_{1}=t,\\tag{11}\n\\]\nthe number of invariant factors of any realising group.\n\n\\textbf{F.  Counting isomorphism classes for a fixed $k$.}  \nBecause of (9) every positive $k\\in\\mathcal{K}_{2}$ corresponds to \\emph{exactly one} $2$–admissible sequence and hence to \\emph{exactly one} isomorphism class of finite abelian $2$–groups.  Adding the trivial group for $k=0$ yields\n\\[\n\\forall\\,k\\in\\mathcal{K}_{2}\\qquad\n\\bigl|\\{\\,G\\mid P(G)=2^{\\,k}\\}/\\!\\!\\cong\\bigr|=1.\\tag{12}\n\\]\nThus the required number never exceeds $1$.\n\n\\textbf{G.  The cyclic case.}  \nFor $t=1$ we have $\\lambda=(h)$ $(h\\ge 1)$ and $S_{m}=\\min\\{m,h\\}$.  Substituting into (3) gives\n\\[\nk=(h-1)\\cdot 2^{h}+1\\qquad(h\\ge 1).\\tag{13}\n\\]\n\n\\textbf{H.  The test value $k=7001$.}  \nCompute $k+1=7002=2\\cdot 3501$, so $v_{2}(k+1)=1$.  By (11) any realising group would have $t=1$ and therefore be cyclic; hence (13) must satisfy\n\\[\n(h-1)\\cdot 2^{h}=7000.\\tag{14}\n\\]\nBecause $v_{2}(7000)=3$ one must have $h\\le 3$.  Direct inspection\n\\[\nh=1\\;\\Rightarrow\\;0,\\quad\nh=2\\;\\Rightarrow\\;4,\\quad\nh=3\\;\\Rightarrow\\;16,\n\\]\nshows (14) has no solution.  Therefore\n\\[\n7001\\notin\\mathcal{K}_{2}.\\tag{15}\n\\]\n\n\\textbf{I.  Final answers.}  \n(1)  The set $\\mathcal{K}_{2}\\setminus\\{0\\}$ is in bijection with $2$–admissible sequences; every admissible $k$ possesses the unique representation $(\\star)$.  \n\n(2)  For every $k\\in\\mathcal{K}_{2}$ precisely one finite abelian group (up to isomorphism) satisfies $P(G)=2^{\\,k}$; the number of such groups never exceeds $1$.  \n\n(3)  One has $7001\\notin\\mathcal{K}_{2}$; consequently no finite abelian group realises the value $2^{7001}$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.622868",
        "was_fixed": false,
        "difficulty_analysis": "The original task asked for the mere existence of a group whose order-product is a fixed power of 2.  The enhanced variant demands\n\n1. A complete characterisation of *all* exponents k that can occur (Problem (1)),  \n2. An exact enumeration of the isomorphism classes for every admissible k (Problem (2)), and  \n3. An explicit decision for a concrete large k, namely 7001 (Problem (3)).\n\nSolving this requires:\n\n• Encoding abelian 2–groups via integer partitions and analysing the cumulative minima S_m;  \n• Deriving and manipulating a non-obvious closed formula (3) through telescoping sums;  \n• Proving the *bijectivity* of the map λ ↔ k(λ), which forces a delicate 2-adic analysis rather than the single congruence used in the original solution;  \n• Interpreting binary strings as concatenations of structured blocks imposed by the partition, an additional combinatorial layer absent from the kernel problem;  \n• Producing a proof that the enumeration function is identically 1, again relying on the injectivity part of the bijection.\n\nThese ingredients combine group theory, valuation theory, partition combinatorics, and binary arithmetic, far exceeding the single congruence argument needed before."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}