{
  "index": "2000-B-1",
  "type": "COMB",
  "tag": [
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $a_j,b_j,c_j$ be integers for $1\\leq j\\leq N$.  Assume for each\n$j$, at least one of $a_j,b_j,c_j$ is odd.  Show that there exist integers\n$r$, $s$, $t$ such that $ra_j+sb_j+tc_j$ is odd for at least $4N/7$ values\nof $j$, $1\\leq j\\leq N$.",
  "solution": "Consider the seven triples $(a,b,c)$ with $a,b,c \\in \\{0,1\\}$ not\nall zero. Notice that if $r_j, s_j, t_j$ are not all even, then four\nof the sums $ar_j + bs_j + ct_j$ with $a,b,c \\in \\{0,1\\}$ are even\nand four are odd. Of course the sum with $a=b=c=0$ is even, so at\nleast four of the seven triples with $a,b,c$ not all zero yield an odd\nsum. In other words, at least $4N$ of the tuples $(a,b,c,j)$ yield\nodd sums. By the pigeonhole principle, there is a triple $(a,b,c)$\nfor which at least $4N/7$ of the sums are odd.",
  "vars": [
    "a",
    "a_j",
    "b",
    "b_j",
    "c",
    "c_j",
    "j",
    "r",
    "r_j",
    "s",
    "s_j",
    "t",
    "t_j"
  ],
  "params": [
    "N"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "scalara",
        "a_j": "sequencea",
        "b": "scalarb",
        "b_j": "sequenceb",
        "c": "scalarc",
        "c_j": "sequencec",
        "j": "indexer",
        "r": "scalarr",
        "r_j": "sequencer",
        "s": "scalars",
        "s_j": "sequences",
        "t": "scalart",
        "t_j": "sequencet",
        "N": "totalnum"
      },
      "question": "Let $sequencea,sequenceb,sequencec$ be integers for $1\\leq indexer\\leq totalnum$.  Assume for each\n$indexer$, at least one of $sequencea,sequenceb,sequencec$ is odd.  Show that there exist integers\n$scalarr$, $scalars$, $scalart$ such that $scalarr sequencea+scalars sequenceb+scalart sequencec$ is odd for at least $4 totalnum/7$ values\nof $indexer$, $1\\leq indexer\\leq totalnum$.",
      "solution": "Consider the seven triples $(scalara,scalarb,scalarc)$ with $scalara,scalarb,scalarc \\in \\{0,1\\}$ not\nall zero. Notice that if $sequencer, sequences, sequencet$ are not all even, then four\nof the sums $scalara sequencer + scalarb sequences + scalarc sequencet$ with $scalara,scalarb,scalarc \\in \\{0,1\\}$ are even\nand four are odd. Of course the sum with $scalara=scalarb=scalarc=0$ is even, so at\nleast four of the seven triples with $scalara,scalarb,scalarc$ not all zero yield an odd\nsum. In other words, at least $4 totalnum$ of the tuples $(scalara,scalarb,scalarc,indexer)$ yield\nodd sums. By the pigeonhole principle, there is a triple $(scalara,scalarb,scalarc)$\nfor which at least $4 totalnum/7$ of the sums are odd."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "pineapple",
        "a_j": "dragonfish",
        "b": "honeycomb",
        "b_j": "tangerine",
        "c": "lighthouse",
        "c_j": "buttercup",
        "j": "waterfall",
        "r": "starlight",
        "r_j": "whisperer",
        "s": "moonstone",
        "s_j": "firebrand",
        "t": "earthworm",
        "t_j": "blueberry",
        "N": "candlestick"
      },
      "question": "Let $dragonfish,tangerine,buttercup$ be integers for $1\\leq waterfall\\leq candlestick$.  Assume for each\n$waterfall$, at least one of $dragonfish,tangerine,buttercup$ is odd.  Show that there exist integers\n$starlight$, $moonstone$, $earthworm$ such that $starlightdragonfish+moonstonetangerine+earthwormbuttercup$ is odd for at least $4candlestick/7$ values\nof $waterfall$, $1\\leq waterfall\\leq candlestick$.",
      "solution": "Consider the seven triples $(pineapple,honeycomb,lighthouse)$ with $pineapple,honeycomb,lighthouse \\in \\{0,1\\}$ not\nall zero. Notice that if $whisperer, firebrand, blueberry$ are not all even, then four\nof the sums $pineapplewhisperer + honeycombfirebrand + lighthouseblueberry$ with $pineapple,honeycomb,lighthouse \\in \\{0,1\\}$ are even\nand four are odd. Of course the sum with $pineapple=honeycomb=lighthouse=0$ is even, so at\nleast four of the seven triples with $pineapple,honeycomb,lighthouse$ not all zero yield an odd\nsum. In other words, at least $4candlestick$ of the tuples $(pineapple,honeycomb,lighthouse,waterfall)$ yield\nodd sums. By the pigeonhole principle, there is a triple $(pineapple,honeycomb,lighthouse)$\nfor which at least $4candlestick/7$ of the sums are odd."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "omegaelem",
        "a_j": "omegaelemindex",
        "b": "terminus",
        "b_j": "terminusindex",
        "c": "conclude",
        "c_j": "concludeindex",
        "j": "entirety",
        "r": "stillness",
        "r_j": "stillnessindex",
        "s": "constant",
        "s_j": "constantindex",
        "t": "timeless",
        "t_j": "timelessindex",
        "N": "minuscule"
      },
      "question": "Let $omegaelemindex,terminusindex,concludeindex$ be integers for $1\\leq entirety\\leq minuscule$.  Assume for each\n$entirety$, at least one of $omegaelemindex,terminusindex,concludeindex$ is odd.  Show that there exist integers\n$stillness$, $constant$, $timeless$ such that $stillnessomegaelemindex+constantterminusindex+timelessconcludeindex$ is odd for at least $4minuscule/7$ values\nof $entirety$, $1\\leq entirety\\leq minuscule$.",
      "solution": "Consider the seven triples $(omegaelem,terminus,conclude)$ with $omegaelem,terminus,conclude \\in \\{0,1\\}$ not\nall zero. Notice that if $stillnessindex, constantindex, timelessindex$ are not all even, then four\nof the sums $omegaelemstillnessindex + terminusconstantindex + concludetimelessindex$ with $omegaelem,terminus,conclude \\in \\{0,1\\}$ are even\nand four are odd. Of course the sum with $omegaelem=terminus=conclude=0$ is even, so at\nleast four of the seven triples with $omegaelem,terminus,conclude$ not all zero yield an odd\nsum. In other words, at least $4minuscule$ of the tuples $(omegaelem,terminus,conclude,entirety)$ yield\nodd sums. By the pigeonhole principle, there is a triple $(omegaelem,terminus,conclude)$\nfor which at least $4minuscule/7$ of the sums are odd."
    },
    "garbled_string": {
      "map": {
        "a": "vxzplmqa",
        "a_j": "yrcdnbtw",
        "b": "qlkestru",
        "b_j": "hvnixoae",
        "c": "pugozalf",
        "c_j": "nsyqerjm",
        "j": "tbkxzmnr",
        "r": "gafqhscv",
        "r_j": "qiuwpztk",
        "s": "wecuragn",
        "s_j": "fjzdyltx",
        "t": "lxvskmho",
        "t_j": "zmqorptw",
        "N": "dodjnswc"
      },
      "question": "Let $yrcdnbtw,hvnixoae,nsyqerjm$ be integers for $1\\leq tbkxzmnr\\leq dodjnswc$.  Assume for each\n$tbkxzmnr$, at least one of $yrcdnbtw,hvnixoae,nsyqerjm$ is odd.  Show that there exist integers\n$gafqhscv$, $wecuragn$, $lxvskmho$ such that $gafqhscv yrcdnbtw+wecuragn hvnixoae+lxvskmho nsyqerjm$ is odd for at least $4dodjnswc/7$ values\nof $tbkxzmnr$, $1\\leq tbkxzmnr\\leq dodjnswc$.",
      "solution": "Consider the seven triples $(vxzplmqa,qlkestru,pugozalf)$ with $vxzplmqa,qlkestru,pugozalf \\in \\{0,1\\}$ not\nall zero. Notice that if $qiuwpztk, fjzdyltx, zmqorptw$ are not all even, then four\nof the sums $vxzplmqa qiuwpztk + qlkestru fjzdyltx + pugozalf zmqorptw$ with $vxzplmqa,qlkestru,pugozalf \\in \\{0,1\\}$ are even\nand four are odd. Of course the sum with $vxzplmqa=qlkestru=pugozalf=0$ is even, so at\nleast four of the seven triples with $vxzplmqa,qlkestru,pugozalf$ not all zero yield an odd\nsum. In other words, at least $4dodjnswc$ of the tuples $(vxzplmqa,qlkestru,pugozalf,tbkxzmnr)$ yield\nodd sums. By the pigeonhole principle, there is a triple $(vxzplmqa,qlkestru,pugozalf)$\nfor which at least $4dodjnswc/7$ of the sums are odd."
    },
    "kernel_variant": {
      "question": "Let \\(\\Lambda\\) be a finite index set with \\(|\\Lambda|=M\\).  For each \\(\\alpha\\in\\Lambda\\) let the triple of integers \\((p_{\\alpha},q_{\\alpha},r_{\\alpha})\\) satisfy the condition that at least one of the three numbers is odd.  Prove that there exist integers \\(x,y,z\\) such that the dot–products\n\\[\n   x p_{\\alpha}+y q_{\\alpha}+z r_{\\alpha}\n\\]\nare odd for at least \\(\\dfrac{4M}{7}\\) different indices \\(\\alpha\\in\\Lambda\\).",
      "solution": "Step 1 — enumerate test triples.\nConsider the seven non–zero triples (u,v,w) with u,v,w∈{0,1}. (There are 2³–1=7 of them.)\n\nStep 2 — a parity observation.\nFix an index α∈Λ. Because not all of pₐ,qₐ,rₐ are even, the triple (pₐ,qₐ,rₐ) is not congruent to (0,0,0) mod 2. For such a fixed triple, exactly four of the eight sums\n   u pₐ + v qₐ + w rₐ  (u,v,w∈{0,1})\nare odd and the remaining four are even. The case (u,v,w)=(0,0,0) produces an even sum, so among the seven non–zero test triples at least four give an odd value.\n\nStep 3 — double counting.\nForm all ordered pairs ((u,v,w),α) with (u,v,w)≠(0,0,0) and α∈Λ; there are 7M such pairs. From Step 2, for each fixed α at least four of the corresponding seven sums are odd. Hence there are at least 4M pairs for which the sum is odd.\n\nStep 4 — pigeonhole principle.\nDistribute those 4M odd pairs among the seven test triples (u,v,w). Some particular triple (u*,v*,w*) must be responsible for at least ⌈4M/7⌉ ≥ 4M/7 of them; denote this number by K ≥ 4M/7.\n\nStep 5 — produce the required coefficients.\nTake\n   (x,y,z) = (u*,v*,w*).\nBy construction the dot‐product x pₐ + y qₐ + z rₐ is odd for those K indices α, and K ≥ 4M/7. Thus such integers x,y,z exist, completing the proof.",
      "_meta": {
        "core_steps": [
          "Enumerate the 2^3−1 = 7 non-zero triples (a,b,c) with entries in {0,1}.",
          "Parity fact: for any fixed (a_j,b_j,c_j) that is not all even, exactly four of the eight sums a·a_j + b·b_j + c·c_j (with a,b,c ∈ {0,1}) are odd; the (0,0,0) case is even, hence ≥4 of the 7 non-zero triples give an odd sum.",
          "Double–count: this yields at least 4N odd (triple, j) pairs among the 7N possible.",
          "Pigeonhole principle: some binary triple (a,b,c) must account for ≥4N⁄7 of those odd sums.",
          "Take r=a, s=b, t=c for that triple to satisfy the problem’s requirement."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Label of the index set of triples (currently 1≤j≤N). Any contiguous or non-contiguous indexing of size N works.",
            "original": "1 ≤ j ≤ N"
          },
          "slot2": {
            "description": "Choice of variable symbols for the three coefficients whose parity we use (now a,b,c / r,s,t). Renaming them does not affect the argument.",
            "original": "a,b,c ; r,s,t"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}