{
  "index": "1973-B-1",
  "type": "NT",
  "tag": [
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "B-1. Let \\( a_{1}, a_{2}, \\ldots, a_{2 n+1} \\) be a set of integers such that, if any one of them is removed, the remaining ones can be divided into two sets of \\( n \\) integers with equal sums. Prove \\( a_{1}=a_{2}=\\cdots=a_{2 n+1} \\).",
  "solution": "B-1. Since the sum of the \\( 2 n \\) integers remaining is always even, no matter which of the \\( a_{i} \\) is taken away, all of the \\( a_{i} \\) must have the same parity. Now a similar argument shows that they are all congruent \\( (\\bmod 4) \\); for the property held by the \\( a_{i} \\) is shared by the integers \\( a_{i} / 2 \\) or \\( \\left(a_{i}-1\\right) / 2 \\) (depending on whether the \\( a_{i} \\) are all even or all odd). Continuing in this manner, all of the \\( a_{i} \\) are congruent \\( \\left(\\bmod 2^{k}\\right) \\) for every \\( k \\). This is possible for integers only if they are equal.",
  "vars": [
    "a_1",
    "a_2",
    "a_2n+1",
    "a_i"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_1": "firstentry",
        "a_2": "secondentry",
        "a_2n+1": "lastentry",
        "a_i": "genericentry",
        "n": "countsize"
      },
      "question": "B-1. Let \\( firstentry, secondentry, \\ldots, lastentry \\) be a set of integers such that, if any one of them is removed, the remaining ones can be divided into two sets of \\( countsize \\) integers with equal sums. Prove \\( firstentry=secondentry=\\cdots=lastentry \\).",
      "solution": "B-1. Since the sum of the \\( 2 countsize \\) integers remaining is always even, no matter which of the \\( genericentry \\) is taken away, all of the \\( genericentry \\) must have the same parity. Now a similar argument shows that they are all congruent \\( (\\bmod 4) \\); for the property held by the \\( genericentry \\) is shared by the integers \\( genericentry / 2 \\) or \\( \\left(genericentry-1\\right) / 2 \\) (depending on whether the \\( genericentry \\) are all even or all odd). Continuing in this manner, all of the \\( genericentry \\) are congruent \\( \\left(\\bmod 2^{k}\\right) \\) for every \\( k \\). This is possible for integers only if they are equal."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_1": "mushroom",
        "a_2": "pineapple",
        "a_2n+1": "salamander",
        "a_i": "porcupine",
        "n": "carnation"
      },
      "question": "B-1. Let \\( mushroom_{1}, pineapple_{2}, \\ldots, salamander_{2 carnation+1} \\) be a set of integers such that, if any one of them is removed, the remaining ones can be divided into two sets of \\( carnation \\) integers with equal sums. Prove \\( mushroom_{1}=pineapple_{2}=\\cdots=salamander_{2 carnation+1} \\).",
      "solution": "B-1. Since the sum of the \\( 2 carnation \\) integers remaining is always even, no matter which of the \\( porcupine_{i} \\) is taken away, all of the \\( porcupine_{i} \\) must have the same parity. Now a similar argument shows that they are all congruent \\( (\\bmod 4) \\); for the property held by the \\( porcupine_{i} \\) is shared by the integers \\( porcupine_{i} / 2 \\) or \\( \\left(porcupine_{i}-1\\right) / 2 \\) (depending on whether the \\( porcupine_{i} \\) are all even or all odd). Continuing in this manner, all of the \\( porcupine_{i} \\) are congruent \\( \\left(\\bmod 2^{k}\\right) \\) for every \\( k \\). This is possible for integers only if they are equal."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_1": "irrationalfirst",
        "a_2": "irrationalsecond",
        "a_2n+1": "irrationalterminal",
        "a_i": "irrationalelement",
        "n": "unboundedindex"
      },
      "question": "B-1. Let \\( irrationalfirst, irrationalsecond, \\ldots, irrationalterminal \\) be a set of integers such that, if any one of them is removed, the remaining ones can be divided into two sets of \\( unboundedindex \\) integers with equal sums. Prove \\( irrationalfirst=irrationalsecond=\\cdots=irrationalterminal \\).",
      "solution": "B-1. Since the sum of the \\( 2\\,unboundedindex \\) integers remaining is always even, no matter which of the \\( irrationalelement \\) is taken away, all of the \\( irrationalelement \\) must have the same parity. Now a similar argument shows that they are all congruent \\( (\\bmod 4) \\); for the property held by the \\( irrationalelement \\) is shared by the integers \\( irrationalelement / 2 \\) or \\( \\left(irrationalelement-1\\right) / 2 \\) (depending on whether the \\( irrationalelement \\) are all even or all odd). Continuing in this manner, all of the \\( irrationalelement \\) are congruent \\( \\left(\\bmod 2^{k}\\right) \\) for every \\( k \\). This is possible for integers only if they are equal."
    },
    "garbled_string": {
      "map": {
        "a_1": "qzxwvtnp",
        "a_2": "hjgrksla",
        "a_2n+1": "mnbvcxzl",
        "a_i": "plmoknij",
        "n": "asdfghjk"
      },
      "question": "B-1. Let \\( qzxwvtnp, hjgrksla, \\ldots, mnbvcxzl \\) be a set of integers such that, if any one of them is removed, the remaining ones can be divided into two sets of \\( asdfghjk \\) integers with equal sums. Prove \\( qzxwvtnp=hjgrksla=\\cdots=mnbvcxzl \\).",
      "solution": "B-1. Since the sum of the \\( 2 asdfghjk \\) integers remaining is always even, no matter which of the \\( plmoknij \\) is taken away, all of the \\( plmoknij \\) must have the same parity. Now a similar argument shows that they are all congruent \\( (\\bmod 4) \\); for the property held by the \\( plmoknij \\) is shared by the integers \\( plmoknij / 2 \\) or \\( \\left(plmoknij-1\\right) / 2 \\) (depending on whether the \\( plmoknij \\) are all even or all odd). Continuing in this manner, all of the \\( plmoknij \\) are congruent \\( \\left(\\bmod 2^{k}\\right) \\) for every \\( k \\). This is possible for integers only if they are equal."
    },
    "kernel_variant": {
      "question": "Let $k\\ge 1$ be an integer and let \\[b_{1},b_{2},\\dots ,b_{6k-1}\\] be integers with the property that, whenever any one of them is deleted, the remaining $6k-2$ integers can be partitioned into two disjoint subsets of $3k-1$ elements each having the same sum.  Prove that \n\\[b_{1}=b_{2}=\\dots =b_{6k-1}.\\]",
      "solution": "Set S = b_{1} + b_{2} + … + b_{6k−1}.\n\nStep 1 (parity).  When b_j is removed the remaining sum is S − b_j = 2T for some integer T.  Hence S ≡ b_j (mod 2) for every j, so all the b_j have the same parity.\n\nStep 2 (halving trick).  There are two cases.\n\n• If every b_j is even, write b_j = 2c_j.  Dividing all b’s by 2 shows that deleting any c_j from {c_i} again leaves 6k−2 numbers splittable into two groups of 3k−1 with equal sum.\n\n• If every b_j is odd, write b_j = 2c_j + 1.  After deleting some b_j, subtract 1 from each of the remaining 6k−2 numbers to get 2c_i.  Each of the two equal subset‐sums drops by (3k−1); dividing by 2 shows the c_j again satisfy the same property.\n\nIn both cases the new integers c_1, …, c_{6k−1} satisfy the original condition.\n\nStep 3 (iteration).  Re‐applying Step 1 to {c_i} shows all b_j ≡ (mod 4).  Repeating the halving argument gives congruence mod 8, then mod 16, and so on.  By induction\n\n    b_1 ≡ b_2 ≡ … ≡ b_{6k−1}  (mod 2^t)  for every t ≥ 1.\n\nStep 4 (conclusion).  A nonzero integer cannot be divisible by 2^t for every t, so the difference of any two b_j must be 0.  Therefore b_1 = … = b_{6k−1}.",
      "_meta": {
        "core_steps": [
          "Parity test:  (total − a_i) is always even ⇒ all a_i are congruent mod 2",
          "Halving trick:  replace each a_i by a_i/2 (even case) or (a_i−1)/2 (odd case); property survives because the two n-element parts lose the same amount",
          "Iterate the halving step ⇒ a_i are congruent mod 2^k for every k≥1",
          "Only equal integers can lie in every residue class mod 2^k ⇒ all a_i are identical"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Expression for the (odd) size of the whole set; any odd count symbolised similarly would work",
            "original": "2n+1"
          },
          "slot2": {
            "description": "Symbol used for the size of each part after one element is removed; the letter can be changed without affecting the logic",
            "original": "n"
          },
          "slot3": {
            "description": "Concrete modulus shown after the second step; any higher power of 2 could be used in its place",
            "original": "4"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}