{
  "index": "1969-B-1",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-1. Let \\( n \\) be a positive integer such that \\( n+1 \\) is divisible by 24 . Prove that the sum of all the divisors of \\( n \\) is divisible by 24 .",
  "solution": "B-1 The condition \\( 24 \\mid n+1 \\) is equivalent to \\( n \\equiv-1(\\bmod 3) \\) and \\( n \\equiv-1(\\bmod 8) \\). Let \\( d \\) be a divisor of \\( n \\), then \\( d \\equiv 1 \\) or \\( 2(\\bmod 3) \\) and \\( d \\equiv 1,3,5 \\) or \\( 7(\\bmod 8) \\). Since \\( d(n / d)=n \\equiv-1(\\bmod 3) \\) or \\( (\\bmod 8) \\), the only possibilities are:\n\\[\n\\begin{array}{llll}\nd \\equiv 1, & n / d \\equiv 2(\\bmod 3) & \\text { or } & \\text { vice versa } \\\\\nd \\equiv 1, & n / d \\equiv 7(\\bmod 8) & \\text { or } & \\text { vice versa } \\\\\nd \\equiv 3, & n / d \\equiv 5(\\bmod 8) & \\text { or } & \\text { vice versa. }\n\\end{array}\n\\]\n\nIn every case, \\( d+n / d \\equiv 0(\\bmod 3) \\) and \\( (\\bmod 8) \\). Thus \\( d+n / d \\) is a multiple of 24. Note that \\( d \\not \\equiv n / d \\) and thus no divisor is used twice in the pairing, so the sum of all the divisors is a multiple of 24 .",
  "vars": [
    "n",
    "d"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "posint",
        "d": "dividr"
      },
      "question": "B-1. Let \\( posint \\) be a positive integer such that \\( posint+1 \\) is divisible by 24 . Prove that the sum of all the divisors of \\( posint \\) is divisible by 24 .",
      "solution": "B-1 The condition \\( 24 \\mid posint+1 \\) is equivalent to \\( posint \\equiv-1(\\bmod 3) \\) and \\( posint \\equiv-1(\\bmod 8) \\). Let \\( dividr \\) be a divisor of \\( posint \\), then \\( dividr \\equiv 1 \\) or \\( 2(\\bmod 3) \\) and \\( dividr \\equiv 1,3,5 \\) or \\( 7(\\bmod 8) \\). Since \\( dividr(posint / dividr)=posint \\equiv-1(\\bmod 3) \\) or \\( (\\bmod 8) \\), the only possibilities are:\n\\[\n\\begin{array}{llll}\ndividr \\equiv 1, & posint / dividr \\equiv 2(\\bmod 3) & \\text { or } & \\text { vice versa } \\\\\ndividr \\equiv 1, & posint / dividr \\equiv 7(\\bmod 8) & \\text { or } & \\text { vice versa } \\\\\ndividr \\equiv 3, & posint / dividr \\equiv 5(\\bmod 8) & \\text { or } & \\text { vice versa. }\n\\end{array}\n\\]\n\nIn every case, \\( dividr+posint / dividr \\equiv 0(\\bmod 3) \\) and \\( (\\bmod 8) \\). Thus \\( dividr+posint / dividr \\) is a multiple of 24. Note that \\( dividr \\not \\equiv posint / dividr \\) and thus no divisor is used twice in the pairing, so the sum of all the divisors is a multiple of 24 ."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pineapple",
        "d": "riverbank"
      },
      "question": "B-1. Let \\( pineapple \\) be a positive integer such that \\( pineapple+1 \\) is divisible by 24 . Prove that the sum of all the divisors of \\( pineapple \\) is divisible by 24 .",
      "solution": "B-1 The condition \\( 24 \\mid pineapple+1 \\) is equivalent to \\( pineapple \\equiv-1(\\bmod 3) \\) and \\( pineapple \\equiv-1(\\bmod 8) \\). Let \\( riverbank \\) be a divisor of \\( pineapple \\), then \\( riverbank \\equiv 1 \\) or \\( 2(\\bmod 3) \\) and \\( riverbank \\equiv 1,3,5 \\) or \\( 7(\\bmod 8) \\). Since \\( riverbank(pineapple / riverbank)=pineapple \\equiv-1(\\bmod 3) \\) or \\( (\\bmod 8) \\), the only possibilities are:\n\\[\n\\begin{array}{llll}\nriverbank \\equiv 1, & pineapple / riverbank \\equiv 2(\\bmod 3) & \\text { or } & \\text { vice versa } \\\\\nriverbank \\equiv 1, & pineapple / riverbank \\equiv 7(\\bmod 8) & \\text { or } & \\text { vice versa } \\\\\nriverbank \\equiv 3, & pineapple / riverbank \\equiv 5(\\bmod 8) & \\text { or } & \\text { vice versa. }\n\\end{array}\n\\]\n\nIn every case, \\( riverbank+pineapple / riverbank \\equiv 0(\\bmod 3) \\) and \\( (\\bmod 8) \\). Thus \\( riverbank+pineapple / riverbank \\) is a multiple of 24. Note that \\( riverbank \\not \\equiv pineapple / riverbank \\) and thus no divisor is used twice in the pairing, so the sum of all the divisors is a multiple of 24 ."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "emptiness",
        "d": "multipleof"
      },
      "question": "B-1. Let \\( emptiness \\) be a positive integer such that \\( emptiness+1 \\) is divisible by 24 . Prove that the sum of all the divisors of \\( emptiness \\) is divisible by 24 .",
      "solution": "B-1 The condition \\( 24 \\mid emptiness+1 \\) is equivalent to \\( emptiness \\equiv-1(\\bmod 3) \\) and \\( emptiness \\equiv-1(\\bmod 8) \\). Let \\( multipleof \\) be a divisor of \\( emptiness \\), then \\( multipleof \\equiv 1 \\) or \\( 2(\\bmod 3) \\) and \\( multipleof \\equiv 1,3,5 \\) or \\( 7(\\bmod 8) \\). Since \\( multipleof(emptiness / multipleof)=emptiness \\equiv-1(\\bmod 3) \\) or \\( (\\bmod 8) \\), the only possibilities are:\n\\[\n\\begin{array}{llll}\nmultipleof \\equiv 1, & emptiness / multipleof \\equiv 2(\\bmod 3) & \\text { or } & \\text { vice versa } \\\\\nmultipleof \\equiv 1, & emptiness / multipleof \\equiv 7(\\bmod 8) & \\text { or } & \\text { vice versa } \\\\\nmultipleof \\equiv 3, & emptiness / multipleof \\equiv 5(\\bmod 8) & \\text { or } & \\text { vice versa. }\n\\end{array}\n\\]\n\nIn every case, \\( multipleof+emptiness / multipleof \\equiv 0(\\bmod 3) \\) and \\( (\\bmod 8) \\). Thus \\( multipleof+emptiness / multipleof \\) is a multiple of 24. Note that \\( multipleof \\not \\equiv emptiness / multipleof \\) and thus no divisor is used twice in the pairing, so the sum of all the divisors is a multiple of 24 ."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "d": "hjgrksla"
      },
      "question": "B-1. Let \\( qzxwvtnp \\) be a positive integer such that \\( qzxwvtnp+1 \\) is divisible by 24 . Prove that the sum of all the divisors of \\( qzxwvtnp \\) is divisible by 24 .",
      "solution": "B-1 The condition \\( 24 \\mid qzxwvtnp+1 \\) is equivalent to \\( qzxwvtnp \\equiv-1(\\bmod 3) \\) and \\( qzxwvtnp \\equiv-1(\\bmod 8) \\). Let \\( hjgrksla \\) be a divisor of \\( qzxwvtnp \\), then \\( hjgrksla \\equiv 1 \\) or \\( 2(\\bmod 3) \\) and \\( hjgrksla \\equiv 1,3,5 \\) or \\( 7(\\bmod 8) \\). Since \\( hjgrksla(qzxwvtnp / hjgrksla)=qzxwvtnp \\equiv-1(\\bmod 3) \\) or \\( (\\bmod 8) \\), the only possibilities are:\n\\[\n\\begin{array}{llll}\nhjgrksla \\equiv 1, & qzxwvtnp / hjgrksla \\equiv 2(\\bmod 3) & \\text { or } & \\text { vice versa } \\\\\nhjgrksla \\equiv 1, & qzxwvtnp / hjgrksla \\equiv 7(\\bmod 8) & \\text { or } & \\text { vice versa } \\\\\nhjgrksla \\equiv 3, & qzxwvtnp / hjgrksla \\equiv 5(\\bmod 8) & \\text { or } & \\text { vice versa. }\n\\end{array}\n\\]\n\nIn every case, \\( hjgrksla+qzxwvtnp / hjgrksla \\equiv 0(\\bmod 3) \\) and \\( (\\bmod 8) \\). Thus \\( hjgrksla+qzxwvtnp / hjgrksla \\) is a multiple of 24. Note that \\( hjgrksla \\not \\equiv qzxwvtnp / hjgrksla \\) and thus no divisor is used twice in the pairing, so the sum of all the divisors is a multiple of 24 ."
    },
    "kernel_variant": {
      "question": "Let n be a positive integer such that  n+1\\equiv 0 \\pmod{112}\\,(=7\\cdot16).  Prove that the sum of all positive divisors of n is divisible by 8.",
      "solution": "Write σ(n)=∑_{d\\mid n} d for the sum of the positive divisors of n.\n\n1.  A first congruence.\n   \n   From the hypothesis n+1≡0 (mod 112) we immediately have n≡−1 (mod 8); that is\n   n≡7 (mod 8) and in particular n is odd.\n\n2.  Every divisor is odd.\n   \n   Because n is odd, each divisor d of n is also odd, so d≡1,3,5, or 7 (mod 8).\n\n3.  How the divisors come in pairs.\n   \n   For every divisor d of n, the quotient n/d is again a divisor of n.  Reduce the\n   defining relation d·(n/d)=n modulo 8:\n        d·(n/d)≡7\\pmod{8}.\n   Using the four possible odd residues for d we find the corresponding residue of\n   n/d:\n        if d≡1 (mod 8)  then n/d≡7 (mod 8),\n        if d≡3 (mod 8)  then n/d≡5 (mod 8),\n        if d≡5 (mod 8)  then n/d≡3 (mod 8),\n        if d≡7 (mod 8)  then n/d≡1 (mod 8).\n   Thus for every divisor d we have\n        d+n/d≡0\\pmod{8}.                    (★)\n\n4.  No divisor is paired with itself.\n   \n   Squares modulo 8 are 0,1,4.  Because n≡7 (mod 8), n is **not** a perfect square;\n   hence d≠n/d for every divisor d.  Formula (★) therefore pairs every divisor of n\n   with a *distinct* divisor whose sum is a multiple of 8.\n\n5.  Adding all the pairs.\n   \n   Because each pair contributes a multiple of 8 and the pairs are disjoint,\n        σ(n)=∑_{d\\mid n} d  \n   is itself a multiple of 8.\n\nConsequently     8\\mid σ(n),  as was to be shown.",
      "_meta": {
        "core_steps": [
          "Rewrite 24|n+1 as the two congruences n ≡ –1 (mod 3) and n ≡ –1 (mod 8).",
          "Pair every divisor d of n with the complementary divisor n/d.",
          "From d·(n/d) ≡ –1 mod 3 and mod 8, list the possible residue pairs and note that in every case d + n/d ≡ 0 mod 3 and mod 8.",
          "Hence each pair–sum is a multiple of 24, and the distinct pairs cover all divisors.",
          "Therefore the total sum of all divisors of n is divisible by 24."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The modulus whose divisibility is ultimately proved for the divisor-sum.",
            "original": "24"
          },
          "slot2": {
            "description": "The odd prime factor of that modulus; it is required that −1 be a non-quadratic residue modulo this factor so that d ≠ n/d.",
            "original": "3"
          },
          "slot3": {
            "description": "The power-of-two factor (≥8) of the modulus, again with −1 a non-quadratic residue.",
            "original": "8"
          },
          "slot4": {
            "description": "The set of possible residues of a divisor modulo the odd factor (all units).",
            "original": "{1,2}"
          },
          "slot5": {
            "description": "The set of possible residues of a divisor modulo the 2-power factor (all odd residues) together with the specific complementary pairs whose sums are 0 modulo that factor.",
            "original": "{1,3,5,7} with pairs (1,7) and (3,5)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}