Permutation Sum

All 3 digit numbers are of the form $100a+10b+c$ . The permutations can be:

$100a+10b+c,\space 100a+10c+b,\space 100b+10a+c,\space 100b+10c+a,\space 100c+10a+b,\space and\space 100c+10b+a$

Adding all these, we get: $222a+222b+222c=111(2a+2b+2c), \text{which is obviously divisible by 111.}$ $\therefore\space \text{All three digit numbers satisfy this property.}$

Let A, B and C be the three numbers. Out of the six possible permutations ABC, ACB, BAC, BCA, CAB, CBA, the sum of the hundreds, tens and one are all 2A+2B+2C. Let 2A+2B+2C be k. The sum of the permutations is then 100k+10k+1k = 111k which always is divisible by 111.

If anyone would like a brute force proof, run this C#:

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class Program
  {
    static void Main(string[] args)
    {
      // cross join, then cross join again
      var numbers = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9};
      var numbersX2 = numbers.SelectMany<int, int, int[]>(x => numbers, (a, b) => new int[2] {a, b});
      var numbersX3 = numbersX2.SelectMany<int[], int, int[]>(x => numbers, (a, b) => new int[3] {a[0], a[1], b});

      foreach (int[] number in numbersX3)
      {
        Console.Write($"Using {number[0]}{number[1]}{number[2]}... ");

        var v1 = (number[0] * 100) + (number[1] * 10) + number[2];
        var v2 = (number[0] * 100) + (number[2] * 10) + number[1];
        var v3 = (number[2] * 100) + (number[1] * 10) + number[0];
        var v4 = (number[2] * 100) + (number[0] * 10) + number[1];
        var v5 = (number[1] * 100) + (number[2] * 10) + number[0];
        var v6 = (number[1] * 100) + (number[0] * 10) + number[2];
        var sum = (decimal)(v1 + v2 + v3 + v4 + v5 + v6);
        var result = sum / 111.0M;
        Console.WriteLine($"{v1} + {v2} + {v3} + {v4} + {v5} + {v6} = {sum} / 111 = {result}");
        if (sum % 111.0M != 0)
        {
          Console.WriteLine("Does not divide evenly by 111!");
          return;
        }
      }
    }
  }

pick any $a$ , $b$ , $c$ because of both symmetry and each of the 3 numbers appear twice at each digit the sum of the permutations would be $200(a + b + c) + 20(a + b + c) + 2(a + b + c) = 222(a+b+c)$

For 3 digit numbers where digits are not repeated we have total 6 permutations and the sum of all such permutations will be divisible by 222 therefore also divisible by 111. For 3 digit numbers where 1 digits is repeated like abb. The sum of permutations (abb ,bab ,bba ) will be 111(a+2b) which is divisible by 111. For 3 digit numbers aaa =a*111,it is obviously divisible by 111 . So this follows for all types of 3 digit numbers .

Let l, m, n be the 3 digits of the number. Then sum of permutations is

each column summation will result into 2 x ( l + m + n) .

Let 2 x ( l + m + n ) = k

then the summation is,

 =   100 k +10 k + k

 =  (100 + 10 + 1) k

 = 111 k

And 111 k is divisible by 111. :)

No matter how many digits the number has: 1 2 3 4 5 Inverse it: 5 4 3 2 1 And add them one by one : 6 6 6 6 6 You will always get a long one made of a specific number, which is obviously divisible by 11111, Another example: 4213 +1342 =5555

We are making a permutation, so on every decimal position we will have the same set of numbers occurring every time.

Let's say the sum over each position is n.

We have then 100n+10n+1n = 111n, which is always divisible by 111 :-)

Yes, 111 is factor of sum of all permutations of any 3-digit numbers because 111 is the factor of the sum derived from units.

Let's assume sum of three numbers A,B and C is X => X = A+B+C Now sum of all permutations should contain Summation of X twice at each digit place ex.

SUM of all Permutations = 100 (2X)+10 (2X)+1*(2X) = 2X(100+10+1) = 111 (2X) =>

Resultant sum always going to have 111 as a factor so it is TRUE that all 3-digit numbers have this property in which the sum of all the permutations of its digits is divisible by 111

Any number which has sum of its digits as 3 will always be divisible by 3 and when we add 3 different numbers they are divisible by 3 so in this case 111 can divide a number which is falling in this category