One may show an addition of two three digits numbers where all the digits from 1 to 9 are used once only.

                                            n₁     n₂         n₃

                                           n₄      n₅        n₆

                                      ---------------------------

                                           n₇      n₈        n₀     

(100 n₁ + 10 n₂+ 9) + (100 n₄  + 10 n₅ + n₆ )  = 100 n₇ + 10 n_{8 +} n₉

The digital root of the sum of the digital roots of adders  D (n₁ + n_{2 +}  n_{3 +}  n₄ + n₅ + n₆ ) must be equal to the digital root of the additive sum D( n_{4 +} n₅ + n₆)  This is possible only if the digital roots of both sides are equal to 9.

Few solutions are

182 + 763 = 945

492 +183 = 675

592 + 146 = 738

782 + 154 = 936

193 + 275 = 468

159 + 327 = 486

349 + 218 = 567

529 + 317 = 846

There are very few solutions where the sum is a square.

143 + 586 = 729 = 27 x27 = 27²

194 + 382 = 576 = 24 x 24 = 24²

369 + 415 = 784 = 28 x 28 = 28²

. The sum of three two digits numbers give a three digits number. Get solutions in the summation all the digits from 1 to 9 are used once only.

(10 n₁ + n₂ ) + ( 10 n₃ + n₄ ) +(10 n₅ +n₆ ) = (100 n₇ + 10 n₈ + n₉ )

The digital roots must be same for both sides of the relation.D (n₁+n_{2 +} n₃ + n_{2 +}  n₄ + n₅ + n₆)                                 = D (n₇+n₈ + n₉)

A typical solution is 83 +54 +79 = 216 = 6³. By interchanging the digits of unit and 10 th decimal place  we get additional solution 84 +59 + 73 = 89 +53 + 74 = 216

  xyz + xyz + xyz = zzz  or xyz + xyz + xyz = yyy  or xyz + xyz + xyz = www, where w,x,y,z

take a digit from 1 to 9.. Find allowed values of w,x,y.z.

xyz + xyz + xyz = zzz

3 ( 100 x + 10 y+ z) = 100 z + 10 z + z = 111 z

      100x +10y + z = 37 z -à  100x +10y = 36 z-à  -à 50 x + 5 y = 18 z

One possible solution is z =5 , x= 1 and y =8  )185, which gives 185 x 3 = 555

xyz + xyz + xyz = yyy

3 ( 100 x + 10 y+ z) = 100 y + 10 y + y = 111 y

     100 x + 10y + z = 37 y --à 100x + z = 27 y

One possible  solution is  y =4 ,x=1 , z= 8 which gives 148 x 3 = 444

( x≠w, as it gives forbidden solutions)

If the identical digit in th sum is different from the digits of the adder

3(100x + 10y +z) = 100 w + 10w + w = 111 w

  (100x + 10y +z) = 37 w

 37 x 3n gives all identical digits.  Two solutions are available\

259 x 3 = 777 and 296 x 3 = 888