Question

Find all three digit numbers abc (with a not equal to 0) such that a 2 +b 2 +c 2 is divisible by 26.

Answer 1

a, b, c are digits.  
so   0 <= c <= 9,    0 <= b <= 9...   1 <= a <= 9

Number N = abc = 100 a + 10 b + c

given,   A = a² + b² + c² = 26 k      for some integer k.
Maximum value of A = 9² + 9² + 9² = 243
Minimum value of A = 1² + 0² + 0² = 1

 Thus  1 <= 26 k <= 243
So, k = 1, 2, 3, 4, 5, 6 , 7, 8, 9

1)  k = 1 :    a² + b² + c² = 26   
          Clearly,   a,b,c <= 5
          a, b, c are from : (1, 3, 4)   or, (1, 0, 5)
          Numbers: abc = N = 134, 143, 341,314, 413, 431

2) k = 2.   a² + b² + c² = 52 = 2² * 13
             clearly,   a,b,c <= 7
            (a,b,c) are :  (0,4,6)
             Numbers : N = 406, 460, 604, 640

3) k = 3.    a² + b² + c² = 78
              clearly,   a,b,c <= 8
              (a,b,c) are   (2,5,7)       NUmbers :  257,275,527,572,725,752

4) k = 4...   a² + b² + c² = 2² * 26 = 104
             (a,b,c) = (2,6,8)  : Numbers: 268, 286,628,682,826,862

5) k =5.    a²+b²+c² = 130 = 2*5*13
             (a,b,c) =  (0,7,9).  So numbers:  709, 790, 907, 970.

6) k = 6.   a² + b² + c² = 156
                  (a,b,c) :  no combinations possible.

7) k = 7.   a²+b²+c² = 182
                 (a,b,c) : no combinations possible.

8) k = 8.   a²+b²+c² = 208
                (a,b,c) : no combinations...

9) k = 9.   a²+b²+c²  = 234
                (a,b,c) = no combinations...