abccba can be written as :
100000a + 10000b + 1000c + 100c + 10b + a
100001a + 10010b + 1100c
We can resolve it as below:
(99995a + 6a) + (10010b) + (1099c + c)
Apart from, 6a and c all of them are multiple of 7.
Now, (6a + c) = 7a + (c - a) must be divided by7.
now c - a must be equal to either -7, 0 or 7
for c-a = -7; (0,7); (2,9); (1,8) i.e. 3 combinations.
for c-a = 0; (1,1)..... (9,9) i.e. 9 combinations.
for c-a = 7; (9,2); (8,1) i.e. 2 combinations.
for (3+9+2 = 14) b can be 1,3,5,7,9
So, 14 * 5 = 70 combinations are possible.