You can solve it as follows: You can write any number as x = \overline{x_nx_{n-1}...x_1x_0}
With the overline meaning concatenation, and x_i, the different digits of x.
This notation is a shorthand for
x = \sum 10^i x_i
We can rewrite this sum using distributivity to be
x = x_0 + 10(x_1 + 10(x_2 + ... )))
If we define our instuction f_{x_i} as
f_{x_i}(a) = x_i + 10a
Then x can be written as
x = f_{x_0}(f_{x_1}(f_{x_2}( ... )))
or
x = f_{x_0} \circ f_{x_1} \circ ... \circ f_{x_{n-1}}(x_n)
But as f_{x_n}(0) = x_n, it is more natural to write it as
x = f_{x_0} \circ f_{x_1} \circ ... \circ f_{x_{n-1}}\odot f_{x_n}(0)
We are only interested in whether or final result is a multiple of 7, ans as we only use multiplication and addition, we can consider the the full system in mod 7.
x \equiv f_{x_0} \circ f_{x_1} \circ ... \circ f_{x_{n-1}}\circ f_{x_n}(0) (\mod 7)
This formula now represents a finite state machine with 7 states, (0,1,2,3,4,5 and 6) and 10 instructions (f_0 to f_9), where the final state is the same a s the remainder of the original number when dividing by 7.
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 3 | 6 | 2 | 5 | 1 | 4 |
| 1 | 1 | 4 | 0 | 3 | 6 | 2 | 5 |
| 2 | 2 | 5 | 1 | 4 | 0 | 3 | 6 |
| 3 | 3 | 6 | 2 | 5 | 1 | 4 | 0 |
| 4 | 4 | 0 | 3 | 6 | 2 | 5 | 1 |
| 5 | 5 | 1 | 4 | 0 | 3 | 6 | 2 |
| 6 | 6 | 2 | 5 | 1 | 4 | 0 | 3 |
| 7 | 0 | 3 | 6 | 2 | 5 | 1 | 4 |
| 8 | 1 | 4 | 0 | 3 | 6 | 2 | 5 |
| 9 | 2 | 5 | 1 | 4 | 0 | 3 | 6 |