Answer D
Let is consider a number as 123456
- In to get $1^{st}$ digit from right $\frac{123456}{10^{0}} = 123456 % 10$ = 6$
- In to get $2^{nd}$ digit from right $\frac{123456}{10^{1}} = 12345 % 10$ = 5$
- In to get $3^{rd}$ digit from right $\frac{123456}{10^{2}} = 1234 % 10 = 4$
- In to get $4^{th}$ digit from right $\frac{123456}{10^{3}} = 123 % 10 = 3$
- In to get $i^{th}$ digit from right $\frac{123456}{10^{i-1}} = 123 % 10 = 3$
So, if the number is N, its $i^{th}$ digit from right would be $\left [ \frac{N}{10^{i-1}} \right ] mod 10$
