2022/011) Find the rightmost digit of $\lfloor\frac{10^{20000}}{10^{100}+3}\rfloor$

To keep the calculation simple let n = 10^{100}$

So $10^{100} + 3 = n + 3$

So $10^{20000} = n^{200}$

We need to find remainder when $n^{200}$ is divided by n+ 3

Now $n^{200} = n^{200} - 3^{200} + 3^{200}$ 

$n^{200} - 3^{200}$ is divisible by n + 3 and $3^{200} = 9^{100}$ whcih is less than $n+3$

So we need to find the unit digit of quotient of $n^{200} - 3^{200}$ divided by n + 3

we have $n^{200} - 3^{200} = (n+3)\sum_{k=0}^{199}(-1)^k n^k 3^{199-k}$

Hence

$\frac{n^{200}-3^{200}}{n+3}= \sum{k=0}^{199}(-1)^k n^k 3^{199-k}$

putting back $n= 10^{100}$

we get $\frac{10^{20000}-3^{200}}{10^{100}+3}= \sum{k=0}^{199}(-1)^k (10^{200})^k 3^{199-k}$

All the trerms except when k = 0 have a factor 100 so we get unit digit is unit digit of $-3{199}$

now $3^4$ ends with 1 so $3^{199} = (3^4)^{49} * 3^3$ ends with unit digit of $3^3$ or 7 so if we negate it it ends with 10 - 7 = 3

so ans 7