Question

Find the second right most non-zero digit in the number $N={2470}^{1613}$.___

Answer 1

Approach 1: Conventional
${2470}^{1613}={247}^{1613}\times {10}^{1613}.$
The second right most non-zero digit in the number would be the digit at the ten's place of the number ${247}^{1613}$.
$\frac{{247}^{1613}}{100}=\frac{{247}^{1613}}{4\times 25}$
The remainder when ${247}^{1613}$ is divided by 4 is 3.
So, ${247}^{1613}=4k+3.$
Now we need to find the remainder when ${247}^{1613}$ is divided by 25.
$\frac{{247}^{1613}}{25}=\frac{\left(\right(250-3{)}^{1613})}{25}$
$R=\frac{(-3^{1613})}{25}=\frac{\left(\right(-3^3)\times (\left(3^{10}{)}^{161}\right))}{25}=\left(\frac{-27\times \left({59049}^{161}\right)}{25}\right)=\left(\frac{(-2)\times (-1{)}^{161}}{25}\right)$
Therefore the remainder is 2. So, ${247}^{1613}=25n+2$.
4k+3 =25n+2.
For k to be an integer, possible values of n are 1, 5, 9, 13...
Corresponding values of k are 6, 31, 56, 81...
Therefore, values of k are of the form: 25m -19.
${247}^{1613}=4(25m-19)+3=100m-73.$
Therefore, the last two digits is 100-73=27.
So, the second right most non-zero digit in the number $N={2470}^{1613}$ is 2.

Approach 2: Using the techique of last two digits
We need to consider only the last two digits.
The number can be written as:
$(47{)}^{1613}=({47}^4{)}^{403}\times 47=({47}^2\times {47}^2{)}^{403}\times 47$
$=(09\times 09{)}^{403}\times 47=(81{)}^{403}\times 47=41\times 47=....27$