Question

# The last digit of the number ${7^{886}}$is$a.{\text{ 9}} \\ b.{\text{ 7}} \\ c.{\text{ 3}} \\ d.{\text{ 1}} \\$

Hint- last digit means we need to check the unit digit in ${n^x}$.For finding the unit digit number we must know the concept of cyclicity of power to a number.

For finding the last digit of the number, we will have to check unit digit in ${n^x}$
So, check the unit digit in ${7^{886}}$
${7^1} = 7$
So, the unit digit is 7.
${7^2} = 49$
So, the unit digit is 9.
${7^3} = 343$
So, the unit digit is 3.
${7^4} = 2401$
So, the unit digit is 1.
${7^5} = 16807$
So, the unit digit is 7.
${7^6} = 117649$
So, the unit digit is 9.
Now we can see that after 4 steps the unit digits are repeating.
Now in the given number the power of 7 is 886.
So divide 886 by 4
$\Rightarrow \frac{{886}}{4} = 221\frac{2}{4}...............\left( 1 \right)$
Now we are generalizing the number of form and the corresponding unit digit can be written as
$4n = 1 \\ 4n + 1 = 7 \\ 4n + 2 = 9 \\ 4n + 3 = 3 \\$

Now as we see from equation (1) that the power of 7 which is 886 is of form
$886 = 221\left( 4 \right) + 2 \\ {\text{i}}{\text{.e}}{\text{. }}4n + 2 \\$
Hence the unit digit would be 9.
So, the correct answer is option (a).

Note- In such types of questions if we asked to find out the last digit of some ${n^x}$number, first we have to check the unit digit and find after how many steps it gets repeated, then divide the power of number with the number of steps after which the unit digit is repeating then generalize the number as above we will get the required last digit of the given number.