Question

What is the last digit of this number? ${2222}^{3333}$

Answer 1

Hint: To obtain the last digit of the number given we will use the recognizing pattern of number method. Firstly we will see what the last digit of starting power of 2 is. Then we will see how the pattern works and then accordingly we will solve the given number without actually multiplying it and get our desired answer.

Complete step by step solution:
The number is given as:
${2222}^{3333}$
As we know that power of 2 have following values:
$\begin{array}{rl}& 2^1=2\\ & 2^2=4\\ & 2^3=8\\ & 2^4=16\end{array}$
$\begin{array}{rl}& 2^5=32\\ & 2^6=64\\ & 2^7=128\\ & 2^8=256\end{array}$
As we can see, the last digit of the first four terms is 2,4,8,6 and then after that the last digit of next four terms is 2,4,6,8.
So we can conclude that the power of any number where the last digit is 2 will have the same pattern.
After a group of 4 the pattern starts again.
So as the power of the number 2222 is 3333 we have to find where 3333 falls in the above pattern.
The pattern repeats after 4 values so we will divide 3333 by 4 as below:
$\begin{array}{rl}& 3333÷4\\ & \Rightarrow {\displaystyle \frac{3333}{4}}\end{array}$
$\Rightarrow 833{\displaystyle \frac{1}{4}}$
So by above value we get the pattern continues for 833 times and then one more time.
So we get that
${2222}^{3332}$ Last digit will be 6
${2222}^{3333}$ Last digit will be 2

Hence last digit of ${2222}^{3333}$ is 2

Note:
There are numbers of tools that are used to solve such types of problems such as modular arithmetic, Chinese remainder theorem and also Euler’s theorem but the recognizing the pattern is much simpler and easy to remember. Every number has a different pattern when we raise their power we just have to be observant in order to get the answer.