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What Is the last digit of ${{6}^{100}}?$

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Hint: In question it is asked that if we raise the power of 6 to the hundred then the value we will get, what will be its last digit. It is simple we will find the cyclicity of the number of six which will help in finding the last digit of the value of ${{6}^{100}}.$

Complete step-by-step solution:
Before we solve the question let us see what is the term ‘ cyclicity ‘ in mathematics.
Now, let’s take the question: it is given that we have to find the last digit of the value of ${{6}^{100}}$. Now, if we start solving the ${{6}^{100}}$ it will take a long time to solve and also not easy to calculate.
There is one method that can be used to solve such questions which is based on the last digit of numbers method is called cyclicity. The concept of cyclicity is used to identify the last digit of the number. It basically tells after how many cycles the last digit repeats itself in a pattern and on that basis we can find out the last digit of any number to an integral power.
Now, let us consider a term ${{6}^{n}}$ where n varies from 1 to 100.
So, at n = 1, ${{6}^{1}}=6$
At n = 2 , ${{6}^{2}}=36$
At n = 3 , ${{6}^{3}}=216$
At n = 4 , ${{6}^{1}}=1296$
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So on up to n = 100.
For n = 1, 2, 3, 4,….. we see that the last unit digit is always 6 . This is what cyclicity says, any integral power to number 6, the last unit digit will always be 6. So, the cyclicity of 6 equals to 1.
Hence the cyclicity of 6 is 1, so the last digit of ${{6}^{100}}$ will be 6.

Note: It is not necessary that cyclicity of all numbers will be 1, it changes for different numbers for example cyclicity of 6 and 5 is 1 but cyclicity of 4 and 9 is 2. While making a pattern, it must be kept in mind that at least the first 5 terms must be calculated and those 5 terms must be correct as it may change the last digit.