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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.