Cyclicity
To find the last digit of $$a^n$$ find the cyclicity of a. For eg. if a=2, we see that
$$2^1 = 2$$
$$2^2 = 4$$
$$2^3 = 8$$
$$2^4 = 16$$
$$2^5 = 32$$
Hence, the last digit of 2 repeats after every 4 th power. Hence cyclicity of 2 =4. Hence if we have to find the last digit of $$a^n$$, the steps are:
Find the cyclicity of a, say it is x
Find the remainder when n is divided by x, say remainder r
Find $$a^{r}$$ if r>0 and $$a^{x}$$ when r=0
For example, find the last digit of $$4^{102}$$ * $$6^{39}$$
The cyclicity of 4 is as follows:
$$4^1 = 4$$
$$4^2 = 16$$
$$4^3 = 64$$
Hence, cyclicity of 4 is 2 and of 6 is 1. Last digit of $$4^{102}$$ is 6 and that of $$6^{39}$$ is 6. Thus 6*6 would give 6 as the last digit.
Cyclicity table for all digits 0-9
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Digit Cycle Length 0,1,5,6 Fixed 1 4,9 4,9,6,1 or 9,1 2 2,3,7,8 Repeats 4
