Question

Last two digits   

     $\frac{14^{14^{14}}}{100}$

How to find it by cyclicity

Comments

last 1 digit can found by taking power of last digit

but donot know it will give correct result for 2 digit too or not

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Yes....Unit digit can be found easily using cyclicity of 4.

Remember $4^{odd}= 4$ in unit place and $4^{even}=6$ at unit place.

Here $14^{14}$ means we will get 6 at unit place now $14^{...6}$, power is again a even number, $14^{even}$  will give 6 in unit place

If someone know better approach to find 10th place digit plz share :)

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4^even is 6 and 4^odd is 4 by this we can conclude that unit place is 6 but how to get tens place ,suggest and answer someone please