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1 solution
For 14^n:
Note: I am not considering the case n=1 (14^1)
If n= 2 (mod 10): last 2 digits are 96
If n=3 (mod 10): last 2 digits are 44
If n=4 (mod 10): last 2 digits are 16
If n=5 (mod 10): last 2 digits are 24
If n=6 (mod 10): last 2 digits are 36
If n=7 (mod 10): last 2 digits are 04
If n=8 (mod 10): last 2 digits are 56
If n=9 (mod 10): last 2 digits are 84
If n=0 (mod 10): last 2 digits are 76
If n=1 (mod 10): last 2 digits are 64
Now in the question we have to find last 2 digits of 14^(14^14)
As 14^14 leaves a remainder of 6 (last 2 digits are 16) .
Therefore 14^(14^14) have 36 as its last 2 digits