A number theory problem by Aditya Das
Number Theory
Level
1
Find the last digit of (2013)^(100)^n where n is a positive integer and not equal to 0....too easy try.
0
6
3
1
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1 solution
3^2=9;3^3=27;3^4=81...this shows that if an exp. has fourth power of 3 or any multiple of it the last digit remains 1. Then it is true for this and firstly look only at last digit 3 not other one..now 100 is exactly div. by 4 and n being positive integer (100)^n is but a multiple of 100 and hence the latter is exactly div. by 4 also.Hence last digit of exp. is 1.....so easy you see