Monday, April 25, 2016

elementary number theory - How to find last two digits of 22016



What should the 'efficient' way of finding the last two digits of 22016 be? The way I found them was by multiplying the powers of 2 because 2016=1024+512+256+128+64+32. I heard that one way would be with the Chinese Remainder Lemma but I don't really know how I should start?


Answer



Essentially we need 22016(mod100)



As (22016,100)=4




let us find 220162(mod100/4)



Now as 2101(mod25)



22014=220110+4=(210)20124(1)201249(mod25)



2222014229(mod2225)


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