Last digit

What is the last number of 2016 power of 2017

Correct result:

n =  1

Solution:

x0=20170=1;l=1 x1=20171=2017;l=7 x2=20172=4068289;l=9 x3=20173=8205738913;l=3 x4=20174=16550975387521;l=1 x5=20175=33383317356629857;l=7 x6=20176=67334151108322421569;l=9 x7=20177=135812982785486324304673;l=3 x8=20178=273934786278325916122525441;l=1 x9=20179=552526463923383372819133814497;l=7 x10=201710=1114445877733464262976192903840449;l=9 x11=201711=2247837335388397418422981087046185633 x12=201712=4533887905478397592959152852572156421761 x13=201713=9144851905349927944998611303638039502691937 x14=201714=18445166293090804665062198999437925676929636929  0,4,8,12,..(4n)l=1 1,5,9,13(4n+1)l=7 2,6,10,14(4n+2)l=9 3,7,11,15.(4n+3).l=3  2016=504 4+0 n=1



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Showing 1 comment:
#
Dr Math
Usually, whenever there is such a question to find the last digit of a^b.., the first step is to find the pattern.

Now 2017^n follows a pattern, with respect to its last digit. The last digits follow a pattern of 1,7,9,3 and this pattern keeps repeating.

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