Last digit

What is the last number of 2016 power of 2017

Correct answer:

n =  1

Step-by-step explanation:

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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