Last digit

What is the last number of 2016 power of 2017

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=1x_{0}=2017^0=1; l=1 \ \\ x_{1}=2017^1=2017; l=7 \ \\ x_{2}=2017^2=4068289; l=9 \ \\ x_{3}=2017^3=8205738913; l=3 \ \\ x_{4}=2017^4=16550975387521; l=1 \ \\ x_{5}=2017^5=33383317356629857; l=7 \ \\ x_{6}=2017^6=67334151108322421569; l=9 \ \\ x_{7}=2017^7=135812982785486324304673; l=3 \ \\ x_{8}=2017^8=273934786278325916122525441; l=1 \ \\ x_{9}=2017^9=552526463923383372819133814497; l=7 \ \\ x_{10}=2017^{10}=1114445877733464262976192903840449; l=9 \ \\ x_{11}=2017^{11}=2247837335388397418422981087046185633 \ \\ x_{12}=2017^{12}=4533887905478397592959152852572156421761 \ \\ x_{13}=2017^{13}=9144851905349927944998611303638039502691937 \ \\ x_{14}=2017^{14}=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 \cdot \ 4+0 \ \\ n=1



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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

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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