# Last digit

What is the last number of 2016 power of 2017

Result

n =  1

#### Solution:

$x_{ 0 } = 2017^0 = 1; l = 1 \ \\ x_{ 1 } = 2017^1 = 2017; l = 7 \ \\ x_{ 2 } = 2017^2 = 4068289; l = 9 \ \\ x_{ 3 } = 2017^3 = 8205738913; l = 3 \ \\ x_{ 4 } = 2017^4 = 16550975387521; l = 1 \ \\ x_{ 5 } = 2017^5 = 33383317356629857; l = 7 \ \\ x_{ 6 } = 2017^6 = 67334151108322421569; l = 9 \ \\ x_{ 7 } = 2017^7 = 135812982785486324304673; l = 3 \ \\ x_{ 8 } = 2017^8 = 273934786278325916122525441; l = 1 \ \\ x_{ 9 } = 2017^9 = 552526463923383372819133814497; l = 7 \ \\ x_{ 10 } = 2017^{10} = 1114445877733464262976192903840449; l = 9 \ \\ x_{ 11 } = 2017^{11} = 2247837335388397418422981087046185633 \ \\ x_{ 12 } = 2017^{12} = 4533887905478397592959152852572156421761 \ \\ x_{ 13 } = 2017^{13} = 9144851905349927944998611303638039502691937 \ \\ x_{ 14 } = 2017^{14} = 18445166293090804665062198999437925676929636929 \ \\ \ \\ 0,4,8,12,.. (4n)... l = 1 \ \\ 1,5,9,13 ... (4n+1)... l = 7 \ \\ 2,6,10,14 ... (4n+2) ... l = 9 \ \\ 3,7,11,15.... (4n+3) .... l = 3 \ \\ \ \\ 2016 = 504 \cdot \ 4+0 \ \\ n = 1$

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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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