Last digit

What is the last number of 2016 power of 2017

Correct result:

n =  1

Solution:

$$\begin{gathered} x_0=2017^0=1;l=1 \\ {x}_1=2017^1=2017;l=7 \\ {x}_2=2017^2=4068289;l=9 \\ {x}_3=2017^3=8205738913;l=3 \\ {x}_4=2017^4=16550975387521;l=1 \\ {x}_5=2017^5=33383317356629857;l=7 \\ {x}_6=2017^6=67334151108322421569;l=9 \\ {x}_7=2017^7=135812982785486324304673;l=3 \\ {x}_8=2017^8=273934786278325916122525441;l=1 \\ {x}_9=2017^9=552526463923383372819133814497;l=7 \\ {x}_{10}=2017^{10}=1114445877733464262976192903840449;l=9 \\ {x}_{11}=2017^{11}=2247837335388397418422981087046185633 \\ {x}_{12}=2017^{12}=4533887905478397592959152852572156421761 \\ {x}_{13}=2017^{13}=9144851905349927944998611303638039502691937 \\ {x}_{14}=2017^{14}=18445166293090804665062198999437925676929636929 \\ 0,4,8,12,\mathrm{.}\mathrm{.}(4n)\dots l=1 \\ 1,5,9,13\dots (4n+1)\dots l=7 \\ 2,6,10,14\dots (4n+2)\dots l=9 \\ 3,7,11,15\dots \mathrm{.}(4n+3)\dots \mathrm{.}l=3 \\ 2016=504\cdot 4+0 \\ n=1 \end{gathered}$$

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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.

2 years ago  Like

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