# FInd the last digit?

by sutupidmath
Tags: digit
 P: 1,633 1. The problem statement, all variables and given/known data Find the last digit of $$7^{123}$$ 2. Relevant equations 3. The attempt at a solution $$7^{123} \equiv x(mod 10)$$ 123=12*10+3 Now, since in Z_10 $$7^{120 }\equiv 1 (mod 10)=> 7^{123} \equiv 7^3=343 mod 10=>343(mod 10)=3$$ SO would the last digit be 3???? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
 P: 1,633 Also, how would one find the last 3 digits of $$7^{9999}$$ I know i have to work mod 1000, but i haven't been able to pull out anything so far.
 P: 1,633 FInd the last digit? So,since we are working mod 1000, i will have to find the order of $$V_{1000}=\phi(1000)$$ so i know for sure that $$7^{\phi(1000)}\equiv 1(mod1000)$$ NOw $$\phi(1000)=\phi(2^3)\phi(5^3)=400=>7^{400}\equiv 1(mod 1000)$$ Now also $$(7^{1000})^{25}\equiv 1(mod 1000)=>7^{10000}=1+k1000=1001+(k-1)1000$$ Now from here i guess, not sure though, we have $$7|(k-1)$$ Now above if we devide both parts by 7 we would get: $$7^{9999}=143+\frac{k-1}{7}1000$$ So $$7^{9999}\equiv 143(mod1000)$$ so the last 3 digits are 143 ?? I thought there might be some more easy way...lol.....