1. The problem statement, all variables and given/known data Find the last digit of [tex]7^{123} [/tex] 2. Relevant equations 3. The attempt at a solution [tex]7^{123} \equiv x(mod 10)[/tex] 123=12*10+3 Now, since in Z_10 [tex]7^{120 }\equiv 1 (mod 10)=> 7^{123} \equiv 7^3=343 mod 10=>343(mod 10)=3[/tex] 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
Also, how would one find the last 3 digits of [tex]7^{9999}[/tex] I know i have to work mod 1000, but i haven't been able to pull out anything so far.
You used 7^4=1 mod 10 to do the first one, right? You want to do the second one the same way. Find a large k such that 7^k=1 mod 1000. Use Euler's theorem and the Euler totient function to find such a k. Once you've done that you may find it useful to know that 7 has a multiplicative inverse mod 1000 (since 7 and 1000 are coprime). Factor 1001.
So,since we are working mod 1000, i will have to find the order of [tex]V_{1000}=\phi(1000)[/tex] so i know for sure that [tex]7^{\phi(1000)}\equiv 1(mod1000)[/tex] NOw [tex]\phi(1000)=\phi(2^3)\phi(5^3)=400=>7^{400}\equiv 1(mod 1000)[/tex] Now also [tex](7^{1000})^{25}\equiv 1(mod 1000)=>7^{10000}=1+k1000=1001+(k-1)1000[/tex] Now from here i guess, not sure though, we have [tex]7|(k-1)[/tex] Now above if we devide both parts by 7 we would get: [tex]7^{9999}=143+\frac{k-1}{7}1000[/tex] So [tex]7^{9999}\equiv 143(mod1000)[/tex] so the last 3 digits are 143 ?? I thought there might be some more easy way...lol.....
That works. I would have just said since 7^400=1 mod 1000, then 7^10000=1 mod 1000. So if you let x=7^9999. Then you want to solve 7*x=1 mod 1000. Since 7 and 1000 are relatively prime, you can do that. And knowing 1001=7*143 give you a cheap way. x=143.
This euler function seems to be very powerful, and i am far behind from being able to properly and easily use it....darn..