What is the Last Digit of 7 to the Power of 123?

[sutupidmath]

Homework Statement

Find the last digit of $$7^{123}$$

Homework Equations

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?

 

[sutupidmath]

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.

 

[Dick]

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.

 

[sutupidmath]

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 divide 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...

 

[Dick]

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.

 

[sutupidmath]

This euler function seems to be very powerful, and i am far behind from being able to properly and easily use it...darn..