what is the remainder if we divide $72^{200}$ by $5$?

Question

What is the remainder if we divide $72^{200}$ by $5$?

I am very new to modular arithmetic!

Please help!

Answer 1

First hint:

$$72^{200} = (70 + 2)^{200} \equiv 2^{200} \pmod 5$$

Second hint: $2^4 \equiv 1 \pmod5$. What does this tell you about $2^{200} = (2^4)^{50}$ modulo $5$?

Answer 2

$$72^{200} \equiv (2^{200})\cdot(36^{200})\equiv2^{200}\cdot1\equiv 4^{-100}\equiv (-1)^{-100}\equiv 1\pmod 5$$

Answer 3

I think you understand that when you multiply $72$ with $72$, the last digit is supposed to be $4$. So, we can use,

$72^2\equiv -1 \mod 5 \implies (72^2)^{100}\equiv(-1)^{100}\mod 5$

Another method:

$72\equiv2\mod 5 \implies 72^4\equiv2^4\equiv1\mod 5\implies (72^4)^{50}\equiv(2^4)^{50}\equiv1^{50}\mod 5$