Modular Arithmetic with Sines Given $$\sin(10^{100})+\sin(n)=0$$ find $n$.
I wrote so far that $$\sin(10^{100})=\sin(10^{100} \mod 360)$$ and I noticed that $10^3 \mod 360=280$ and $10^4 \mod 360=280$ so I (correctly) assumed that $$10^{100} \mod 360=280$$
but why is this the case?
 A: And regarding your problem, considering $n$ in degrees, you want to know why we have,
$$10^{100}\equiv 280\pmod{360}$$
Right? Well, you can simply verify the result using Euler's Totient Theorem.
$$\phi(360)=\phi(3^2\times 2^3\times 5)=360\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=96$$
where $\phi(n)$ is the Euler Totient function that counts the totatives of $n$.
Now, what the theorem tells us is,
$$10^{96}\equiv 1\pmod{360}\implies 10^{100}\equiv 10^4\pmod{360}$$
$10^4=1000\times 10\equiv (-80)\times 100\equiv (-8)\times 1000\equiv (-8)\times (-80)\equiv 640\equiv 280\pmod{360}$
Hence, we have,
$$\boxed{10^{100}\equiv 280\pmod{360}}$$
A: It seems your argument is in degrees, as you are worried with residue modulo 360, so I will make this assumption. Modular rings work like a cartesian product of the modulos of the prime factor powers. So, $\mathbb{Z}_{360}$ behaves exactly like $\mathbb{Z}_8\times\mathbb{Z}_9\times\mathbb{Z}_5$. Writing 10 in the latter ring, we have $2\times 1\times 0$. Then cube that to get $10^3$: $0\times 1\times 0$. Multiplying this (component-wise) by 10 ($2\times 1\times 0$) does not change it. So, any power of 10 three and above is equal to any other.
