Working out modular arithmetic of a modulus that is coprime

Question

Ever since I started doing work on Bézout Identity and Extended Euclidean Algorithm, I have tried to understand how to do modular arithmetic of big numbers etc. Basically, I have come up with the following problem of my own that I wish to solve:

$t = 24^3 \mod 35$

We know that $35 = 7\cdot 5$ and both $7$ and $5$ are prime numbers. I believe that there is a way of doing modular arithmetic of this nature if the modulus number is coprime like in this case.

I got as far as this: $$t = 24^3 \mod 7$$ $$t = 24^3 \mod 5$$

What do I do after? It would be great if someone could help me compute this answer as I am very keen to better understand all of this using my own examples.

Please I am new to CRT and any help of steps of calculating this would mean a lot!

Answer 1

Because $24=21+3\equiv3\pmod7,$

$t\equiv24^3 \pmod 7\implies t\equiv3^3 \pmod 7\implies t\equiv27\equiv\color{purple}{-1}\pmod 7.$

Because $24=25-1\equiv-1\pmod5,$

$ t\equiv24^3 \pmod 5\implies t\equiv(-1)^3 \pmod 7\implies t\equiv\color{purple}{-1}\pmod 5.$

Therefore, since $7$ and $5$ are coprime, by the constant case of the Chinese remainder theorem,

$t\equiv\color{purple}-1\equiv34\pmod {7\times5=35}.$

Answer 2

I can rewrite tour congruence system as: $$t\equiv 6 \mod 7$$ $$t\equiv 4 \mod 5$$

Because $24^3>7$ then I can calculate the true remainder by (for $7$): $\left (\frac{24^3}{7}-\left \lfloor \frac{24^3}{7} \right \rfloor \right )\cdot 7=6$; and (for $5$): $\left (\frac{24^3}{5}-\left \lfloor \frac{24^3}{5} \right \rfloor \right )\cdot 5=4$.

From the first congruences I obtain: $t\equiv 34 \mod 35$. The solutions are: $t=35k-1$ with $k \in Z$.