Calculate $3\cdot 4+ 4$ in $\mathbb{Z}_7$ and $\mathbb{Z}_{10}$. Question: calculate $3\cdot 4+ 4$ in $\mathbb{Z}_7$  and $\mathbb{Z}_{10}$.
I don't really understand how to approach this problem, any ideas are appreciated.
Thanks.
 A: In order to calculate this, simply do the calculation and then reduce the answer mod 7 or 10:
$$
3\cdot 4+4=16\equiv 2 \text{ (mod }7\text{)} \\ 3\cdot 4+4=16\equiv 6 \text{ (mod }10\text{)}
$$
Regarding your question in the comments, $\frac{1}{5}$ generally means the inverse of 5 which of course depends on what ring you are in. The inverse of 5 in $\mathbb{Z}/11\mathbb{Z}$ can be found naively by multiplying 5 by every element of $\mathbb{Z}/11\mathbb{Z}$ until you find one that multiplies to give 1. You can do it systematically by finding solutions to $11x+5y=1$ using the Euclidean algorithm. The value you obtain for $y$ will be the multiplicative inverse of 5 in $\mathbb{Z}/11\mathbb{Z}$. In this case you get $y=-2\equiv 9$.
A: There are two ways to go about it. You can do each operation in $\mathbb{Z}$ and convert to $\mathbb{Z}_7$ or $\mathbb{Z}_{10}$ as needed, or do all the operations (addition, subtraction, multiplication, as the case may be) in $\mathbb{Z}$ and then convert to $\mathbb{Z}_7$ or $\mathbb{Z}_{10}$ at the end.
For example, you can first do $3 \times 4 \equiv 5 \pmod 7$ and then $5 + 4 \equiv 2 \pmod 7$. Or you can do $3 \times 4 + 4 = 16 \equiv 2 \pmod 7$, you get the same result either way.
