A factory produced an original calculator that performs two operations,

(a) the usual addition, denoted by $+\mathrm{e}$

(b) an operation denoted by $\circledast$.

We know that, for any natural number $a$, the following hold

$$
\text { (i) } \quad a \circledast a=a \quad \text { and } \quad \text { (ii) } \quad a \circledast 0=2 a
$$

and, for any four natural numbers $a, b, c$ and $d$, the following holds

$$
\text { (iii) } \quad(a \circledast b)+(c \circledast d)=(a+c) \circledast(b+d)
$$

What are the results of the operations $(2+3) \circledast(0+3)$ and $1024 \circledast 48$?
To calculate $(2+3) \circledast(0+3)$, we will use property (iii), obtaining $(2+3) \circledast(0+3)=(2 \circledast 0)+(3 \circledast 3)$. Now, by (ii), we have $2 \circledast 0=2 \times 2=4$ and, by (i), we have $3 \circledast 3=3$. Therefore, $(2+3) \circledast(0+3)=4+3=7$.

To calculate $1024 \circledast 48$ we will use the same strategy, noting that $1024=976+48$ and $48=0+48$. Thus,

$$
\begin{aligned}
1024 \circledast 48 & =(976+48) \circledast(0+48) \\
& =(976 \circledast 0)+(48 \circledast 48) \\
& =2 \times 976+48=1952+48=2000 .
\end{aligned}
$$