Determine the outcomes of the following computations by the "Long Division method": i) 3478 / 38 ii) 6048 / 48

The quotient is $1_8$.

Divide $4_8$ by $3_8$. The quotient is $1_8$ with a remainder of $1_8$.

The new number is $17_8$. Divide $17_8$ by $3_8$.
In decimal: $17_8 = 1 \times 8 + 7 = 15_{10}$. $3_8 = 3_{10}$. $15 \div 3 = 5$. So, $3_8 \times 5_8 = 17_8$. The quotient is $5_8$.

The remainder is $0_8$.

The outcome is $\boxed{115_8}$.

The quotient is $1_8$ with a remainder of $2_8$.

The new number is $20_8$. Divide $20_8$ by $4_8$.
In decimal: $20_8 = 2 \times 8 + 0 = 16_{10}$. $4_8 = 4_{10}$. $16 \div 4 = 4$. So, $4_8 \times 4_8 = 20_8$. The quotient is $4_8$.

The new number is $04_8$. Divide $4_8$ by $4_8$. The quotient is $1_8$.

The remainder is $0_8$.

The outcome is $\boxed{141_8}$.

Consider $33_{16}$.
In decimal: $33_{16} = 3 \times 16 + 3 = 51_{10}$. $7_{16} = 7_{10}$.
$51 \div 7 = 7$ with a remainder of 2. So, $7_{16} \times 7_{16} = 49_{10} = 31_{16}$. The quotient is $7_{16}$.

The new number is $23_{16}$. Divide $23_{16}$ by $7_{16}$.
In decimal: $23_{16} = 2 \times 16 + 3 = 35_{10}$. $7_{16} = 7_{10}$.
$35 \div 7 = 5$. So, $7_{16} \times 5_{16} = 35_{10} = 23_{16}$. The quotient is $5_{16}$.

The remainder is $0_{16}$.

The outcome is $\boxed{75_{16}}$.

Consider $1B_{16}$.
In decimal: $1B_{16} = 1 \times 16 + 11 = 27_{10}$. $C_{16} = 12_{10}$.
$27 \div 12 = 2$ with a remainder of 3. So, $C_{16} \times 2_{16} = 12_{10} \times 2_{10} = 24_{10} = 18_{16}$. The quotient is $2_{16}$.

The new number is $39_{16}$. Divide $39_{16}$ by $C_{16}$.
In decimal: $39_{16} = 3 \times 16 + 9 = 57_{10}$. $C_{16} = 12_{10}$.
$57 \div 12 = 4$ with a remainder of 9. So, $C_{16} \times 4_{16} = 12_{10} \times 4_{10} = 48_{10} = 30_{16}$. The quotient is $4_{16}$.

The new number is $90_{16}$. Divide $90_{16}$ by $C_{16}$.
In decimal: $90_{16} = 9 \times 16 + 0 = 144_{10}$. $C_{16} = 12_{10}$.
$144 \div 12 = 12$. So, $C_{16} \times C_{16} = 12_{10} \times 12_{10} = 144_{10} = 90_{16}$. The quotient is $C_{16}$.