## Elementary Technical Mathematics

The result of the hexadecimal addition is $1D02$.
Find the first number in the left vertical column, find the second number in the top horizontal row, and find their sum at the intersection in the table. These numbers can be written as, \underline{\begin{align} & 2B5 \\ & 1A4D \\ \end{align}} Now, use the table for the addition of $\text{D+5}$, which is found to be $12$ and the addition of $\text{B+4}$, which is $\text{F}$ and the addition of $\text{A+2}$, which is $\text{C}$. Thus, \begin{align} & \underline{\begin{matrix} \text{Carry,} & \text{1 1 }\,\,\,\, & \text{D+5=12} \\ {} & \text{2 B 5} & \text{B+4=F} \\ + & \text{1 A 4 D} & \text{A+2=C} \\ \end{matrix}} \\ & \underline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{ 1 D 0 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \end{align} Thus, the result of the addition is $\text{1 D 0 2}$. Now, check this addition using decimal addition. First, add the provided hexadecimal number. Now, since $\text{B}=11$, the hexadecimal number $2B5$ is written in decimal form as, \begin{align} & 2B5=2\left( {{16}^{2}} \right)+11\left( {{16}^{1}} \right)+5\left( {{16}^{0}} \right) \\ & =512+176+5 \\ & =693 \end{align} Also, convert the number $1A4D$ into decimal form as, \begin{align} & 1A4D=1\left( {{16}^{3}} \right)+10\left( {{16}^{2}} \right)+4\left( {{16}^{1}} \right)+13\left( {{16}^{0}} \right) \\ & =4096+2560+64+13 \\ & =6733 \end{align} Now, $693+6733=7426$ Now, solve the resultant hexadecimal addition $\text{1 D 0 2}$, \begin{align} & \text{1 D 0 2}=1\left( {{16}^{3}} \right)+13\left( {{16}^{2}} \right)+0\left( {{16}^{1}} \right)+2\left( {{16}^{0}} \right) \\ & =4096+3328+2 \\ & =7426 \end{align} Thus, the obtained result is correct. Hence, the result of the hexadecimal addition is $\text{1D02}$.