## Thinking Mathematically (6th Edition)

Multiplication of two given numbers in base sixteen is $\text{1F1}{{\text{C}}_{\text{sixteen}}}$.
Since the computation involves base sixteen, the only digit symbols which are allowed are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. The procedure to multiply two numbers in base sixteen is same as in base ten. $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ Hence, first multiply ${{\text{C}}_{\text{sixteen}}}$ in first column from right, with ${{5}_{\text{sixteen}}}$ which is above it in the same column: ${{\text{C}}_{\text{sixteen}}}\times {{5}_{\text{sixteen}}}={{60}_{\text{ten}}}=\left( 3\times 16 \right)+\left( \text{C}\times 1 \right)=3{{\text{C}}_{\text{sixteen}}}$. Now, write ${{\text{C}}_{\text{sixteen}}}$ in the first column from right, below the horizontal line and carry ${{\text{3}}_{\text{sixteen}}}$ to the second column from right: $\overset{3}{\mathop{\text{B}}}\,{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ ${{\text{C}}_{\text{sixteen}}}$ Now, multiply ${{\text{C}}_{\text{sixteen}}}$ in first column from right, with ${{\text{B}}_{\text{sixteen}}}$ which is in the second column from right, and add ${{3}_{\text{sixteen}}}$to the product: $\left( {{\text{C}}_{\text{sixteen}}}\times {{\text{B}}_{\text{sixteen}}} \right)+{{3}_{\text{sixteen}}}=132+3={{135}_{\text{ten}}}=\left( 8\times 16 \right)+\left( 7\times 1 \right)={{87}_{\text{sixteen}}}$. Write ${{87}_{\text{sixteen}}}$ in front of ${{\text{C}}_{\text{sixteen}}}$ below the horizontal line: $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ $\text{87}{{\text{C}}_{\text{sixteen}}}$ Repeat the whole procedure, but with ${{2}_{\text{sixteen}}}$, after placing the symbol $\times$ below ${{\text{C}}_{\text{sixteen}}}$ in $\text{87}{{\text{C}}_{\text{sixteen}}}$: $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ $\text{87}{{\text{C}}_{\text{sixteen}}}$ $\times$ Now, multiply ${{2}_{\text{sixteen}}}$ in second column from right, with ${{5}_{\text{sixteen}}}$ in the first column from right: ${{2}_{\text{sixteen}}}\times {{5}_{\text{sixteen}}}={{10}_{\text{ten}}}=\left( \text{A}\times 1 \right)={{\text{A}}_{\text{sixteen}}}$. Now, write ${{\text{A}}_{\text{sixteen}}}$ below ${{7}_{\text{sixteen}}}$in $\text{87}{{\text{C}}_{\text{sixteen}}}$: $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ $\text{87}{{\text{C}}_{\text{sixteen}}}$ $\text{A}\times$ Now, multiply ${{2}_{\text{sixteen}}}$ in second column from right, with ${{\text{B}}_{\text{sixteen}}}$ which is in the same column: ${{2}_{\text{sixteen}}}\times {{\text{B}}_{\text{sixteen}}}={{22}_{\text{ten}}}=\left( 1\times 16 \right)+\left( 6\times 1 \right)={{16}_{\text{sixteen}}}$. Now, write ${{16}_{\text{sixteen}}}$ in front of ${{\text{A}}_{\text{sixteen}}}$: $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ $\text{87}{{\text{C}}_{\text{sixteen}}}$ $+\underline{\text{16A}\times }$ Now, add $\text{87}{{\text{C}}_{\text{sixteen}}}$ and $\text{16}{{\text{A}}_{\text{sixteen}}}$ in the manner shown above. First, drop down ${{\text{C}}_{\text{sixteen}}}$ below the symbol $\times$: $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ $\text{87}{{\text{C}}_{\text{sixteen}}}$ $+\underline{\text{16A}\times }$ ${{\text{C}}_{\text{sixteen}}}$ Now, ${{7}_{\text{sixteen}}}+{{\text{A}}_{\text{sixteen}}}={{17}_{\text{ten}}}=\left( 1\times 16 \right)+\left( 1\times \text{1} \right)=\text{1}{{\text{1}}_{\text{sixteen}}}$. $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ $\overset{1}{\mathop{\text{8}}}\,\text{7}{{\text{C}}_{\text{sixteen}}}$ $+\underline{\text{16A}\times }$ $\text{1}{{\text{C}}_{\text{sixteen}}}$ Now, ${{1}_{\text{sixteen}}}+{{8}_{\text{sixteen}}}+{{6}_{\text{sixteen}}}={{15}_{\text{ten}}}=\left( \text{F}\times \text{1} \right)={{\text{F}}_{\text{sixteen}}}$. $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ $\text{87}{{\text{C}}_{\text{sixteen}}}$ $+\underline{\text{16A}\times }$ $\text{F1}{{\text{C}}_{\text{sixteen}}}$ Now, drop down ${{1}_{\text{sixteen}}}$ as it is: $\text{B}{{5}_{\text{sixteen}}}$ $\underline{\times 2{{\text{C}}_{\text{sixteen}}}}$ $\text{87}{{\text{C}}_{\text{sixteen}}}$ $+\underline{\text{16A}\times }$ $\text{1F1}{{\text{C}}_{\text{sixteen}}}$ Now, to check whether the above obtained solution is correct, perform the multiplication by converting each number to base ten: $\text{B}{{5}_{\text{sixteen}}}=181$, $2{{\text{C}}_{\text{sixteen}}}=44$and $\text{1F1}{{\text{C}}_{\text{sixteen}}}=7964$. Since, $181\times 44$ indeed equals 7964, the solution obtained is correct. Hence, multiplication of two given numbers in base sixteen is $\text{1F1}{{\text{C}}_{\text{sixteen}}}$.