## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 4 - Number Representation and Calculation - 4.3 Computation in Positional Systems - Exercise Set 4.3 - Page 235: 45

#### Answer

Multiplication of two given numbers in base sixteen is $\text{71B}{{\text{E}}_{\text{sixteen}}}$.

#### Work Step by Step

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 sixteenis same as in base ten. $\text{D}{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ Hence, first multiply ${{\text{A}}_{\text{sixteen}}}$ in first column from right, with ${{3}_{\text{sixteen}}}$ which is above it in the same column: ${{\text{A}}_{\text{sixteen}}}\times {{3}_{\text{sixteen}}}={{30}_{\text{ten}}}=\left( 1\times 16 \right)+\left( \text{E}\times 1 \right)=1{{\text{E}}_{\text{sixteen}}}$. Now, write ${{\text{E}}_{\text{sixteen}}}$ in the first column from right, below the horizontal line and carry ${{\text{1}}_{\text{sixteen}}}$ to the second column from right: $\overset{1}{\mathop{\text{D}}}\,{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ ${{\text{E}}_{\text{sixteen}}}$ Now, multiply ${{\text{A}}_{\text{sixteen}}}$ in first column from right, with ${{\text{D}}_{\text{sixteen}}}$ which is in the second column from right, and add ${{1}_{\text{sixteen}}}$to the product: $\left( {{\text{A}}_{\text{sixteen}}}\times {{\text{D}}_{\text{sixteen}}} \right)+{{1}_{\text{sixteen}}}=130+1={{131}_{\text{ten}}}=\left( 8\times 16 \right)+\left( 3\times 1 \right)={{83}_{\text{sixteen}}}$. Write ${{83}_{\text{sixteen}}}$ in front of ${{\text{E}}_{\text{sixteen}}}$ below the horizontal line: $\text{D}{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ $83{{\text{E}}_{\text{sixteen}}}$ Repeat the whole procedure, but with ${{8}_{\text{sixteen}}}$, after placing the symbol $\times$ below ${{\text{E}}_{\text{sixteen}}}$ in $83{{\text{E}}_{\text{sixteen}}}$: $\text{D}{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ $83{{\text{E}}_{\text{sixteen}}}$ $\times$ Now, multiply ${{8}_{\text{sixteen}}}$ in second column from right, with ${{3}_{\text{sixteen}}}$ in the first column from right: ${{8}_{\text{sixteen}}}\times {{3}_{\text{sixteen}}}={{24}_{\text{ten}}}=\left( 1\times 16 \right)+\left( 8\times 1 \right)={{18}_{\text{sixteen}}}$. Now, write${{8}_{\text{sixteen}}}$ below ${{3}_{\text{sixteen}}}$in $83{{\text{E}}_{\text{sixteen}}}$, and carry ${{1}_{\text{sixteen}}}$ to the second column from right: $\overset{1}{\mathop{\text{D}}}\,{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ $83{{\text{E}}_{\text{sixteen}}}$ $8\times$ Now, multiply ${{8}_{\text{sixteen}}}$ in second column from right, with ${{\text{D}}_{\text{sixteen}}}$ which is in the same column, and add ${{1}_{\text{sixteen}}}$to the product: $\left( {{8}_{\text{sixteen}}}\times {{\text{D}}_{\text{sixteen}}} \right)+{{1}_{\text{sixteen}}}=104+1={{105}_{\text{ten}}}=\left( 6\times 16 \right)+\left( 9\times 1 \right)={{69}_{\text{sixteen}}}$. Now, write ${{69}_{\text{sixteen}}}$ in front of ${{8}_{\text{sixteen}}}$: $\text{D}{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ $83{{\text{E}}_{\text{sixteen}}}$ $+\underline{698\times }$ Now, add $83{{\text{E}}_{\text{sixteen}}}$ and ${{698}_{\text{sixteen}}}$ in the manner shown above. First, drop down ${{\text{E}}_{\text{sixteen}}}$ below the symbol $\times$: $\text{D}{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ $83{{\text{E}}_{\text{sixteen}}}$ $+\underline{698\times }$ ${{\text{E}}_{\text{sixteen}}}$ Now, ${{3}_{\text{sixteen}}}+{{8}_{\text{sixteen}}}={{11}_{\text{ten}}}=\left( \text{B}\times \text{1} \right)={{\text{B}}_{\text{sixteen}}}$. $\text{D}{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ $83{{\text{E}}_{\text{sixteen}}}$ $+\underline{698\times }$ $\text{B}{{\text{E}}_{\text{sixteen}}}$ Now, ${{8}_{\text{sixteen}}}+{{9}_{\text{sixteen}}}={{17}_{\text{ten}}}=\left( 1\times 16 \right)+\left( 1\times \text{1} \right)=\text{1}{{\text{1}}_{\text{sixteen}}}$. $\text{D}{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ $83{{\text{E}}_{\text{sixteen}}}$ $+\underline{\overset{1}{\mathop{6}}\,98\times }$ $\text{1B}{{\text{E}}_{\text{sixteen}}}$ Now, ${{1}_{\text{sixteen}}}+{{6}_{\text{sixteen}}}={{7}_{\text{ten}}}=\left( 7\times \text{1} \right)={{7}_{\text{sixteen}}}$: $\text{D}{{3}_{\text{sixteen}}}$ $\underline{\times 8{{\text{A}}_{\text{sixteen}}}}$ $83{{\text{E}}_{\text{sixteen}}}$ $+\underline{698\times }$ $\text{71B}{{\text{E}}_{\text{sixteen}}}$ Now, to check whether the above obtained solution is correct, perform the multiplication by converting each number to base ten: $\text{D}{{3}_{\text{sixteen}}}=211$, $8{{\text{A}}_{\text{sixteen}}}=138$and $\text{71B}{{\text{E}}_{\text{sixteen}}}=29118$. Since, $211\times 138$ indeed equals 29118, the solution obtained is correct. Hence, multiplication of two given numbers in base sixteen is $\text{71B}{{\text{E}}_{\text{sixteen}}}$.

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