Question

Convert the hexadecimal expansion of each of these in- tigersto a binary expansion.

a) (80E)16 b) (135AB)16

c) (ABBA)16 d) (DEFACED)1

Short answer

$$\begin{gathered} a\left)\right(100000001110{)}_2 \\ b\left)\right(10011010110101011{)}_2 \\ c\left)\right(1010101110111010{)}_2 \\ d\left)\right(1101111011111010110011101101{)}_2 \end{gathered}$$

Step-by-step solution

Step 1:

$n=a_k{b}^k+a_{k1}+b^{k1}+\dots .+a_{1}b+a_0$

in hexadecimal notation

$\begin{array}{r}A=10\\ B=11\\ C=12\\ D=13\\ E=14\\ F=15\end{array}$

The hexadecimal expansion has base

$\begin{array}{r}(80E{)}_{16}={8.16}^2+{0.16}^1+{14.16}^0\\ \\ =2048+14\\ \\ =2062\end{array}$

You obtain binary expansion of the decimal expansion of an integer by consecutively dividing the integer by 2 until you obtain 0

localid="1668487048178" $\begin{array}{r}2062=2.1031+0\\ \\ 1031=2.515+1\\ \\ 515=2.257+1\\ \\ 128=2.64\end{array}$

$\begin{array}{r}64=2.32+0\\ \\ 32=2.16+0\\ \\ 16=2.8+0\\ \\ 8=2.4+0\\ \\ 4=2.2+0\\ \\ 2=2.1+0\\ \\ 1=2.0+1\end{array}$

The successive remainders of each division represents the binary expansion from right left

$(100000001110{)}_2$

Step 2: 

The hexadecimal expansion has base

$\begin{array}{r}(135AB{)}_{16}={1.16}^3+{3.16}^2+{5.16}^1+{10.16}^1+{11.16}^0\\ \\ =655536+12288+1280+160+11\\ \\ =79275\end{array}$

You obtain binary expansion of the decimal expansion of an integer by consecutively dividing the integer by 2 until you obtain 0

$$\begin{gathered} \begin{array}{r}79275=2.39637+1\\ \\ 39637=2.19818+1\\ \\ 19818=2.9909+1\\ \\ 9909=2.4954+1\\ \\ 4954=2.2477+0\\ \\ 2477=2.1238+1\\ \\ 1238=2.619+0\\ \end{array} \\ \begin{array}{r}619=2.309+1\\ \\ 309=2.154+1\\ \\ 154=2.77+0\\ \\ 77=2.38+1\\ \\ 38=2.19+0\\ \\ 19=2.9+1\\ \\ 9=2.4+1\\ \\ 4=2.2+0\\ \\ 2=2.1+0\\ \\ 1=2.0+1\end{array} \end{gathered}$$

The successive remainders of each division represents the binary expansion from right left

$(10011010110101011{)}_2$

Step 3:

The hexadecimal expansion has base

$\begin{array}{r}(ABBA{)}_{16}={10.16}^3+{11.16}^2+{11.16}^1+{10.16}^1+{10.16}^0\\ \\ =40960+2816+176+10\\ \\ =43962\end{array}$

You obtain binary expansion of the decimal expansion of an integer by consecutively dividing the integer by 2 until you obtain 0

$\begin{array}{r}43962=2.21981+0\\ \\ 21981=2.10990+1\\ \\ 10990=2.5495+0\\ \\ 5495=2.4954+1\end{array}$

$\begin{array}{r}2747=2.1373+1\\ \\ 1373=2.686+1\\ \\ 686=2.343+0\\ \\ 343=2.171+1\\ \\ 171=2.85+1\\ \\ 85=2.42+1\\ \\ 42=2.21+0\\ \\ 21=2.10+1\\ \\ 10=2.5+0\\ \\ 5=2.2+1\\ \\ 2=2.1+0\\ \\ 1=2.0+1\end{array}$

The successive remainders of each division represents the binary expansion from right left

$(1010101110111010{)}_2$

Step 4: 

The hexadecimal expansion has base

$\begin{array}{r}(\text{DEFACED}{)}_{16}=13\cdot {16}^6+{14.16}^5+15\cdot {16}^4+{10.16}^3+{12.16}^2+{14.16}^1+{13.16}^0\\ \\ =218103808+14680064+983040+40960+3072+224+13\\ \\ =233811181\end{array}$

You obtain binary expansion of the decimal expansion of an integer by consecutively dividing the integer by 2 until you obtain 0

$$\begin{gathered} 233811181=2.116905590+1 \\ \begin{array}{r}116905590=2.58452795+0\\ \\ 58452795=2.29226397+1\\ \\ 29226397=2.14613198+1\\ \\ 14613198=2.7306599+0\\ \\ 7306599=2.3653299+1\\ \\ 3653299=2.1826649+1\\ \\ 1826649=2.913324+1\\ \\ 913324=2.456662+0\\ \\ 456662=2.228331+0\\ \\ 228331=2.114165+1\\ \\ 114165=2.57082+1\\ \\ 57082=2.28541+0\\ \\ 28541=2.14720+1\\ \\ 14720=2.7135+0\\ \\ 7135=2.3567+1\\ \\ 3567=2.1783+1\\ \\ 1783=2.891+1\\ \\ 891=2.445+1\\ \\ 445=2.222+1\\ \\ 222=2.111+0\\ \\ 111=2.55+1\\ \end{array} \\ \begin{array}{r}27=13+1\\ \\ 6=2.3+0\\ \\ 3=2.1+1\\ \\ 1=2.0+1\end{array} \end{gathered}$$

The successive remainders of each division represents the binary expansion from right left

$(1101111011111010110011101101{)}_2$