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Solved Examples on the Conversion of Number Systems

Verify your answers as applicable with the Modulo Arithmetic and Algorithms Calculators
Prerequisites:
(1.) Exponents
(2.) Factoring

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
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(1.) Calculate the decimal expansion of $11100111_2$

This means that $11100111$ in base two needs to be converted to a number in base ten.

$ 11100111_2 \\[3ex] = 1 * 2^7 + 1 * 2^6 + 1 * 2^5 + 0 * 2^4 + 0 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 \\[3ex] = 1 * 128 + 1 * 64 + 1 * 32 + 0 * 16 + 0 * 8 + 1 * 4 + 1 * 2 + 1 * 1 \\[3ex] = 128 + 64 + 32 + 0 + 0 + 4 + 2 + 1 \\[3ex] $ $11100111_2 = 231$

(2.) Calculate the binary expansion of $231$

This means that $231$ needs to be converted to a number in base two.

$ \begin{array}{c|c} 2 & 231 \\ \hline 2 & 115 \:R\: 1 \\ \hline 2 & 57 \:R\: 1 \\ \hline 2 & 28 \:R\: 1 \\ \hline 2 & 14 \:R\: 0 \\ \hline 2 & 7 \:R\: 0 \\ \hline 2 & 3 \:R\: 1 \\ \hline 2 & 1 \:R\: 1 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 231 = 11100111_2 \\[3ex] $ $231 = 11100111_2$

(3.) Calculate the decimal expansion of $11000100100011000_2$

This means that $11000100100011000$ in base two needs to be converted to a number in base ten.

$ 11000100100011000_2 \\[3ex] = 1 * 2^{16} + 1 * 2^{15} + 0 * 2^{14} + 0 * 2^{13} + 0 * 2^{12} + 1 * 2^{11} + 0 * 2^{10} + 0 * 2^9 + 1 * 2^8 + 0 * 2^7 + 0 * 2^6 + 0 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 0 * 2^1 + 0 * 2^0 \\[3ex] = 1 * 65536 + 1 * 32768 + 0 * 16384 + 0 * 8192 + 0 * 4096 + 1 * 2048 + 0 * 1024 + 0 * 512 + 1 * 256 + 0 * 128 + 0 * 64 + 0 * 32 + 1 * 16 + 1 * 8 + 0 * 4 + 0 * 2 + 0 * 1 \\[3ex] = 65536 + 32768 + 0 + 0 + 0 + 2048 + 0 + 0 + 256 + 0 + 0 + 0 + 16 + 8 + 0 + 0 + 0 \\[3ex] = 100632 \\[3ex] $ $11100111_2 = 100632$

(4.) Calculate the binary expansion of $100632$

This means that $100632$ needs to be converted to a number in base two.

$ \begin{array}{c|c} 2 & 100632 \\ \hline 2 & 50316 \:R\: 0 \\ \hline 2 & 25158 \:R\: 0 \\ \hline 2 & 12579 \:R\: 0 \\ \hline 2 & 6289 \:R\: 1 \\ \hline 2 & 3144 \:R\: 1 \\ \hline 2 & 1572 \:R\: 0 \\ \hline 2 & 786 \:R\: 0 \\ \hline 2 & 393 \:R\: 0 \\ \hline 2 & 196 \:R\: 1 \\ \hline 2 & 98 \:R\: 0 \\ \hline 2 & 49 \:R\: 0 \\ \hline 2 & 24 \:R\: 1 \\ \hline 2 & 12 \:R\: 0 \\ \hline 2 & 6 \:R\: 0 \\ \hline 2 & 3 \:R\: 0 \\ \hline 2 & 1 \:R\: 1 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 100632 = 11000100100011000_2 \\[3ex] $ $100632 = 11000100100011000_2$

(5.) Calculate the hexadecimal expansion of $11101011100_2$
Solve this question using at least three different methods.

This means that $11101011100$ in base two needs to be converted to a number in base sixteen.

First Method:$\:BIN \rightarrow DEC \rightarrow HEX$
This is a long method.

$ BIN \rightarrow DEC \\[3ex] 11101011100_2 \\[3ex] = 1 * 2^{10} + 1 * 2^9 + 1 * 2^8 + 0 * 2^7 + 1 * 2^6 + 0 * 2^5 + 1 * 2^4 + 1 * 2^3 + 1 * 2^2 + 0 * 2^1 + 0 * 2^0 \\[3ex] = 1 * 1024 + 1 * 512 + 1 * 256 + 0 * 128 + 1 * 64 + 0 * 32 + 1 * 16 + 1 * 8 + 1 * 4 + 0 * 2 + 0 * 1 \\[3ex] = 1024 + 512 + 256 + 0 + 64 + 0 + 16 + 8 + 4 + 0 + 0 \\[3ex] \therefore 11101011100_2 = 1884 \\[3ex] DEC \rightarrow HEX \\[3ex] \begin{array}{c|c} 16 & 1884 \\ \hline 16 & 117 \:R\: 12 \\ \hline 16 & 7 \:R\: 5 \\ \hline & 0 \:R\: 7 \end{array} Count \:the\: remainders\: upwards \\[3ex] 1884 = 7\: 5\: {12} \\[3ex] 12 = C_{16} \\[3ex] 1884 = 75C \\[3ex] $ $11101011100_2 = 75C_{16}$

Second Method:$\:BIN \rightarrow DEC \rightarrow HEX \rightarrow OCT \rightarrow QUA \:$ Conversion Table
This is a short method.

Each digit in $HEX$ is equivalent to four digits in $BIN$
We split the digits in a set of four digits each.

$ 11101011100_2 111\: 0101\: 1100 \\[3ex] $ We have two complete sets of four digits each. But we have one incomplete set.
We will need to complete the first incomplete set by adding a $0$
$0111\: 0101\: 1100$
We begin from behind

$ 1100 = C \\[3ex] 0101 = 5 \\[3ex] 0111 = 7 \\[3ex] $ We write it upwards (beginning from the bottom)
$7 \:5\: C$

$\therefore 11101011100_2 = 75C_{16}$

Third Method:$\:BIN \rightarrow HEX\:$ Method
This is also a short method.

Each digit in $HEX$ is equivalent to four digits in $BIN$
We split the digits in a set of four digits each.
$11101011100_2$
$111\: 0101\: 1100$
We have two complete sets of four digits each. But we have one incomplete set.
We will need to complete the first incomplete set by adding a $0$
$0111\: 0101\: 1100$
These are the face values.
Let us write the place values beneath the face values.
$ 0\ \ \ 1\ \ \ 1\ \ \ 1\ \ \ \ \ \ \ \ 0\ \ \ 1\ \ \ 0\ \ \ 1\ \ \ \ \ \ \ 1\ \ \ 1\ \ \ 0\ \ \ 0\ \ \ \ \\ 2^3\ 2^2\ 2^1\ 2^0\ \ \ \ \ \ 2^3\ 2^2\ 2^1\ 2^0\ \ \ \ \ \ \ 2^3\ 2^2\ 2^1\ 2^0\ \ \ \ $

$ 0\ 1\ 1\ 1\ \ \ \ \ 0\ 1\ 0\ 1\ \ \ \ \ 1\ 1\ 0\ 0\ \ \ \ \\ 8\ 4\ 2\ 1\ \ \ \ \ 8\ 4\ 2\ 1\ \ \ \ \ 8\ 4\ 2\ 1\ \ \ \ $

Draw a line beneath the place values (as you would do with addition, subtraction, division, etc.)

$ \begin{align} 0\ 1\ 1\ 1\ \ \ \ \ 0\ 1\ 0\ 1\ \ \ \ \ 1\ 1\ 0\ 0\ \ \ \ \\ 8\ 4\ 2\ 1\ \ \ \ \ 8\ 4\ 2\ 1\ \ \ \ \ 8\ 4\ 2\ 1\ \ \ \ \\ \hline \end{align} $

For each face value of $0$, write a $0$ under the corresponding place value
For each face value that is not a $0$, write the value of the corresponding place "as is"

$ \begin{align} 0\ 1\ 1\ 1\ \ \ \ \ 0\ 1\ 0\ 1\ \ \ \ \ 1\ 1\ 0\ 0\ \ \ \ \\ 8\ 4\ 2\ 1\ \ \ \ \ 8\ 4\ 2\ 1\ \ \ \ \ 8\ 4\ 2\ 1\ \ \ \ \\ \hline \\ 0\ 4\ 2\ 1\ \ \ \ \ 0\ 4\ 0\ 1\ \ \ \ \ 8\ 4\ 0\ 0\ \ \ \ \end{align} $

Add each set of four digits and write the sum below them.
$8 + 4 + 0 + 0 = 12$
But, $12 = C_{16}$
$0 + 4 + 0 + 1 = 5$
$0 + 4 + 2 + 1 = 7$

$ \begin{align} 0\ 1\ 1\ 1\ \ \ \ \ 0\ 1\ 0\ 1\ \ \ \ \ 1\ 1\ 0\ 0\ \ \ \ \\ 8\ 4\ 2\ 1\ \ \ \ \ 8\ 4\ 2\ 1\ \ \ \ \ 8\ 4\ 2\ 1\ \ \ \ \\ \hline \\ \quad \underbrace{0\ 4\ 2\ 1}\ \ \ \ \underbrace{0\ 4\ 0\ 1}\ \ \ \underbrace{8\ 4\ 0\ 0}\ \ \ \\ \underbrace{7}\ \qquad \underbrace{5}\ \qquad \underbrace{C}\ \ \ \ \end{align} \\[3ex] $ $\therefore 11101011100_2 = 75C_{16}$

(6.) Calculate the hexadecimal expansion of $1001001101001010_2$

This means that $1001001101001010$ in base two needs to be converted to a number in base sixteen.

We shall use the $\:BIN \rightarrow DEC \rightarrow HEX \rightarrow OCT \rightarrow QUA \:$ Conversion Table

Each digit in $HEX$ is equivalent to four digits in $BIN$
We split the digits in a set of four digits each.
$1001001101001010_2$
$1001 \:0011\: 0100\: 1010$
We have a complete set of four digits each. We are good.
We begin from behind
$1010 = A$
$0100 = 4$
$0011 = 3$
$1001 = 9$
We write it upwards (beginning from the bottom)
$9 \:3\: 4\: A$
$\therefore 1001001101001010_2 = 934A_{16}$

(7.) Calculate the binary expansion of $ABBA_{16}$
Solve this question using at least two different methods.

This means that $ABBA$ in base sixteen needs to be converted to a number in base two.

First Method:$\:HEX \rightarrow DEC \rightarrow BIN$
This is a long method.

$HEX \rightarrow DEC$

$ ABBA_{16} \\[3ex] = 10 * 16^3 + 11 * 16^2 + 11 * 16^1 + 10 * 16^0 \\[3ex] = 10 * 4096 + 11 * 256 + 11 * 16 + 10 * 1 \\[3ex] = 40960 + 2816 + 176 + 10 \\[3ex] \therefore ABBA_{16} = 43962 \\[3ex] DEC \rightarrow BIN \\[3ex] \begin{array}{c|c} 2 & 43962 \\ \hline 2 & 21981 \:R\: 0 \\ \hline 2 & 10990 \:R\: 1 \\ \hline 2 & 5495 \:R\: 0 \\ \hline 2 & 2747 \:R\: 1 \\ \hline 2 & 1373 \:R\: 1 \\ \hline 2 & 686 \:R\: 1 \\ \hline 2 & 343 \:R\: 0 \\ \hline 2 & 171 \:R\: 1 \\ \hline 2 & 85 \:R\: 1 \\ \hline 2 & 42 \:R\: 1 \\ \hline 2 & 21 \:R\: 0 \\ \hline 2 & 10 \:R\: 1 \\ \hline 2 & 5 \:R\: 0 \\ \hline 2 & 2 \:R\: 1 \\ \hline 2 & 1 \:R\: 0 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 43962 = 1010101110111010_2 \\[3ex] $ $ABBA_{16} = 1010101110111010_2$

Second Method:$\:BIN \rightarrow DEC \rightarrow HEX \rightarrow OCT \rightarrow QUA \:$ Conversion Table
This is a short method.

Remember that each digit in $HEX$ is equivalent to four digits in $BIN$
$ABBA_{16}$
We begin from behind
$A = 1010$
$B = 1011$
$B = 1011$
$A = 1010$
We write it upwards (beginning from the bottom)
$1010 \:1011\: 1011\: 1010$
Remove any leading zero(s) if any.
Leading zeros are zeros that may be in front of the first set of digits.
In this case, there are no leading zeros.

$\therefore ABBA_{16} = 1010101110111010_2$

(8.) Calculate the binary expansion of $934A_{16}$

This means that $934A$ in base sixteen needs to be converted to a number in base two.

We shall use the $\:BIN \rightarrow DEC \rightarrow HEX \rightarrow OCT \rightarrow QUA \:$ Conversion Table

Each digit in $HEX$ is equivalent to four digits in $BIN$
$934A_{16}$
We begin from behind
$A = 1010$
$4 = 0100$
$3 = 0011$
$9 = 1001$
We write it upwards (beginning from the bottom)
$1001 \:0011\: 0100\: 1010$
Remove any leading zero(s) if any.
Leading zeros are zeros that may be in front of the first set of digits.
In this case, there are no leading zeros.

$\therefore 934A_{16} = 1001001101001010_2$

(9.) Calculate the binary expansion of $CAF57_{16}$
Solve this question using at least two different methods.

This means that $CAF57$ in base sixteen needs to be converted to a number in base two.

First Method:$\:HEX \rightarrow DEC \rightarrow BIN$
This is a long method.

$ HEX \rightarrow DEC \\[3ex] CAF57_{16} \\[3ex] = 12 * 16^4 + 10 * 16^3 + 15 * 16^2 + 5 * 16^1 + 7 * 16^0 \\[3ex] = 12 * 65536 + 10 * 4096 + 15 * 256 + 5 * 16 + 7 * 1 \\[3ex] = 786432 + 40960 + 3840 + 80 + 7 \\[3ex] \therefore CAF57_{16} = 831319 \\[3ex] DEC \rightarrow BIN \\[3ex] \begin{array}{c|c} 2 & 831319 \\ \hline 2 & 415659 \:R\: 1 \\ \hline 2 & 207829 \:R\: 1 \\ \hline 2 & 103914 \:R\: 1 \\ \hline 2 & 51957 \:R\: 0 \\ \hline 2 & 25978 \:R\: 1 \\ \hline 2 & 12989 \:R\: 0 \\ \hline 2 & 6494 \:R\: 1 \\ \hline 2 & 3247 \:R\: 0 \\ \hline 2 & 1623 \:R\: 1 \\ \hline 2 & 811 \:R\: 1 \\ \hline 2 & 405 \:R\: 1 \\ \hline 2 & 202 \:R\: 1 \\ \hline 2 & 101 \:R\: 0 \\ \hline 2 & 50 \:R\: 1 \\ \hline 2 & 25 \:R\: 0 \\ \hline 2 & 12 \:R\: 1 \\ \hline 2 & 6 \:R\: 0 \\ \hline 2 & 3 \:R\: 0 \\ \hline 2 & 1 \:R\: 1 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 831319 = 11001010111101010111_2 \\[3ex] $ $CAF57_{16} = 11001010111101010111_2$

Second Method:$\:BIN \rightarrow DEC \rightarrow HEX \rightarrow OCT \rightarrow QUA \:$ Conversion Table
This is a short method.

Remember that each digit in $HEX$ is equivalent to four digits in $BIN$
$CAF57_{16}$
We begin from behind
$7 = 0111$
$5 = 0101$
$F = 1111$
$A = 1010$
$C = 1100$
We write it upwards (beginning from the bottom)
$1100 \:1010\: 1111\: 0101\: 0111$
Remove any leading zero(s) if any.
Leading zeros are zeros that may be in front of the first set of digits.
In this case, there are no leading zeros.

$\therefore CAF57_{16} = 11001010111101010111_2$

(10.) Calculate the binary expansion of $75C_{16}$
Solve this question using at least two different methods.

This means that $75C$ in base sixteen needs to be converted to a number in base two.

First Method:$\:HEX \rightarrow DEC \rightarrow BIN$
This is a long method.

$ HEX \rightarrow DEC \\[3ex] 75C_{16} \\[3ex] = 7 * 16^2 + 5 * 16^1 + 12 * 16^0 \\[3ex] = 7 * 256 + 5 * 16 + 12 * 1 \\[3ex] = 1792 + 80 + 12 \\[3ex] \therefore 75C_{16} = 1884 \\[3ex] DEC \rightarrow BIN \\[3ex] \begin{array}{c|c} 2 & 1884 \\ \hline 2 & 942 \:R\: 0 \\ \hline 2 & 471 \:R\: 0 \\ \hline 2 & 235 \:R\: 1 \\ \hline 2 & 117 \:R\: 1 \\ \hline 2 & 58 \:R\: 1 \\ \hline 2 & 29 \:R\: 0 \\ \hline 2 & 14 \:R\: 1 \\ \hline 2 & 7 \:R\: 0 \\ \hline 2 & 3 \:R\: 1 \\ \hline 2 & 1 \:R\: 1 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 1884 = 11101011100_2 \\[3ex] $ $75C_{16} = 11101011100_2$

Second Method:$\:BIN \rightarrow DEC \rightarrow HEX \rightarrow OCT \rightarrow QUA \:$ Conversion Table
This is a short method.

Remember that each digit in $HEX$ is equivalent to four digits in $BIN$
$75C_{16}$
We begin from behind
$C = 1100$
$5 = 0101$
$7 = 0111$
We write it upwards (beginning from the bottom)
$0111 \:0101\: 1100$
Remove any leading zero(s) if any.
Leading zeros are zeros that may be in front of the first set of digits.
In this case, there is only one leading zero.
$111 \:0101\: 1100$

$\therefore 75C_{16} = 11101011100_2$

(11.) Calculate the binary expansion of $3AC1_{16}$
Solve this question using at least two different methods.

This means that $3AC1$ in base sixteen needs to be converted to a number in base two.

First Method:$\:HEX \rightarrow DEC \rightarrow BIN$
This is a long method.

$ HEX \rightarrow DEC \\[3ex] 3AC1_{16} \\[3ex] = 3 * 16^3 + 10 * 16^2 + 12 * 16^1 + 1 * 16^0 \\[3ex] = 3 * 4096 + 10 * 256 + 12 * 16 + 1 * 1 \\[3ex] = 12288 + 2560 + 192 + 1 \\[3ex] \therefore 3AC1_{16} = 15041 \\[3ex] DEC \rightarrow BIN \\[3ex] \begin{array}{c|c} 2 & 15041 \\ \hline 2 & 7520 \:R\: 1 \\ \hline 2 & 3760 \:R\: 0 \\ \hline 2 & 1880 \:R\: 0 \\ \hline 2 & 940 \:R\: 0 \\ \hline 2 & 470 \:R\: 0 \\ \hline 2 & 235 \:R\: 0 \\ \hline 2 & 117 \:R\: 1 \\ \hline 2 & 58 \:R\: 1 \\ \hline 2 & 29 \:R\: 0 \\ \hline 2 & 14 \:R\: 1 \\ \hline 2 & 7 \:R\: 0 \\ \hline 2 & 3 \:R\: 1 \\ \hline 2 & 1 \:R\: 1 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 15041 = 11101011000001_2 \\[3ex] $ $3AC1_{16} = 11101011000001_2$

Second Method:$\:BIN \rightarrow DEC \rightarrow HEX \rightarrow OCT \rightarrow QUA \:$ Conversion Table
This is a short method.

Remember that each digit in $HEX$ is equivalent to four digits in $BIN$
$3AC1_{16}$
We begin from behind
$1 = 0001$
$C = 1100$
$A = 1010$
$3 = 0011$
We write it upwards (beginning from the bottom)
$0011 \:1010\: :1100\: 0001$
Remove any leading zero(s) if any.
Leading zeros are zeros that may be in front of the first set of digits.
In this case, there are two leading zeros.
$11 \:1010\: :1100\: 0001$

$\therefore 3AC1_{16} = 11101011000001_2$

(12.) Calculate the binary expansion of $100632$

(13.)

(14.) WASSCE Given that $110_x = 40_{five}$, find the value of $x$

$ 110_x = 40_{five} \\[3ex] Convert\:\:both\:\:bases\:\:to\:\:base\:\:ten \\[3ex] 1 * x^2 + 1 * x^1 + 0 * x^0 = 4 * 5^1 + 0 * 5^0 \\[3ex] x^2 + x + 0 = 20 + 0 \\[3ex] x^2 + x = 20 \\[3ex] x^2 + x - 20 = 0 \\[3ex] (x + 5)(x - 4) = 0 \\[3ex] x = -5 \;\;OR\;\; x = 4 \\[3ex] The\;\;base\;\;cannot\;\;be\;\;negative \\[3ex] \therefore x = 4 $