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Solved Examples on the Conversion of Number Systems

Samuel Dominic Chukwuemeka (SamDom For Peace) Verify your answers as applicable with the Modulo Arithmetic and Algorithms Calculators
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
Use at least two methods for each question as applicable.
Show all work

(1.) Calculate the decimal expansion of 110001001000110002


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] $ $11000100100011000_2 = 100632$
(2.) 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] $ $100632 = 11000100100011000_2$
(3.) 10001112 * 11110002


First Method: Operate in Base Two
We shall do our multiplication the usual way we multiply numbers.
Remember that Multiplication is Repeated Addition.

$ \begin{align} 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ \\ * \ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ \\ \hline \end{align} \\[3ex] $ The minuend = $1000111$
It is seven digits
The subtrahend = $1111000$
It is also seven digits
Line them up
Do the multiplications
Do the additions

$ \begin{align} 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ \\ * \ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ \\ \hline 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ \\ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0~~~~\ \ \\ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0~~~~~~~~\ \ \\ + \ \ \ \ \ \ \ \ \ \ 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1~~~~~~~~~~~~\ \ \\ 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1~~~~~~~~~~~~~~~~\ \ \\ 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1~~~~~~~~~~~~~~~~~~~~\ \ \\ 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1~~~~~~~~~~~~~~~~~~~~~~~~\ \ \\ \hline 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 0\ \ 1\ \ 0\ \ 0\ \ 1\ \ 0\ \ 0\ \ 0~~ \\ \hline \\[3ex] \end{align} $ $1000111_2 * 1111000_2 = 10000101001000_2$

Second Method: Convert to Base Ten, Operate in Base Ten, Convert to Base Two

$ \underline{Convert\;\;to\;\;Base\;\;Ten} \\[3ex] 1000111_2 \\[3ex] = 1 * 2^6 + 0 * 2^5 + 0 * 2^4 + 0 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 \\[3ex] = 1(64) + 0 + 0 + 0 + 1(4) + 1(2) + 1(1) \\[3ex] = 64 + 4 + 2 + 1 \\[3ex] = 71 \\[5ex] 1111000_2 \\[3ex] = 1 * 2^6 + 1 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 0 * 2^1 + 0 * 2^0 \\[3ex] = 1(64) + 1(32) + 1(16) + 1(8) + 0 + 0 + 0 \\[3ex] = 64 + 32 + 16 + 8 \\[3ex] = 120 \\[5ex] \underline{Operate\;\;in\;\;Base\;\;Ten} \\[3ex] 71 * 120 = 8520 \\[5ex] \underline{Convert\;\;to\;\;Base\;\;Two} \\[3ex] \begin{array}{c|c} 2 & 8520 \\ \hline 2 & 4260 \:R\: 0 \\ \hline 2 & 2130 \:R\: 0 \\ \hline 2 & 1065 \:R\: 0 \\ \hline 2 & 532 \:R\: 1 \\ \hline 2 & 266 \:R\: 0 \\ \hline 2 & 133 \:R\: 0 \\ \hline 2 & 66 \:R\: 1 \\ \hline 2 & 33 \:R\: 0 \\ \hline 2 & 16 \:R\: 1 \\ \hline 2 & 8 \:R\: 0 \\ \hline 2 & 4 \:R\: 0 \\ \hline 2 & 2 \:R\: 0 \\ \hline 2 & 1 \:R\: 0 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 8520 = 11000100100011000_2 \\[3ex] $ $1000111_2 * 1111000_2 = 10000101001000_2$
(4.) 10001112 + 11110002


First Method: Operate in Base Two

$ \begin{align} 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ \\ + \ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ \\ \hline \end{align} \\[3ex] $ The augend = $1000111$
It is seven digits
The addend = $1111000$
It is also seven digits
Line them up
Add each digit of the augend to the corresponding digit of the addend
Begin from the left

$ 1 + 0 = 1 \\[3ex] 1 + 0 = 1 \\[3ex] 1 + 0 = 1 \\[3ex] 0 + 1 = 1 \\[3ex] 0 + 1 = 1 \\[3ex] 0 + 1 = 1 \\[3ex] 1 + 1 = 2 = 10 \: (Remember \:that\: 2 = 10_2) \\[3ex] $ Write the result upwards beginning from the bottom
This gives: $10111111$

$ \begin{align} \ \ 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ \\ + \ \ \ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ \\ \hline 1\ \ 0\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1~~ \\ \hline \\[3ex] \end{align} $ $1000111_2 + 1111000_2 = 10111111_2$

Second Method: Convert to Base Ten, Operate in Base Ten, Convert to Base Two

$ \underline{Convert\;\;to\;\;Base\;\;Ten} \\[3ex] 1000111_2 \\[3ex] = 1 * 2^6 + 0 * 2^5 + 0 * 2^4 + 0 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 \\[3ex] = 1(64) + 0 + 0 + 0 + 1(4) + 1(2) + 1(1) \\[3ex] = 64 + 4 + 2 + 1 \\[3ex] = 71 \\[5ex] 1111000_2 \\[3ex] = 1 * 2^6 + 1 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 0 * 2^1 + 0 * 2^0 \\[3ex] = 1(64) + 1(32) + 1(16) + 1(8) + 0 + 0 + 0 \\[3ex] = 64 + 32 + 16 + 8 \\[3ex] = 120 \\[5ex] \underline{Operate\;\;in\;\;Base\;\;Ten} \\[3ex] 71 + 120 = 191 \\[5ex] \underline{Convert\;\;to\;\;Base\;\;Two} \\[3ex] \begin{array}{c|c} 2 & 191 \\ \hline 2 & 95 \:R\: 1 \\ \hline 2 & 47 \:R\: 1 \\ \hline 2 & 23 \:R\: 1 \\ \hline 2 & 11 \:R\: 1 \\ \hline 2 & 5 \:R\: 1 \\ \hline 2 & 2 \:R\: 1 \\ \hline 2 & 1 \:R\: 0 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 191 = 10111111_2 \\[3ex] $ $1000111_2 + 1111000_2 = 10111111_2$
(5.) WASCCE Find the product of 1101two and 111two

$ A.\;\; 1101011_{two} \\[3ex] B.\;\; 1011101_{two} \\[3ex] C.\;\; 1110011_{two} \\[3ex] D.\;\; 1011011_{two} \\[3ex] $

First Method: Operate in Base Two

$ \begin{align} 1\ \ 1\ \ 0\ \ 1\ \ \\ * \ \ \ \ \ \ 1\ \ 1\ \ 1\ \ \\ \hline 1\ \ 1\ \ 0\ \ 1\ \ \\ + \ \ \ \ \ \ \ \ \ \ 1\ \ 1\ \ 0\ \ 1~~~~\ \ \\ 1\ \ 1\ \ 0\ \ 1~~~~~~~~\ \ \\ \hline 1\ \ 0\ \ 1\ \ 1\ \ 0\ \ 1\ \ 1~~ \\ \hline \end{align} $

$1101_{two} * 111_{two} = 1011011_{two}$

Second Method: Convert to Base Ten, Operate in Base Ten, Convert to Base Two

$ \underline{Convert\;\;to\;\;Base\;\;Ten} \\[3ex] 1101_{two} \\[3ex] = 1 * 2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0 \\[3ex] = 1(8) + 1(4) + 0 + 1(1) \\[3ex] = 8 + 4 + 0 + 1 \\[3ex] = 13 \\[5ex] 111_{two} \\[3ex] = 1 * 2^2 + 1 * 2^1 + 1 * 2^0 \\[3ex] = 1(4) + 1(2) + 1(1) \\[3ex] = 4 + 2 + 1 \\[3ex] = 7 \\[5ex] \underline{Operate\;\;in\;\;Base\;\;Ten} \\[3ex] 13 * 7 = 91 \\[5ex] \underline{Convert\;\;to\;\;Base\;\;Two} \\[3ex] \begin{array}{c|c} 2 & 91 \\ \hline 2 & 45 \:R\: 1 \\ \hline 2 & 22 \:R\: 1 \\ \hline 2 & 11 \:R\: 0 \\ \hline 2 & 5 \:R\: 1 \\ \hline 2 & 2 \:R\: 1 \\ \hline 2 & 1 \:R\: 0 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 91 = 1011011_{two} \\[3ex] $ $1101_{two} * 111_{two} = 1011011_{two}$
(6.) JAMB The sum of four numbers is 1214five
What is the average expressed in base five?

$ A.\;\; 411 \\[3ex] B.\;\; 401 \\[3ex] C.\;\; 114 \\[3ex] D.\;\; 141 \\[3ex] $

$ \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \Sigma x = 1214_{five}...convert\;\;to\;\;base\;\;ten \\[3ex] = 1 * 5^3 + 2 * 5^2 + 1 * 5^1 + 4 * 5^0 \\[3ex] = 1 * 125 + 2 * 25 + 1 * 5 + 4 * 1 \\[3ex] = 125 + 50 + 5 + 4 \\[3ex] = 184 \\[3ex] n = 4 \\[3ex] \bar{x} = \dfrac{184}{4} = 46 \\[5ex] Convert\;\;back\;\;to\;\;base\;\;five \\[3ex] \begin{array}{c|c} 5 & 46 \\ \hline 5 & 9 \:R\: 1 \\ \hline 5 & 1 \:R\: 4 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 46 = 141_5 $
(7.)


(8.) JAMB Simplify 2134 * 234

$ A.\;\; 411 \\[3ex] B.\;\; 401 \\[3ex] C.\;\; 114 \\[3ex] D.\;\; 141 \\[3ex] $

We shall solve this question in at least two ways.
Use any method you prefer.

$ 213_4 * 23_4 \\[5ex] \underline{First\:\:Method:}\;\;Multiply\;\;in\;\;base\;\;four \\[3ex] \begin{align} 2\ \ 1\ \ 3\ \ \\ * \ \ 2\ \ 3\ \ \\ \hline 1\ \ 2\ \ 2\ \ 3\ \ 1~~ \\ \hline \end{align} \\[5ex] 3 * 3 = 9 = 21_4 \\[3ex] 1\;\;carry\;\;2 \\[3ex] 3 * 1 = 3 \\[3ex] 3 + 2 = 5 = 11_4 \\[3ex] 1\;\;carry\;\;1 \\[3ex] 3 * 2 = 6 \\[3ex] 6 + 1 = 7 = 13_4 \\[3ex] ********************* \\[3ex] 2 * 3 = 6 = 12_4 \\[3ex] 2\;\;carry\;\;1 \\[3ex] 2 * 1 = 2 \\[3ex] 2 + 1 = 3 \\[3ex] 2 * 2 = 4 = 10_4 \\[3ex] ********************* \\[3ex] 3 + 3 = 6 = 12_4 \\[3ex] ********************* \\[3ex] \underline{Second\:\:Method:} \\[3ex] Convert\;\;to\;\;base\;\;ten \\[3ex] Multiply\;\;in\;\;base\;\;ten \\[3ex] Convert\;\;back\;\;to\;\;base\;\;four \\[3ex] 213_4 \\[3ex] = 2(4)^2 + 1(4)^1 + 3(4)^0 \\[3ex] = 2(16) + 1(4) + 3(1) \\[3ex] = 32 + 4 + 3 \\[3ex] = 39 \\[3ex] 23_4 \\[3ex] = 2(4)^1 + 3(4)^0 \\[3ex] = 2(4) + 3(1) \\[3ex] = 8 + 3 \\[3ex] = 11 \\[3ex] 213_4 * 23_4 \implies 39 * 11 = 429 \\[3ex] Convert\;\;back\;\;to\;\;base\;\;four \\[3ex] \begin{array}{c|c} 4 & 429 \\ \hline 4 & 107 \:R\: 1 \\ \hline 4 & 26 \:R\: 3 \\ \hline 4 & 6 \:R\: 2 \\ \hline 4 & 1 \:R\: 2 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 429 = 12231_4 $