Solved Examples: Arithmetic Operations on Number Systems

Samuel Dominic Chukwuemeka (SamDom For Peace) For ACT Students
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For the Questions:
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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 $