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Grade 7 Mathematics Learner’s Book

INNODEMS

Section 1.2 Factors

A factor is a number that can be multiplied to create a specific number in math.
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For example, A group of \(12\) grade \(7\) students wanted to share \(24\) cookies equally. To find it out, they looked at the factors of \(24\text{:}\)1,2,3,4,6,8 and 24. They realized they could split the cookies into:
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  • \(2\) cookies each for the \(12\) people.
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  • \(4\) cookies each for the \(6\) people.
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  • \(6\) cookies each for the \(2\) people.
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Factors helped them share the cookies fairly among themselves.
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Subsection 1.2.1 Divisibility test for 2

A number is divisible by \(2\) if the digit in the ones place value is an even number or zero.
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Activity 1.2.1.

Work in groups
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(a)
Form groups of \(10\) learners. Each learner should write down a number between \(500\) and \(5000\text{.}\)
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(b)
Identify which of the numbers have an even number as their last digit.
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(c)
Identify which of the numbers end with a zero.
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(d)
Find out which numbers are divisible by \(2.\)
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(e)
Share your findings with the rest of the class.
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Example 1.2.1.

Which of the following numbers are divisible by \(2\text{?}\)
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\(34\,436, 78\,135, 23\,521, 56\,490\)
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Solution.
To identify if a number is divisible by \(2\text{,}\) we check if its last digit (the digit in the ones place) is either \(0\) or an even number \((2, 4, 6, 8)\text{.}\)
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  1. \(34\,436\)
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    The last digit of \(34\,43{\color{red}6}\) is \(6\text{,}\) which is an even number.
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    Therefore, \(34\,436\) is divisible by \(2.\)
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  2. \(78\,135\)
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    The last digit of \(78\,13{\color{red}5}\) is \(5\text{,}\) which is an odd number.
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    Therefore, \(78\,135\) is not divisible by \(2.\)
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  3. \(23\,521\)
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    The last digit of \(23\,52{\color{red}1}\) is \(1\text{,}\) which is an odd number.
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    Therefore, \(23\,521\) is not divisible by \(2.\)
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  4. \(56\,490\)
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    The last digit of \(56\,49{\color{red}0}\) is \(0\)
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    Therefore, \(56\,490\) is divisible by \(2.\)
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Hence, \(34\,436, 56\,490\) are divisible by \(2\text{.}\)
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Checkpoint 1.2.2.

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Subsection 1.2.2 Divisibility test for 3

A number is divisible by \(3\) if the sum of its digits gives a number that is divisible by \(3\text{.}\)
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Esther bought \(12\) mangoes and needs to divide them amoung her \(3\) children. She wondered if she can share the mangoes equally amoung her children.
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Since adding \(12\) gives \(3\) i.e
\begin{equation*} 1+2 = 3 \end{equation*}
hence \(12\) is divisible by \(3\) and Esther will share the mangoes equally among the children.
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Activity 1.2.2.

1. Count the number of blue pens in your class and one of you to write it down.
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2. Count also the number of red pens in your class and indicate the number.
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3. Find the total number of blue and red pens in the class.
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4. Write down the number.
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5. Add the digits in that number and identify if the result is divisible by \(3\text{.}\)
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Example 1.2.3.

Which of the following numbers are divisible by 3?
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  1. \(\displaystyle 42\,879\)
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  2. \(\displaystyle 259\,470\)
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  3. \(\displaystyle 687\,852\)
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  4. \(\displaystyle 2\,257\,123\)
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Solution.
  1. \(42\, 879 \)
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    First, we add up all the single digits in the number
    \begin{equation*} 4+2+8+7+9 = 30 \end{equation*}
    Now, we check if \(30\) can be divided by \(3\) without any remainder
    \begin{equation*} 30 \,\div 3 \,= 10 \end{equation*}
    \(30\) is divisible by \(3\) hence \(42\,879\) is divisible by \(3\text{.}\)
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  2. \(259\,470\)
    \begin{align*} 2+5+9+4+7+0 \amp= 27 \\ 27 \,\div 3 \,\amp= 9 \end{align*}
    \(27\) is divisible by \(3\) hence \(247\,329\) is divisible by \(3\)
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  3. \(687\,852\)
    \begin{align*} 6+8+7+8+5+2\amp= 36 \\ 36\,\div 3 \,\amp= 12 \end{align*}
    \(36\) is divisible by \(3\) hence \(687\, 852\) is divisible by \(3\)
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  4. \(3 \,752 \,212\)
    \begin{align*} 3+7+5+2+2+1+2\amp= 22 \\ 22 \,\div 3 \,\amp= 7\text{ remainder 1} \end{align*}
    \(22\) is not divisible by \(3\) hence \(3\,752\,212\) is not divisible by \(3\text{.}\)
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Checkpoint 1.2.4.

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Subsection 1.2.3 Divisibility test for 4

A number is divisible by 4 if its digits in the ones and tens place value are both zero or the number formed by the digits in the ones and tens place value is divisible by 4
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Activity 1.2.3.

1. Each student to write a number between \(10\) and \(100\) and put it on a table at the center of the class.
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2. P lace your math textbooks in the center of the classroom.
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3. In turns, pick a number from the table without choosing.
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4. Check if the number is divisible by \(4\) by looking if the digits at the ones and tens place value are both zero or the number formed by the digits in the ones and tens place value is divisible by 4.
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5. If the number is divisible by \(4\) then pick \(4\) textbooks and remain standing and if the number is not divisible by \(4\) pass and go take a seat.
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6. Share with your fellow learners the numbers that are divisible by \(4\text{.}\)
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Example 1.2.5.

Which of the following numbers are divisible by \(4\text{?}\)
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  1. \(\displaystyle 362\,982\)
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  2. \(\displaystyle 1\,732\,240\)
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  3. \(\displaystyle 289\,367\)
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  4. \(\displaystyle 964\,200\)
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Solution.
  1. \(362\,982\)
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    The number formed by the digits in ones and tens place value is \(62\)
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    Now divide \(82\) by \(4\)
    \begin{equation*} 82 \div 4 = 20 \text{ remainder } 2 \end{equation*}
    Since \(82\) is not divisible by \(4\) hence \(362\,982\) is not divisible by \(4\)
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  2. \(1\,732\,240\)
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    The number formed by the digits in ones and tens place value is \(40\)
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    Now divide \(40\) by \(4\)
    \begin{equation*} 40 \div 4 = 10 \end{equation*}
    Since \(40\) is divisible by \(4\) hence \(1\,732\,240\) is divisible by \(4\)
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  3. \(289\,367\)
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    The number formed by the digits in ones and tens place value is \(67\)
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    Now divide \(67\) by \(4\)
    \begin{equation*} 67\div 4 = 16 \text{ remainder } 3 \end{equation*}
    Since \(67\) is not divisible by \(4\) hence \(289\,367\) is not divisible by \(4\)
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  4. \(964\,200\)
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    The last two numbers are zeros hence \(964\,200\) is divisible by \(4\)
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Checkpoint 1.2.6.

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Subsection 1.2.4 Divisibility test for 5

A number is divisible by \(5\) if the digit in the ones place value is \(5\) or zero.
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Activity 1.2.4.

1. Working in groups of 5, think of numbers between \(50\) and \(200\) and select five numbers from the range .
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2. Check which of your selected numbers are divisible by \(5\) and write them down.
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3. Look at the last digits of the numbers that are divisible by \(5\text{.}\)
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4. What do you notice?
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5. Present your observation to the class.
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Example 1.2.7.

Which of the following numbers are divisible by \(5\text{?}\)
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  1. \(\displaystyle 18\,765\)
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  2. \(\displaystyle 901\,070\)
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  3. \(\displaystyle 135\,932\)
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  4. \(\displaystyle 420\,520\)
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Solution.
To identify the numbers divisible by \(5\text{:}\)
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Check the numbers whose digits in the ones place value is either \(5\) or \(0\text{.}\)
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\(18\,765\text{,}\) \(901\,070\text{,}\) \(420\,520\) are divisible by \(5\) since the digits in the ones place value is either \(5\) or zero.
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Checkpoint 1.2.8.

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Subsection 1.2.5 Divisibility test for 6

A number is divisible by \(6\) if it is divisible by both \(2\) and \(3\text{.}\)
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Activity 1.2.5.

1. Working in groups, find \(1\) mathematical textbook and put it at the table.
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2. In turns, each group member to close their eyes and open a page in the textbook.
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3. One of the goup members to write down the number pages.
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4. Test if the numbers are divisible by \(2\) that is if the last digit is an even number and also test if the numbers are divisible by \(3\) that is add the digits and check if the sum is divisible by \(3\text{.}\)
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5. If the number is divisible by both \(2\) and \(3\) then it is divisble by \(6\text{.}\)
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6. Share your group’s findings with the class. What did you notice about the numbers divisible by \(6\text{.}\)
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Example 1.2.9.

Which of the following numbers are divisible by 6?
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\(17\,832\text{,}\) \(80\,258\text{,}\) \(632\,842\)
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Solution.
  1. \(17\,832\)
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    First we check if \(17\,832\) is divisible by both \(2 \text{ and } 3.\)
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    The digit in the ones place value of \(17\,832\) is even, hence it is divisible by \(2.\)
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    Add up all the single digits in the number \(17\,832\) then divide the sum by \(3.\)
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    \(1+7+8+3+2 = 21\)
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    \(21 \div 3 = 7\) which is divisible by 3.
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    \(17\,832\) is divisible by \(6\) because it is divisible by both \(2 \text{ and } 3.\)
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  2. \(80\,258\)
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    First we check if \(80\,258\) is divisible by both \(2 \text{ and } 3.\)
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    The digit in the ones place value of \(80\,258\) is even, hence it is divisible by \(2.\)
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    Add up all the single digits in the number \(80\,258\) then divide the sum by \(3.\)
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    \(8+0+2+5+8 = 23\)
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    \(23 \div 3 = 7 \text{ remainder } 2\) which is not divisible by 3.
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    \(28\,085\) is not divisible by \(6\) because it is not divisible by both \(2 \text{ and } 3.\)
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  3. \(236\,484\)
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    First we check if \(632\,842\) is divisible by both \(2 \text{ and } 3.\)
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    The digit in the ones place value of \(632\,842\) is even, hence it is divisible by \(2.\)
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    Add up all the single digits in the number \(632\,842\) then divide the sum by \(3.\)
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    \(6+3+2+8+4+2 = 25\)
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    \(25 \div 3 = 8\) remainder \(1\) hence it is not divisible by 3.
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    \(632\,842\) is not divisible by \(6\) because it is not divisible by both \(2 \text{ and } 3.\)
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Checkpoint 1.2.10.

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Subsection 1.2.6 Divisibility test for 8

A number is divisible by \(8\) if the number formed by the digits in the ones, tens and hundreds place value is divisible by \(8.\)
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Activity 1.2.6.

1. Working in small groups with your classmates, use shoe sizes to generate numbers and write them down.
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2. Combine two shoe sizes to create new numbers.
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3. Look at the last three digits of the numbers formed and check if these last three digits are divisible by \(8\) by dividing it by \(8\text{.}\) If it is divisible by \(8\) it should divide the number without leaving a remainder.
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4. Write down whether each combined number is divisible by \(8\) or not.
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Example 1.2.11.

Which of the following numbers are divisible by 8?
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\(16\,232\text{,}\) \(56\,468\text{,}\) \(124\,472\)
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Solution.
  1. \(16\,232\)
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    \(232 \div 8 = 29\)
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    \(16\,232\) is divisible by \(8.\)
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  2. \(56\,468\)
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    \(468 \div 8 = 58 \text{ remainder } 4\)
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    \(56\,468\) is not divisible by \(8.\)
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  3. \(124\,472\)
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    \(472 \div 8 = 59\)
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    \(124\,472\) is divisible by \(8.\)
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\(16\,232\) and \(124\,472\) are divisible by \(8\) because the numbers formed by the digits in the ones, tens and hundreds place value are divisible by \(8\text{.}\)
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Checkpoint 1.2.12.

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Subsection 1.2.7 Divisibility test for 9

A number is divisible by \(9\) if the sum of the digits in the number is divisible by \(9.\)
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Exploration 1.2.7.

Here is an example on how to find out if \(243\) and \(154\) are divisible by \(9\text{:}\)
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\(243 \rightarrow 2 + 4 + 3 = 9 \rightarrow 9 \) is divisible by \(9 \) , so \(243\) is divisible by \(9\text{.}\)
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\(154 \rightarrow 1 + 5 + 4 = 10 \rightarrow 10 \) is not divisible by \(9 \) , so \(154\) is not divisible by \(9\text{.}\)
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Activity 1.2.8.

1. Working in groups of \(10\text{,}\) each learner to form a 3 digit number using numbers from \(1-9\text{.}\)
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2. Each student to write down their 3-digit number on a piece of paper.
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3. Outside the classroom, each group members to line up holding their numbers in the air for everyone to see.
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4. Play the shooting game where the first person in the line "shoots" (points) at everyone in positions that are multiples of 2 (e.g., 2nd, 4th, 6th,...).
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5. Check if the numbers from the students who were shot(were in positions which are multiples of 2) are divisible by 9.
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6. Repeat the shooting game but here, the second person in the line shoots everyone in odd positions ( e.g., 1st, 3rd, 5th,...).
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7. Check if the numbers obtained from the students in odd positions are divisible by 9.
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8. Write down all numbers divisible by 9 from both rounds. Share your results with other groups.
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9. Discuss what you realized about the numbers divisible by 9. Were there patterns in the results?
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Example 1.2.13.

Which of the following numbers are divisible by 9?
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\(14\,578\text{,}\) \(68\,373\text{,}\) \(103\,689\)
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Solution.
  1. \(14\,578\)
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    \(1+4+5+7+8=25\)
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    \(25 \div 9 = 2 \text{ remainder } 7\)
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    \(14\,578\) is not divisible by \(9\)
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  2. \(68\,373\)
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    \(6+8+3+7+3=27\)
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    \(27 \div 9 = 3 \)
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    \(68\,373\) is divisible by \(9\)
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  3. \(103\,689\)
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    \(1+0+3+6+8+9=27\)
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    \(27 \div 9 = 3 \)
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    \(103\,689\) is divisible by \(9\)
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\(68\,373\) and \(103\,689\) are divisible \(9\) because the sum of the digits in each number is divisible by \(9\)
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Checkpoint 1.2.14.

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Checkpoint 1.2.15.

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Subsection 1.2.8 Divisibility test for 10

A number is divisible by \(10\) if the digit in the place value is zero.
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Activity 1.2.9.

Work in groups
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(a)
Write down all the years between \(1995\) and \(2025\text{.}\)
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(b)
Identify which years have a zero as the last digit.
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(c)
Determine which years are divisible by \(10\text{.}\)
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(d)
Discuss how you can tell if a year is divisible by \(10.\)
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(e)
Share your findings with other learners in the class.
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Example 1.2.16.

Which of the following numbers are divisible by \(10\text{?}\)
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\(43\,890\text{,}\) \(51\,014\text{,}\) \(230\,170\text{,}\) \(62\,511\,105\)
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Solution.
\(43\,890\) and \(230\,170\) are divisible by \(10\) because their digits in the ones place value is zero.
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\(51\,014\text{,}\) \(62\,511\,105\) are not divisible by \(10\) because their digits in the ones place value is not zero.
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Checkpoint 1.2.17.

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Subsection 1.2.9 Divisibility test for 11

To check if a number is divisible by \(11\) , subtract the sum of the digits in odd positions from the sum of the digits in even positions. If the result is zero or a multiple of \(11\) , the number is divisible by 11.
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Activity 1.2.10.

Work in groups
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(a)
Working in groups of \(5\) learners, each learner to create a 4-digit number using digits from \(1\) to \(9\) and writes it on a piece of paper.
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(b)
In turns, each learner to read out the number they formed for the group to check if it is divisible by \(11\text{.}\)
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(c)
Write down numbers you found to be divisible by \(11\text{.}\)
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(d)
Reflect on and discuss the strategy you used to determine if a number is divisible by \(11\text{.}\)
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(e)
Share your findings with the rest of the class.
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Example 1.2.18.

Is \(375\,149\) divisible by \(11\text{?}\)
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Solution.
  1. First, find the sum of the digits in odd number positions.
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    \(3+5+4=12\)
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    Next we find the sum of digits in even positions.
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    \(7+1+9=17\)
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  2. Finding their difference we get:
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    \(17-12 = 5\)
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Since the difference is \(5\text{,}\) then \(375\,149\) is not divisible by \(11.\)
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Checkpoint 1.2.19.

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Subsection 1.2.10 Expressing composite numbers as a product of prime factors

Composite numbers are numbers that has more than two factors,meaning it can be divided evenly by numbers other than \(1\) and itself.
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Process of breaking a composite number is prime factorization and prime factorization expresses the number as the product of prime numbers.
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Subsubsection 1.2.10.1 Factors of composite numbers

Factors are numbers that can divide a given number without leaving a remainder. Factors are also called divisors.
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Tumaini school is planning a sports day and Kamau needs to organize students into equal-sized teams. There are \(24\) students in his class. Kamau needs factors of \(24\) to divide the students into equal teams so that no one is left out.
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Since the factors of \(24\) are \(1,2,3,4,6,8,12\) and \(24\text{.}\) It gives Kamau possible ways to divide the students into equally.
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Example 1.2.20.
List all the factors of each of the following numbers.
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  1. \(\displaystyle 40\)
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  2. \(\displaystyle 84\)
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Solution.
  1. \(40\)
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    To determine the factors of \(40\text{,}\) we need to identify all numbers between \(1\) and \(40\) that divide \(40\) exactly, leaving no remainder.
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    We check the divisibility of each number step by step:
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    • \(1\) divides \(40\) exactly: \(40 \div 1 = 40\text{.}\) Therefore, \(1\) is a factor of \(40\text{.}\)
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    • \(2\) divides \(40\) exactly: \(40 \div 2 = 20\text{.}\) Therefore, \(2\) is a factor of \(40\text{.}\)
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    • \(3\) does not divide \(40\) exactly: \(40 \div 3 = 13\) with a remainder of \(1\text{.}\) Therefore, \(3\) is not a factor of \(40\text{.}\)
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    • \(4\) divides \(40\) exactly: \(40 \div 4 = 10\text{.}\) Therefore, \(4\) is a factor of \(40\text{.}\)
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    We continue this process for other numbers to check if they divide \(40\) without leaving a remainder.
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    After completing the checks, we find that \(1, 2, 4, 5, 8, 10, 20,\) and \(40\) all divide \(40\) exactly.
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    Therefore, the factors of \(40\) are \(1, 2, 4, 5, 8, 10, 20,\) and \(40\text{.}\)
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  2. \(84\)
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    Another way to find the factors of a number is by identifying pairs of numbers that, when multiplied together, produce that number. If \(a \times b = c\text{,}\) then both \(a\) and \(b\) are factors of \(c\text{.}\)
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    To determine the factors of \(84\text{,}\) we find all pairs of numbers whose product equals \(84\text{:}\)
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    • \(\displaystyle 1 \times 84 = 84\)
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    • \(\displaystyle 2 \times 42 = 84\)
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    • \(\displaystyle 3 \times 28 = 84\)
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    • \(\displaystyle 4 \times 21 = 84\)
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    • \(\displaystyle 6 \times 14 = 84\)
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    • \(\displaystyle 7 \times 12 = 84\)
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    Since we have identified all factor pairs of \(84\text{,}\) we can list its factors as follows:
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    The factors of \(84\) are \(1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42,\) and \(84\text{.}\)
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Checkpoint 1.2.21.
Exercise for Factors goes here.
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Subsubsection 1.2.10.2 Prime factors

Let’s start by defining what are prime numbers. Prime numbers are positive whole numbers that are greater than 1, that cannot be formed by multiplying two smaller numbers. Examples of prime numbers are \(2,3,5,7,11,...\)
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A prime factor is a prime number that can divide a given number without a remainder.
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Example in real life:
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Okuse, an architect want to design a strong and beautiful classroom. To make sure the classroom is stable, Okuse use blocks of the strongest materials. Now, think of prime numbers as the strongest building block in mathematics such that if we want to know what the number \(60\) is made of we break it down into it’s prime factors.
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Start with \(60 = 2 \times 30\)
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\(30 = 2 \times 15\)
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\(15 = 3 \times 5\)
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Therefore, prime factors of \(60\) are \(2 \times 2 \times 3 \times 5\)
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Activity 1.2.11.
(a)
Observe the groupings carefully and determine the total number of groupings as well as the number of circles or dots in each group.
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(b)
Click on the blank boxes provided. Type in the correct factors that multiply to give the number shown. Ensure at least one of the numbers entered is a prime factor.
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(c)
Once you have entered the factors, click the CHECK button to verify your response.
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If your factors are correct, a confirmation message will appear or another prompt will appear for continuation if you haven’t exhausted all the prime factors.
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If your factors are incorrect, reanalyze the groupings and enter the new factors again.
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(d)
What did you notice about the clusters of dots?
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How did you use the dots to help find the factors?
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(e)
After completing successfully in finding the factors of the given number , click the NEW PROBLEM button to attempt another visualization.
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Example 1.2.23.
Express \(48\) as a product of its prime factors.
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Solution.
To break down \(48\) into its prime factors, we need to find which prime numbers multiply together to give us \(48\text{.}\)
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Remember: Always start with the smallest prime number (which is \(2\)). If the number isn’t divisible by \(2\text{,}\) move to the next prime number (\(3\text{,}\) then \(5\text{,}\) and so on) until you can’t divide further.
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We start with the smallest prime number, which is \(2\text{.}\) Let’s divide \(48\) by \(2\text{:}\) \(48 \div 2 = 24\text{.}\) Since it divides exactly, \(2\) is our first prime factor.
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Next, we take \(24\) and divide it by \(2\) again: \(24 \div 2 = 12\text{.}\) It divides exactly, so \(2\) is our second prime factor.
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Now, we take \(12\) and divide it by \(2\) once more: \(12 \div 2 = 6\text{.}\) It divides exactly, so \(2\) is our third prime factor.
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Next, we take \(6\) and divide it by \(2\) again: \(6 \div 2 = 3\text{.}\) It divides exactly, so \(2\) is our fourth prime factor.
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Finally, we have \(3\text{,}\) which is already a prime number and cannot be divided further.
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So, the prime factors of \(48\) are: \(48 = 2 \times 2 \times 2 \times 2 \times 3\text{.}\)
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\({\color{blue}2}\) \(48\)
\({\color{blue}2}\) 24
\({\color{blue}2}\) 12
\({\color{blue}2}\) 6
\({\color{blue}2}\) 3
\({\color{blue}3}\) 1
\(48 = 2 \times 2 \times 2 \times 2 \times 3\)
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Example 1.2.24.
Express \(88\) as a product of its prime factors.
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Solution.
To find the prime factors of \(88\text{,}\) we will use a factor tree. A factor tree helps us break down a composite number into its prime factors step by step.
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Here’s how the factor tree for \(88\) looks:
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We start by breaking \(88\) into \(2\) and \(44\text{.}\) Then, \(44\) is broken into \(2\) and \(22\text{,}\) and finally, \(22\) is broken into \(2\) and \(11\text{.}\) Since \(11\) is a prime number, we stop here.
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We can’t factor any more, so we have found the prime factors.
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Therefore, the product of prime factors of \(88\) is : \(88 = 2 \times 2 \times 2 \times 11 \)
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Subsection 1.2.11 Greatest Common Divisor

GCD is the largest positive integer number that divides evenly into two or more numbers with zero remainder. It is also known as the Highest Common Factor (HFC).
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Example in real life:
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Esther and Shillah have packets of snacks and want to share them equally amoung the top mathematics performers in grade \(7\) without leaving leftovers. Esther have \(36\) crackers and Shillah has \(54\) cookies.
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The largest number of top performers Shillah and Esther can gift is solved using GCD.
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Since the factors of \(36\) are \(1,2,3,4,6,9,12,18,36\) and the factors of \(54\) are \(1,2,3,6,9,18,27,54\text{.}\) This implies that the higher number of top performers to be gifted is \(18\text{.}\)
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Activity 1.2.12.

(a)
Form groups of five learners , count and record the total number of girls and boys in your class.
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(b)
List all the factors of each number.
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(c)
Identify and circle the factors that are common to all the numbers.
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(d)
Determine the Greatest Common Factor (GCD) by selecting the largest common factor.
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(e)
Present your group’s findings to the rest of the class.
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Example 1.2.25.

Find the GCD of \(12,48 \text{ and } 30.\)
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Solution.
To find the GCF by factoring, list out all of the factors of each number
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The divisors of \(12\) are \({\color{red}1}, {\color{red}2}, {\color{red}3}, 4, {\color{red}6}, 12\text{.}\)
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The divisors of \(48\) are \({\color{red}1}, {\color{red}2}, {\color{red}3}, 4, {\color{red}6}, 8, 12, 16, 24, 48\text{.}\)
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The divisors of \(30\) are \({\color{red}1}, {\color{red}2}, {\color{red}3}, 5, {\color{red}6}, 10, 15, 30\text{.}\)
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The common factors of \(12 , 48 \text{ and } 30\) are \(1,2,3,6\text{.}\)
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The greatest common factor of \(12 , 48 \text{ and } 30\) is \(6\text{.}\)
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Example 1.2.26.

Three sticks of lengths \(30\) cm, \(45\) cm and \(60\) cm are cut into smaller pieces of the length. Find the greatest length which can be cut without leaving a remainder.
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Solution.
To determine the greatest length which can be cut without leaving a remainder, we need to find the Greatest Common Divisor (GCD) of the three numbers.
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Figure 1.2.27. prime factors for 30
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Figure 1.2.28. prime factors for 45
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Figure 1.2.29. prime factors for 60
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The prime factorization of \(30\) is \(2 \times {\color{blue}3} \times {\color{red}5}\text{.}\)
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The prime factorization of \(45\) is \({\color{blue}3} \times 3 \times {\color{red}5}\text{.}\)
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The prime factorization of \(60\) is \(2 \times 2 \times {\color{blue}3} \times {\color{red}5}\text{.}\)
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The common prime factors of \(30\text{,}\) \(45\) and \(60\) are \({\color{blue}3}\) and \({\color{red}5}\text{.}\)
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Thus, the greatest common factor of \(30\text{,}\) \(45\) and \(60\) is \(3 \times 5 = 15\text{.}\)
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The greatest length which can be cut from each stick without leaving a remainder is \(15\) cm.
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Checkpoint 1.2.30.

Exercise for GCD goes here.
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Subsection 1.2.12 Least Common Multiple

The Least Common Multiple of two or more numbers is the smallest number that is divisible by two or more numbers without a reminder. The Least Common Multiple (LCM) is also referred to as the Lowest Common Multiple (LCM) and Least Common Divisor (LCD).
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Example:
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Murunga and Onyimbo have busy schedules but want to meet regularly for a football game. Murunga have free time every \(12\) days and Onyimbo has free time every \(18\) days. To find when they will both be free on the same day to play, we solve using LCM.
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Therefore, we list the multiples of Murunga’s free days: \(12,24,36,48,60,72,...\) and Onyimbo’s free days: \(18,36,54,72,90,...\)
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The common multiples are \(36,72\) and the smallest commom multiple is \(36\text{.}\)
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Hence Muranga and Onyimbo will both be free on the \(36\) th day to play football game.
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Activity 1.2.13.

(a)
Click on the first dropdown next to Multiples of and scroll to select the any number.
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After selecting a number, - and + buttons will appear. Click the + button to add multiples of the selected number. To remove a listed multiple, click the - button.
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Keep clicking the + button until you have added at least \(10\) multiples for the chosen number.
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(b)
Click on the second dropdown next to Multiples of and scroll to select a different number from the one you selected on the first dropdown.
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After selecting a number, - and + buttons will appear. Click the + button to add multiples of the selected number. To remove a listed multiple, click the - button.
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Keep clicking the + button until you have added at least \(10\) multiples for the chosen number.
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(c)
Select the checkbox next to Common Multiples.
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If none of the listed multiples are highlighted in blue, keep clicking the + button to generate more multiples.
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(d)
Select the checkbox next to Least Common Multiple.
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In the top right corner, click the circular arrow icon to reset the activity. Then, select new numbers from the dropdowns and repeat the process.
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(e)
Write down what you observed and share with other learner in class.
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Figure 1.2.31.
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Example 1.2.32.

Find the LCM of \(5 \text{ and } 7\)
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Solution.
To find the LCM of \(5 \text{ and } 7\text{,}\) list the multiples of each number until at least one of the multiples appears on all lists
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Multiples of \(5:\) are \(5, 10, 15, 20, 25, 30, {\color{blue}35}, 40, 45,50,55,60,65,{\color{blue}70},75,...\)
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Multiples of \(7:\) are \(7, 14, 21, 28, {\color{blue}35}, 42, 49,56,63,{\color{blue}70},77,84,91,98,...\)
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The common multiples are \({\color{blue}35},{\color{blue}70},...\)
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The least common multiple is \({\color{blue}35}\)
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Therefore, LCM of \(5\) and \(7\) is \(35\text{.}\)
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Example 1.2.33.

Grade 7 pupils was preparing for a party and used a ribbon to decorate their classroom. What is the shortest possible length of a ribbon that can be cut into equal pieces of \(10\) cm, \(12\) cm or \(18\) cm without a remainder?
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Solution.
To determine the shortest possible length of the ribbon that can be cut into equal pieces of \(10\) cm, \(12\) cm, and \(18\) cm without a remainder, we need to find the Least Common Multiple (LCM) of these three numbers.
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  • Write down your numbers in a top table row
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  • Starting with the lowest prime numbers, divide the row of numbers by a prime number that is evenly divisible into at least one of your numbers and bring down the result into the next table row.
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  • If any number in the row is not evenly divisible just bring down that number.
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  • Continue dividing rows by prime numbers that divide evenly into at least one number
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  • When the last row of results is all \(1\)’s you are done.
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\({\color{blue}2}\) \({\color{red}10}\) \({\color{red}12}\) \({\color{red}18}\)
\({\color{blue}2}\) \(5\) \(6\) \(9\)
\({\color{blue}3}\) \(5\) \(3\) \(9\)
\({\color{blue}3}\) \(5\) \(1\) \(3\)
\({\color{blue}5}\) \(5\) \(1\) \(1\)
\(1\) \(1\) \(1\)
Find the product of the prime numbers in the first column to get the LCM:
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\({\color{blue}2} \times {\color{blue}2} \times {\color{blue}3} \times {\color{blue}3} \times {\color{blue}5} = 180\)
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The LCM of \(10, 12 \text{ and } 18\) is \(180.\)
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Therefore, the shortest possible length of the ribbon should be \(180\) cm long.
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Checkpoint 1.2.34.

Exercise for LCM goes here
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