Examples and questions on how to find factors and multiples of whole numbers, with detailed solutions and explanations, for grade 6 students are presented.
Division and multiplication are related operations.
Divide 30 by all numbers starting from 1 and select the division that gives a remainder equal to 0 then write the factors.
We stop here because all factors greater than 5 have already been found. The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30.
Take a whole number and multiply it by 1, 2, 3, 4, ... to obtain the multiples as follows:
The results of the multiplication of 5 by 1, 2, 3, 4, ... which are 5, 10, 15, 20, ... are called the multiples of 5.
\(2 \div 1 = 2\) with remainder 0, or \(2 = 2 \times 1\). Hence the factors of 2 are 1 and 2.
\(6 \div 1 = 6\) with remainder 0, or \(6 = 6 \times 1\). Two factors 1 and 6.
\(6 \div 2 = 3\) with remainder 0, or \(6 = 3 \times 2\). Two factors 2 and 3.
All factors greater than 2 already found, we stop the division process and write the list of factors of 6: 1, 2, 3 and 6.
\(10 \div 1 = 10\) with remainder 0, or \(10 = 10 \times 1\). Two factors 1 and 10.
\(10 \div 2 = 5\) with remainder 0, or \(10 = 5 \times 2\). Two factors 2 and 5.
\(10 \div 3 = 3\) with remainder 1; 3 is not a factor of 10.
\(10 \div 4 = 2\) with remainder 2; 4 is not a factor 10.
All factors greater than 4 already found, we stop the division process and write the list of factors 10: 1, 2, 5 and 10.
\(24 \div 1 = 24\) with remainder 0, or \(24 = 24 \times 1\). Two factors 1 and 24.
\(24 \div 2 = 12\) with remainder 0, or \(24 = 12 \times 2\). Two factors 2 and 12.
\(24 \div 3 = 8\) with remainder 0, or \(24 = 8 \times 3\). Two factors 3 and 8.
\(24 \div 4 = 6\) with remainder 0, or \(24 = 6 \times 4\). Two factors 4 and 6.
\(24 \div 5 = 4\) with remainder 4; no factors.
All factors greater than 5 already found, we stop the division process and write the list of factors 24: 1, 2, 3, 4, 6, 8, 12 and 24.
We divide 120 by each of the numbers in the list. If the remainder is equal to zero, the divisor (number in the list) is a factor of 120.
In the given list all are factors of 120 except 11 and 50.
We first find all factors of 12:
All factors greater than 3 already found, we stop the division process and write the list of factors of 12: 1, 2, 3, 4, 6 and 12.
The factors of 24 were found in solution to 2) part d) and are given by: 1, 2, 3, 4, 6, 8, 12 and 24.
Common factors to 12 and 24 are: 1, 2, 3, 4, 6 and 12.
The first 5 multiples of a number are obtained by multiplying that number by 1, 2, 3, 4 and 5.
\(2 \times 1 = 2\) ; \(2 \times 2 = 4\) ; \(2 \times 3 = 6\) ; \(2 \times 4 = 8\) ; \(2 \times 5 = 10\)
\(11 \times 1 = 11\) ; \(11 \times 2 = 22\) ; \(11 \times 3 = 33\) ; \(11 \times 4 = 44\) ; \(11 \times 5 = 55\)
\(25 \times 1 = 25\) ; \(25 \times 2 = 50\) ; \(25 \times 3 = 75\) ; \(25 \times 4 = 100\) ; \(25 \times 5 = 125\)
The first 5 multiples of 6 and 8 are obtained by multiplying 6 and 8 by 1, 2, 3, 4 and 5.
\(6 \times 1 = 6\) ; \(6 \times 2 = 12\) ; \(6 \times 3 = 18\) ; \(6 \times 4 = 24\) ; \(6 \times 5 = 30\)
\(8 \times 1 = 8\) ; \(8 \times 2 = 16\) ; \(8 \times 3 = 24\) ; \(8 \times 4 = 32\) ; \(8 \times 5 = 40\)
24 is a common multiple to 6 and 8.