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From observations made in the process of multiplication, we have seen that
$(\text{factor})\cdot \text{}(\text{factor})=\text{product}$
We show this in the following examples:
3 is a factor of 27, since $\text{27}\xf73=9$ , or $3\cdot 9=\text{27}$ .
7 is a factor of 56, since $\text{56}\xf77=8$ , or $7\cdot 8=\text{56}$ .
4 is not a factor of 10, since $\text{10}\xf74=\mathrm{2R2}$ . (There is a remainder.)
We can use the tests for divisibility from [link] to determine all the factors of a whole number.
Find all the factors of 24.
$\begin{array}{ccc}\text{Try 1:}\hfill & \text{24}\xf71=\text{24}\hfill & \text{1 and 24 are factors}\hfill \\ \text{Try 2:}\hfill & \text{24 is even, so 24 is divisible by 2.}\hfill & \\ & \text{24}\xf72=\text{12}\hfill & \text{2 and 12 are factors}\hfill \\ \text{Try 3:}\hfill & 2+4=6\phantom{\rule{8px}{0ex}}\text{and 6 is divisible by 3, so 24 is divisible by 3.}\hfill & \\ & \text{24}\xf73=8\hfill & \text{3 and 8 are factors}\hfill \\ \text{Try 4:}\hfill & \text{24}\xf74=6\hfill & \text{4 and 6 are factors}\hfill \\ \text{Try 5:}\hfill & \text{24}\xf75=\mathrm{4R4}\hfill & 5\text{is not a factor.}\hfill \end{array}$
The next number to try is 6, but we already have that 6 is a factor. Once we come upon a factor that we already have discovered, we can stop.
All the whole number factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
Find all the factors of each of the following numbers.
Notice that the only factors of 7 are 1 and 7 itself, and that the only factors of 3 are 1 and 3 itself. However, the number 8 has the factors 1, 2, 4, and 8, and the number 10 has the factors 1, 2, 5, and 10. Thus, we can see that a whole number can have only two factors (itself and 1) and another whole number can have several factors.
We can use this observation to make a useful classification for whole numbers: prime numbers and composite numbers.
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