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# Write the set of even factors of $124$.  Verified
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Hint:$124$ is a composite number which is a positive integer that can be formed by multiplying two smaller positive integers.To write the set of even factors for a composite number and then we write all possible factors of 124 and finally we collect all even factors from it.

It is given that the number $124$, we find the set of even factors of $124$
Firstly, take the set of factors of $124 = \left\{ {1,2,4,31,6\left. {2,124} \right\}} \right.$
$\begin{gathered} 1 \times 124 = 124 \\ 2 \times 62 = 124 \\ 4 \times 31 = 124 \\ 62 \times 2 = 124 \\ 124 \times 1 = 124 \\ \end{gathered}$
Take the set $\left\{ {1,2,4,31,6\left. {2,124} \right\}} \right.$
The numbers $2,4,62,124$ are even
Therefore $\left\{ {2,4,62,124} \right\}$ are the set of even factors of $14$.

Additional Information:To check the above step is right or not, use prime factorization method, that is, for any number prime numbers multiply together to make the original number,
Here $124 = {2^2} \times 31$ are the prime numbers multiply together to make $124$
For any number, the prime factorization will be $= {2^p} \times {3^q} \times {5^r} \times {7^s}...$ where ${\mathbf{p}},{\mathbf{q}},{\mathbf{r}},{\mathbf{s}}$ are the powers (can be different powers)
To find the number of factors for any number,
$\left( {p + 1} \right) \times \left( {q + 1} \right) \times \left( {r + 1} \right) \times \left( {s + 1} \right)...$ then $\left( {2 + 1} \right)\left( {1 + 1} \right) = 6$,
There are totally 6 factors, which is equal to $\left\{ {1,2,4,31,62,124} \right\}$.
So, the number of even factors to the number $= p\;\left( {q + 1} \right)\;\left( {r + 1} \right)$
$= 2\left( {1 + 1} \right) = 4$,
There are totally $4$ even factors, that is $\left\{ {2,4,62,124} \right\}$.
Hence the set of all even factors of $124$ is $\left\{ {2,4,62,124\} } \right.$

Note:For any number, there will be many factor for the number, that may be odd or even factor depending on the number, but we can notice that,$\;1$ is factor for all numbers, and then$\;2\;$ is factor for all even numbers, example: $4 = {2^2}$, $6 = 2 \times 3$ and so on. So that for even numbers $1$ and $2$ is an essential factor for even numbers, at the same time, $2$ is the only essential factor for even numbers.