Answer
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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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
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