Question 214411
Recall that if 'k' is an integer, then {{{2k+1}}} is ALWAYS odd. Why? The number {{{2k}}} is ALWAYS even (because all even integers are a multiple of 2). If you add 1 to an even integer, you will get an odd integer. So this shows us that {{{2k+1}}} is ALWAYS odd where 'k' is an integer.



So the goal is to try to transform each choice into either the form {{{2k}}} (even form) or {{{2k+1}}} (odd form).



a)


{{{4a+2}}} Start with the given expression



{{{2(a+1)}}} Factor out the GCF 2



{{{2k}}} Replace 'a+1' with 'k' (which is an integer)



So {{{4a+2}}} is ALWAYS even (try plugging in some integers for 'a' to test this).


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b) 


We can rewrite {{{10a}}} as {{{2*5a}}} which can come to {{{2k}}} (where {{{k=5a}}}. So {{{10a}}} is ALWAYS even.


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c) 


We can rewrite {{{8a+1}}} as {{{2*4a+1}}} and then let {{{k=4a}}}. Now replace '4a' with 'k' to get {{{2k+1}}}. So {{{8a+1=2k+1}}} (where {{{k=4a}}})


This shows us that {{{8a+1}}} is ALWAYS odd.


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d) 


The value 0.83 is neither even nor odd. Why is this? A number can only be even or odd when it is a whole number or integer. Since 0.83 is NOT a whole number, it is neither even nor odd.




So to recap, the only odd integer of the group is {{{8a+1}}}