Question

Which is correct regarding the statement: “If x is an odd integer, then the median of x, x + 2, x + 6, and x + 10 is an odd number”

Answer 1

x is odd
x+2 is also odd (eg: x = 3, so x+2 = 3+2 = 5)
x+6 is also odd for similar reasons above
x+10 is also odd

We have four values. The median is found in the middle, so exactly halfway between x+2 and x+6

Add up x+2 and x+6 to get
(x+2)+(x+6) = 2x+8

Then divide that result by 2. In other words, divide everything by 2
2x/2+8/2 = x+4
The result x+4 is odd since adding any even number to an odd number is going to result in an odd value

So the claim is true

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Alternative method:
x is odd so x = 2k+1 where k is any integer
x+2 = 2k+1+2 = 2k+3
x+6 = 2k+1+6 = 2k+7
x+10 = 2k+1+10 = 2k+11

Add up the two middle terms
(x+2)+(x+6) = (2k+3)+(2k+7) = 4k+10
Divide that result in half
4k+10 ---> 4k/2 + 10/2 = 2k+5 = 2k+4+1 = 2(k+2)+1

The expression 2(k+2)+1 fits the form 2m+1 where m = k+2 is an integer

So 2(k+2)+1 is odd making the median to be odd. 

This further helps support why the claim is true

-Ekansh Nimbalkar