Consider an ordered set of six consecutive integers in increasing order. To create a new set of six integers, the first, third, and fifth elements are each multiplied by two, and the second, fourth, and sixth elements are each increased by two. The median of this new set of four more than the median of the first set. What is the sum is the median of the first set and the median of the second set?
If the first number is $a$, then the six numbers are $a, a+1, a+2, a+3, a+4, a+5$. I am thinking that we should have two cases, the first number is odd and the first number is even. However, in those cases, I can't figure out the medians of each of the numbers, so that's where I need help.
The original sequence of numbers is $a,a+1,a+2,a+3,a+4,a+5$, where $a$ is a positive integer. Let $M_1$ be the median of the old sequence. Note that $M_1=a+2.5$.
The new sequence of numbers is $2a,a+3,2a+4,a+5,2a+8,a+7$. Let $M_2$ be the median of the new sequence.
Note that $2a$ will eventually be bigger than $a+7$, in which case the median will be the mean of $2a$ and $a+7$. Note that
$$2a\ge a+7\quad\text{iff}\quad a\ge7.$$
So if $a\ge7$, then $M_2$ is the mean of $a+7$ and $2a$. So in this case, we'd have that $M_2=1.5a+3.5$. Note that in this case we have that
$$M_2-M_1=0.5a+1\ge4.5.$$
It follows that we must have $a\le6$.
It is now straightforward to check that:
$$\begin{align*} \text{If }a=1,\text{ then }M_1=3.5&\text{ and }M_2=6.\\ \text{If }a=2,\text{ then }M_1=4.5&\text{ and }M_2=7.5.\\ \text{If }a=3,\text{ then }M_1=5.5&\text{ and }M_2=9.\\ \text{If }a=4,\text{ then }M_1=6.5&\text{ and }M_2=10.\\ \text{If }a=5,\text{ then }M_1=7.5&\text{ and }M_2=11.\\ \text{If }a=6,\text{ then }M_1=8.5&\text{ and }M_2=12.5. \end{align*}$$
It follows that $a=6$, $M_1=8.5$, $M_2=12.5$, and $M_1+M_2=21$.