Ordering of positive integers Three positive integers: m. n and q, their average is 10, median is m+6. what is the greatest possible value of Q ?
I understand how to find the solution which I am outlying below. My question is this: when can I and when can I not assume q is the maximum of the set?
The equation, median=m+6, implies among the positive integers that m is not the median and is smaller. However that leaves us with the situation the n or q could be the median. 
Solution: M+N+Q = 30
Median = M+6, but since we have 3 integers and the median is the middle integer, N=M+6
Now try to solve using the smallest possible numbers for M and N.
M=1
N=(1+6) = 7
Q=30-(1+7)=22
{1,7,22} where Q = 22.
 A: You know that the median is the middle number of the three, and you know that it’s $m+6$. Call it $x$ and the largest number $y$, without deciding yet which of them is $n$ and which is $q$. Then $m<m+6=x<y$, and $\frac13(m+x+y)=10$.
Suppose that we find values of $m,x$, and $y$ that satisfy these conditions. Then we can set $n=x$ and $q=y$, or we can set $n=y$ and $q=x$; both will give us values of $m,n$, and $q$ satisfying the stated conditions. Since we want to maximize $q$, we should clearly set $n=x$ and $q=y$: the other solution gives a smaller value of $q$, $x$ instead of the larger number $y$. This is the argument that justifies assuming that $q$ is the largest of the three numbers: if you have a solution in which it isn’t the largest, you can interchange $n$ and $q$ to get a solution in which it is larger.
For completeness I’ll finish off the solution the way I’d work the problem.
Now that we’ve established that, we know that we can rewrite the relationships in the first paragraph as $m<m+6=n<q$ and $\frac13(m+n+q)=10$, or $m+(m+6)+q=30$. That equation simplifies to $2m+q=24$. We want to make $q$ large, so we must make $m$ small. The smallest possible value of $m$ is $1$, which yields $q=22$, and we can go back and fill in $n=m+6=7$.
