smallest possible sum of b and c 
In the set of positive integers {$2, 6, 8, 9, 11, b, c$}, the mean is larger than the unique mode, which is larger than the median. What is the smallest possible sum of $b$ and $c$?

So I just started plugging in a few values to try to see what works, but values close to 6, 8 and 9 don't seem to work, and I think there would be a much more efficient and mathematical way to see this... Any suggestions?
 A: mode (o)
Since there is an unique mode, there are three possibilites:
(i) $b=c\;\; (o = b = c)$
(ii) $b\neq c;\; b\in\{2,6,8,9,11\}$ and $c\notin\{2,6,8,9,11\}\;\; (o = b)$
(iii) $b\neq c;\;c\in\{2,6,8,9,11\}$ and $b\notin\{2,6,8,9,11\}\;\; (o = c)$
Since (ii) and (iii) are "the same", you can assume that the mode is $b$, and consider only (i) and (ii)
(Note that if $b\notin\{2,6,8,9,11\}$, then $c=b$)

mean (m)
$m = \frac{36+b+c}{7}$

median (d)
$b,c<6 \Rightarrow d=6$
$b,c>9 \Rightarrow d=9$

It was given that
$m>o>d$
Therefore,
$36+b+c>7\,b>7\,d$

Write the rest of possible cases for the median, keep in mind (i) and (ii) and you should be able to solve it!
(also: $b,c\in \mathbb{N^*}$)

I have got $\boxed{b=11}$ and $\boxed{c=31}$, so the smallest possible sum is $42$ (the answer to life, the universe and everything)
$\frac{78}{7} > 11 > 9$
A: First off, either $b = c$ or not. First, let's assume $b = c$. Then $b$ is the unique mode, so we must make sure that it is smaller than the mean and larger than the median. If $b \leq 9$, then the mode is not larger than the median. At the same time, the mean is equal to $\frac{2+6+8+9+11+2b}{7} = \frac{36 +  2b}{7}$ and we want this to be larger than $b$, which gives
$$
b < \frac{36+2b}{7}\\
7b<36+2b\\
5b<36\\
b<\frac{36}7<6
$$
which clearly contradicts what we found above.
So, $b$ and $c$ must be two different numbers. Then exactly one of them (say $b$) must be equal to one of the given numbers, because we want the mode to be unique. Again, we check what happens to the mode vs. the median first. The smallest we can possibly make the median is $6$, so clearly $b>6$. That leaves three possibilities:


*

*$b = 8$. Even in this case, the median is at least $b$, so it's not feasible

*$b = 9$. In this case, the median is less than $9$ iff $c$ is less than $9$. However, the mean is $\frac{2+6+8+9+9+11+c}{7} = \frac{45 + c}{7}$, and if we want this to be larger than $9$, then we get $c>18$, so this is not feasible

*$b = 11$. In this case, the mode is larger than the median no matter what happens, so we just have to choose as low a $c$ as we can to make $\frac{2+6+8+9+11+11+c}{7}<11$, which gives $c>30$.
Thus the lowest $c$ we can pick when picking the only $b$ we can pick is $31$, so $b+c$ is at least $42$.
