Maximising largest integer on a list given mean, median and mode A list of $10$ positive integers has a mean of $11$, a median of $10$ and a unique mode of $7$. What is the largest possible value of an integer in the list?
In an ascending (albeit not strictly ascending) list of ten positive integers ($n_1$ to $n_{10}$), the last will be the highest, and you want to make the earlier ones as small as possible. The given conditions force $\sum_{i=1}^{10}n_i = 110$ and $n_5 + n_6 = 20$. In addition, at least two values have to be $7$.
With a bit of trial and error I deduced the optimal list is probably $1, 2, 7, 7, 9, 11, 12, 13, 14, 34$ with the answer being $34$. But I'm dissatisfied with this as I can't be sure it's the best solution and I'm wondering if there's a more systematic way of approaching the problem (short of an exhaustive search).
Thank you.
 A: We wish to minimize the sum of the other $9$ numbers $a_1\leq \dots \leq a_9$ with the condition $a_5+a_6 = 20$ and the condition that $7$ appears more times than the others, the sum we got is $76$, lets prove less is not possible.
If $a_5$ is $7$ then $a_6$ is $13$ and so $a_6+a_7+a_8+ a_9 \geq 4\times 13 = 52$, so the sum of $a_1+a_2+\dots+a_5$ would have to be less than  $24$. If $a_3,a_4,a_5= 7$ the minimum sum of $a_1+\dots + a_5$ is $1+1+7+7+7 = 23$ but since we can't have all of $a_6,\dots,a_{9} = 13$ this becomes impossible. Hence we must have $a_3\neq 7$ and the minimum is greedily $1+2+3+7+7+13+14+15+16 = 78$.
Hence $a_5> 7$. It can now be seen that if we select how many $7$'s we want then the solution can be formed greedily by taking the smallest numbers possible and setting $a_4 = 7$.
If there are two $7$'s the solution is $1,2,7,7,9,10,11,12,13=76$
If there are three $7$'s the solution is $1,7,7,7,10,10,11,11,12 = 76$
If there are four $7$'s the solution is $7,7,7,7,10,10,10,11,11 = 80$
