$10$ Distinct Integers from a set and their sum equals to $954$ $10$ distinct integers from the set $ \left \{1;2;...;100  \right \} $ are chosen such that their sum is $954$. What is the smallest of the $10$ integers?
How do I start this question? I have no idea what to do unless by using brute force.
 A: The sum $100+99+\dotsb+91$ is $955$ so the $10$ integers can be only $100,99,\dotsc,92,90$.
A: Try starting with the sum $100+99+98+...$.
A: Let's call the integers $a_1 < \ldots < a_{10}$. We now $1 \le a_i \le 100$ and $\sum_i a_i = 954$. That is the mean of the integers is $\frac 1{10} \sum_{i=1}^{10} a_i = 95.4$. As the integers are distinct and smaller than $100$, choosing only for of the $a_i$ larger than the mean will not work, as 
$$ 100 + 99 + 98 + 97 + 95 + 94 + 93+ 92 + 91 + 90 = 949 < 954 $$
so $5$ of the $a_i$ must be larger then $95.4$. Hence $a_{10} =100, \ldots, a_6 = 95$. The largest possible sum now is
$$ 100 + 99 + 98 + 97 + 96+ 95 + 94 + 93+ 92 + 91  = 955 $$
Hence, we must reduce one $a_i$ by 1, but there is only one possibility, as the $a_i$ have to be distinct. Hence $a_1 = 90$.  
A: You can use an extension of Goldbach Conjecture to reach your sum. 
Every number between $1$ to $1000$ can be represented as a sum of distinct '$n$' primes > 2 and less than 100. That is the most easiest and the shortest way of reaching $954$, because we have around 30 of them between 1 and 100.
