If sum of seven distinct natural numbers is 100 How to prove that there exist at least one group of three numbers whose sum is 50 There are $7$ distinct natural numbers whose sum is $100$. From these 7 numbers 3 numbers can be selected in $C(7,3)=210$ ways How to prove that at least one of these groups will have sum at least  $50$ ?
I started in a complicated way.
I want to choose seven distinct natural numbers $x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7}$    
such that sum of any three numbers do not exceed 50.(And sum of these 7 numbers $=100$)
Then I can choose first two numbers $x_{1},x_{2}$ according to my wish with only one assumption -later we can re-think about this one assumption and include more-
Sum of these two numbers $x_{1}+x_{2}= a <50$ 
 from third number on wards the number should be $x_{3}<50-a$
$x_{4}<50-a$ 
$x_{5}<50-a $
$x_{6}<50-a$
$x_{7}<50-a$.
But $x_{4} $ have two more restrictions $x_{4} <50-(x_{1}+x_{3})$ and $x_{4} <50-(x_{2}+x_{3})$ 
But this argument will not lead to any where and more than that ,it will make the things complicated.So how to proceed ?
 A: Arrange the numbers in increasing order. If the largest three don't sum to more than $50$, then the biggest they can be is $15+17+18=50$. Then, the other $4$ numbers are at most $14+13+12+11=50$. But this shows that the choice of $7$ numbers $11, 12, 13, 14, 15, 17, 18$ add to $100$ and have no three of them summing to a number larger than $50$. So your statement is barely false, but it can be made true by asking for three numbers whose sum is at least $50$.
A: As others have noted, you cannot guarantee that there are three numbers whose sum is exactly $50$, but you can show that there are three whose sum is at least $50$, and I suspect that this is what was intended.
I find it easier to work with smaller numbers. The mean of the $7$ numbers is $\frac{100}7=14\frac27$. If we subtract $14$ from each of the $7$ numbers, we get $7$ distinct integers whose sum is $2$, and we’d like to show that the sum of the largest $3$ of them is at least $50-3\cdot14=8$. Suppose that the largest $3$ are $a<b<c$. Consider two cases.


*

*$a\ge 2$. This is easy.  

*$a\le 1$. What is the largest possible sum of the other $4$ numbers (the smallest $4$)? What does this tell you about $a+b+c$? Remember that the sum of all $7$ numbers is $2$.

