The sum of eight positive even integers is $50$. If no integer can appear more than twice in the set, what is the greatest possible value of one of the integers?
This was a question I encountered on an online SAT test, I got stuck on it because if I tried the least even integer and moved $7$ more I'd end up with a value more than $50$. For example $2,4,6,8,10,12,14,16$ would have a sum of $72$ not $50$,
Smallest sum is $$2+2+4+4+6+6+8+8=40\\$$ Since you need one element to be the largest rest should be the smallest so we change the biggest element $8$ to $18$(since everything else is the smallest) which than is $$2+2+4+4+6+6+8+18$$
You said numbers could be repeated (but only once), so there could be two $2$'s, two $4$'s etc.
7 minimal even numbers are $2,2,4,4,6,6,8$ give sum of $32$. So the 8th will be $18$
You have a sum
$$x_1+x_2+...+x_8 = 50$$
with $x_i$ an even natural number and no three $x_i$ equal to each other, for all $i$.
Now, take $x_1$ as the greatest possible integer in that sum. Since $x_1$ is maximal, it also means that $x_2+x_3+...+x_8$ is minimal.
It's quite easy to find the latter, and then you can find $x_1$.
One number can be as large as possible if the others are chosen as small as possible. 2 + 2 + 4 + 4 + 6 + 6 + 8 + x = 50.
Evaluate the sum of $2 + 2 + 4 + 4 + \dots$ until the sum is an even number away from $50$.
That might get you to the correct answer.
