If $1/a + 1/b +1/c +1/d = 1/2$ what is the maximum value of $a + b + c + d$ given that $a,b,c,d$ are positive integers. If $1/a + 1/b +1/c +1/d = 1/2$ what is the maximum value of $a + b + c + d$ given that $a,b,c,d$ are positive integers.
I have tried to solve this but kept on finding larger answers and most of the time fractions. I am not sure how to find the largest possible integer.
 A: My brute-force solution confirms $(a,b,c,d) = (3,7,43,1806)$ to be optimal (as suggested in the comments).
To make it a finite problem a computer could solve, assume $a \le b \le c \le d$. Then, we have lower and upper bounds on $a$:

*

*$a$ must be at least $3$. $\frac1a = \frac12$ would leave nothing for $b,c,d$.

*$a$ can be at most $8$. $\frac1a$ is the largest of four fractions that add up to $\frac12$, so it is at least their average, which is $\frac18$.

For each of these, we have a smaller problem to solve: solve $\frac1b + \frac1c + \frac1d = \frac12 - \frac1a$ for $b,c,d$. We can use the same strategy for the smaller problem.

One of the properties of Sylvester's sequence, which begins $2, 3, 7, 43, 1807, \dots$, is that the sum of the reciprocals of the first $n$ terms gets you as close as possible to $1$ with $n$ unit fractions. From this, it immediately follows that $(3, 7, 43, 1806)$ is the solution to $\frac1a + \frac1b + \frac1c + \frac1d = \frac12$ that maximizes $d$.
This should make us immediately suspect this sequence as a solution to maximizing $a+b+c+d$, but a bit more work would need to be done to make this a proof that does not require the brute force I did...
