Do there exist positive integers $a, b, c$ such that $1/a+1/b+1/c=47/48$? So far the only conclusion i have drawn is that $a+b+c$ is greater than 48. However, I cannot make anymore connections that would help me solve the problem. Am I just supposed to "brute force" this by bashing the algebra, or is there some logic I am missing?
 A: The hint.
Let $a\geq b\geq c$.
Hence, $$\frac{47}{48}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq\frac{3}{c},$$
which gives $$c\leq\frac{3\cdot48}{47}$$  and from here
$$c\leq3.$$
Also, it's obvious that $c\geq2$.
Now, prove that for $c=2$ and for $c=3$ it's impossible.
For example, let $c=3$.
Hence, $$\frac{31}{48}=\frac{1}{a}+\frac{1}{b}\leq\frac{2}{b},$$
which gives $b\leq3$ and since $b\geq c$, we obtain $b=3$, which is impossible. 
A: I will take a different path from Michael. Here we have $\nu_2(48)=4$, hence at least one number among $a,b,c$ has to be a multiple of $16$, and we may assume it is $c$ without loss of generality. We also have $\nu_3(48)=1$. If $c$ is a multiple of both $3$ and $16$ then $\frac{1}{a}+\frac{1}{b}$ is between $\frac{46}{48}$ and $\frac{47}{48}$, but $\frac{1}{a}+\frac{1}{b}$ is either $\geq 1$ or $\leq\frac{5}{6}$, so, no way. We may assume that $c$ is a multiple of $16$ and $b$ is a multiple of $3$. In such a case $\frac{1}{a}$ has to be between $\frac{28}{48}$ and $\frac{47}{48}$, but sadly there are no unit fractions in such interval.
