<s>[INST] Sum of fractions. Let $a_1 < a_2 <a_3 < a_4$ be positive integers such that $\sum_{i=1}^{4} \frac {1}{ a_i}$=$ 11/6$. Then $a_4-a_2=?$
I've tried to find the $a_i$'s but couldn't.
Is there a general way to solve such equations?
 [/INST] Note that if $2\leq a_1<a_2<a_3<a_4$ then 
$$\frac{77}{60}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\geq \sum_{i=1}^{4} \frac {1}{ a_i}=\frac{11}{6}$$
which is a contradiction. Hence $a_1=1$. 
Now we are left with $2\leq a_2<a_3<a_4$ such that
$$\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}=\frac{5}{6}.$$
Similarly if $3\leq a_2<a_3<a_4$ then
$$\frac{47}{60}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\geq \frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}=\frac{5}{6}.$$
which is a contradiction. Hence $a_2=2$. 
Then we have to solve
$$\frac{1}{a_3}+\frac{1}{a_4}=\frac{1}{3}$$
with $3\leq a_3<a_4$. With $a_3=3$ and $a_3\geq 6$ we have again a contradiction, so $a_3$ can be $4$ or $5$.
Can you take it from here?
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