<s>[INST] $a_1Give that $a_1<a_2<a_3<a_4$ are positive integers such that $$\sum_{i=1}^4\frac{1}{a_i}=\frac{11}{6}$$, find the value of $a_4-a_2$.
My try:
Since $11$ is prime, atleast one of $a_1,a_2,a_3,a_4$ should be divisible by $6$.
Now we should express the fraction $\frac{11}{6}$ as $\frac{A+B+C+D}{E}$ such that $E$ is divisible by $A,B,C,D,E$.
I tried $$\frac{11}{6}=1+\frac{1}{2}+\frac{1}{3}$$ But fourth fraction is not possible.
Any hint?
 [/INST] The only choice of $a_1,a_2$ is $a_1=1,a_2=2$ with the reasoning given by Hagen von Eitzen.
So we have $$\frac{1}{a_3}+\frac{1}{a_4}=\frac{11}{6}-\frac{3}{2}=\frac{1}{3}$$
$\implies$
$$(3-a_3)(3-a_4)=9$$
The choice of $a_3=3$ is now ruled out.
So  $a_4>a_3>3$ and hence number $9$ can be factored out only in one way viz $-9 \times -1$ which gives $a_3=4,a_4=12$.
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