<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] If $a_2>2$, then
$$ \sum_{k=1}^4\frac1{a_k}\le1+\frac13+\frac14+\frac15=\frac{107}{60}<\frac{11}{6}.$$
Hence we must have $a_2\le 2$ and therefore $a_1=1$, $a_2=2$.
Solving for $a_4$ in terms of $a_3$, we now find
$$ a_4=\frac1{\frac{11}6-1-\frac12-\frac1{a_3}}=\frac{3a_3}{a_3-3}$$
which forbids $a_3=3$, is no integer for $a_3=5$, and is $\le \frac{3a_3}3=a_3$ for $a_3\ge 6$. Remains only $a_3=4$ and $a_4=12$.
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