$a_1<a_2<a_3<a_4$ such that $\sum_{i=1}^4\frac{1}{a_i}=\frac{11}{6}$

Question

Give 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?

Answer 1

Use the identity $\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$ with $n=3$

Answer 2

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 $a_4=12$.

Answer 3

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 $a_3=4,a_4=12$.