Find the value of $a_4-a_2$

Question

Let $a_1<a_2<a_3<a_4$ be positive integers such that $\sum_{i=1}^{4}\frac{1}{a_i}=\frac{11}{6}$. Find the value of $a_4-a_2.$

I do not know how to proceed. I have tried to simplify the summation but did not obtain anything useful. Please help.

Answer 1

First, we establish $a_1=1$. If $a_1\geq 2$, then $a_2\geq 3$, $a_3\geq 4$ and $a_4\geq 5$, giving $$ \frac{11}6=\frac1{a_1}+\frac1{a_2}+\frac1{a_3}+\frac1{a_4} \leq\frac12+\frac13+\frac14+\frac15=\frac{77}{60} $$

Can you see how to finish from here?

Answer 2

$\dfrac{11}{6}=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4} \nleq\dfrac12+\dfrac13+\dfrac14+\dfrac15=\dfrac{77}{60} \quad \Longrightarrow \quad a_1=1$

$\dfrac{5}{6}=\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4} \nleq\dfrac13+\dfrac14+\dfrac15=\dfrac{47}{60} \quad \Longrightarrow \quad a_2=2$

$\dfrac{1}{3}=\dfrac{1}{a_3}+\dfrac{1}{a_4} \quad \Longrightarrow \quad a_3 a_4=3(a_3+a_4)$

In positive integers solutions of last equation is only $(4,12),(12,4),(6,6)$.

$a_3<a_4 \quad \Longrightarrow \quad (a_3,a_4)=(4,12)$.

$a_4-a_2=10$