Sum of fractions.

Question

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?

Answer 1

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?