Fractions everywhere

Question

$\begin{align} \underbrace{\dfrac23 + \dfrac23 + \cdots + \dfrac23}_{ \text{A copies of }\frac23 }&= \underbrace{\dfrac45 + \dfrac45 + \cdots + \dfrac45}_ {\text{B copies of }\frac45} \\\\\\ 20 \leq A + B &\leq 24\\\\ A + B &=\, ? \end{align}$

I have $A$ copies of $\frac23$ on the left-hand side of the equality, and $B$ copies of $\frac45$ on the right. In total, I've written down between $20$ and $24$ fractions. Exactly how many fractions are there?

Answer 1

For both fractions I made their denominator equal, so:

$\frac{2}{3} = \frac{10}{15} $ and $\frac{4}{5} = \frac{12}{15} $ .

The smallest multiple they have on common is $60$, so to get equality this means we need $6$ times $\frac{2}{3} $ and $5$ times $\frac{4}{5} $. Hence the answer must be a multiple of $11$, so the answer is $22$.

Answer 2

Well you have $$\frac23A=\frac45B\implies B=\frac56A$$ so $$A+B=\frac{11}6A$$ Since $A+B$ is an integer and is between $20$ and $24$, we must have that $$A+B=\frac{11}6(12)=22.$$

Answer 3

We have $$\frac{2}{3}A=\frac{4}{5}B,$$ which gives $$B=\frac{5A}{6},$$ which says that $A$ is divided by $6$.

In another hand, $$20\leq\frac{5}{6}A+A\leq24$$ or $$\frac{120}{11}\leq A\leq\frac{144}{11}$$ or $$10\frac{10}{11}\leq A\leq13\frac{1}{11},$$ which gives $$A=12,$$ $$B=10$$ and $$A+B=22.$$