Problem: Let A, B, C, D, and E be positive integers such that $\frac{2}{3}A=\frac{3}{4}B=\frac{5}{6}C$ and $\frac{7}{8}B=\frac{9}{10}D=\frac{11}{12}E$. Find the minimum value of A, B, C, D, E.

Solution:

We know that:

$\frac{2}{3}A=\frac{3}{4}B=\frac{5}{6}C=k$ (say)

$\frac{7}{8}B=\frac{9}{10}D=\frac{11}{12}E=l$ (say)

$\therefore A = \frac{3k}{2} , B = \frac{4k}{3} , C = \frac{6k}{5} , D = \frac{40l}{21} , E = \frac{60l}{77}$

Let's find the value of k and l.

$\frac{2}{3}A=\frac{3}{4}B$

$\Rightarrow \frac{2}{3} \cdot \frac{3k}{2}=\frac{3}{4} \cdot \frac{4k}{3}$

$\Rightarrow k=\frac{3}{2}$

Similarly, $\frac{5}{6}C=\frac{3}{4}B$

$\Rightarrow \frac{5}{6} \cdot \frac{6k}{5}=\frac{3}{4} \cdot \frac{4k}{3}$

$\Rightarrow k=\frac{3}{2}$

Therefore, $k=\frac{3}{2}$.

Similarly, $\frac{7}{8}B=\frac{9}{10}D$

$\Rightarrow \frac{7}{8} \cdot \frac{4k}{3}=\frac{9}{10} \cdot \frac{40l}{21}$

$\Rightarrow l=\frac{21}{32}$

Similarly, $\frac{9}{10}D=\frac{11}{12}E$

$\Rightarrow \frac{9}{10} \cdot \frac{40l}{21}=\frac{11}{12} \cdot \frac{60l}{77}$

$\Rightarrow l=\frac{21}{32}$

Therefore, $l=\frac{21}{32}$.

Now, let's find the values of A, B, C, D, and E.

$A=\frac{3k}{2}=\frac{9}{4}$

$B=\frac{4k}{3}=2$

$C=\frac{6k}{5}= \frac{9}{5}$

$D=\frac{40l}{21}=\frac{40 \cdot 21}{21 \cdot 32}=\frac{10}{8}=\frac{5}{4}$

$E=\frac{60l}{77}=\frac{60 \cdot 21}{77 \cdot 32}=\frac{15}{16}$

Therefore, the minimum values of A, B, C, D, and E are $\frac{9}{4}$, 2, $\frac{9}{5}$, $\frac{5}{4}$, and