Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced $m$ in the fractions as $nk$. And after a bit of simple manipulation, I obtained $$k+\frac{1-k}{n+1},$$ $$k+\frac{2-k}{n+2},$$ $$k+\frac{3-k}{n+3},$$ $$k+\frac{4-k}{n+4},$$ $$k+\frac{5-k}{n+5}$$ Now I do not know how to proceed any further.

Fixed $n$ a solution is given by

$$m=n+n(n+1)(n+2)(n+3)(n+4)(n+5)=n+\frac{(n+5)!}{(n-1)!}$$

indeed
$$\frac{m}{n}=\frac{n+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n}\\=1+(n+1)(n+2)(n+3)(n+4)(n+5)$$
$$\frac{m+1}{n+1}=\frac{n+1+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+1}\\=1+n(n+2)(n+3)(n+4)(n+5)$$
$$\frac{m+2}{n+2}=\frac{n+2+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+2}\\=1+n(n+1)(n+3)(n+4)(n+5)$$
$$\frac{m+3}{n+3}=\frac{n+3+n(n+1)(n+3)(n+3)(n+4)(n+5)}{n+3}=\\1+n(n+1)(n+2)(n+4)(n+5)$$
$$\frac{m+4}{n+4}=\frac{n+4+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+4}\\=1+n(n+1)(n+2)(n+3)(n+5)$$
$$\frac{m+5}{n+5}=\frac{n+5+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+5}\\=1+n(n+1)(n+2)(n+3)(n+4)$$
