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.

If $m=30a$ and $n=30b$, then 
$${m\over n}={30a\over30b}\\{m+2\over n+2}={2(15a+2)\over2(15b+1)}\\{m+3\over n+3}={3(10a+1)\over3(10b+1)}\\{m+4\over n+4}={2(15a+2)\over2(15b+2)}\\\text{and}\\{m+5\over n+5}={5(6a+1)\over5(6b+1)}$$ are all reducible.  It remains to choose $a$ and $b$ so that 
$$30a+1\over30b+1$$
is reducible.  If we try $b=1$, we see we need $31\mid30a+1$, or $a\equiv1$ mod $31$, so $a=32$ does the trick. Thus $m/n=960/30$ has the desired property.
