# For positive integer m,n $\frac{m}{n}$ start with 0.711. What is the minimum possible value of n?

For positive integer $m,n$ $\frac{m}{n}$ start with 0.711. What is the minimum possible value of $n$?
My Attempt
As $\frac{2}{3} < 0.711.. < \frac{3}{4}$ so fraction lie in between these values. If we multiply numerator and denominator by 10 , 100 and 1000 we can conclude $[\frac{10m}{n}]=7,[\frac{100m}{n}]=71, [\frac{1000m}{n}]=711$. So if $n|10, n|100 , n|1000$ then $n=5$ or $n=10$ as n must be greater than 2. From this point I am clueless. Can anyone help? Thanks in advance.

• Continued fractions ? – Xoff Aug 2 '17 at 8:27

I would use the Stern Brocot Tree. Descending in the tree and stopping as soon you are in the interval $[0.711,0.712[$ gives $\frac{32}{45}$.
• You can also try brute force. Compute $[0.711\cdot n]/n$ until you find something beginning with 0.711 (where $[ ]$ is the round function). You'll find the same result. – Xoff Aug 2 '17 at 8:49