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.

  • 1
    $\begingroup$ Continued fractions ? $\endgroup$ – Xoff Aug 2 '17 at 8:27

I found 45.

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 $\frac{32}{45}$.

  • $\begingroup$ I think there is an easier and elementary method as this problem appeared in a school competition. $\endgroup$ – rugi Aug 2 '17 at 8:44
  • $\begingroup$ 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. $\endgroup$ – Xoff Aug 2 '17 at 8:49
  • $\begingroup$ However I don't think there is a more general and easy answer. That's mainly why this tree exists : to answer such a question and find the solution as quickly as possible. $\endgroup$ – Xoff Aug 2 '17 at 8:51

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