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#### sweatingbear

##### Member

- May 3, 2013

- 91

**Q:**Find a fraction between \(\displaystyle \frac {96}{35}\) and \(\displaystyle \frac {97}{36}\) that has the smallest possible denominator.

**My thoughts:**Let's assume this fraction is of the form \(\displaystyle \frac pq\) where \(\displaystyle \gcd(p,q) = 1\). Since \(\displaystyle \frac{96}{35} > \frac{97}{36}\), we can write

\(\displaystyle \frac{97}{36} < \frac {p}{q} < \frac{96}{35} \, .\)

By multiplying by an appropriate factor in each given fraction respectively and thus obtaining common denominators, we equivalently have

\(\displaystyle \frac{3395}{1260} < \frac {p}{q} < \frac{3456}{1260} \, .\)

In total, there exist 60 different fractions in the interval above which \(\displaystyle \frac {p}{q}\) can equal. Therefore, intuitively, if we can find a \(\displaystyle p\) between \(\displaystyle 3395\) and \(\displaystyle 3456\) which cancels as many prime factors in \(\displaystyle 1260\), we will be able to determine the fraction with the smallest denominator.

This is where I am stuck; surely one could resort to a brute-force method, but I am hoping to solve it in more algebraic terms. Anyone got a clue?