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I saw this problem in an Australian maths olympiad:

$6/10 < a/b < 10/15$

The problem asked for the lowest possible value of $b\in \mathbb{Z}$.

I tried manipulating but couldnt derive one of their answers.

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1 Answer 1

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That's how I did it. First of all, note that \begin{equation} \frac{11}{15} - \frac{7}{10}= \frac{1}{30} \end{equation} thus b=31 certainly works, i.e. is a higher bound to the answer.

Then notice that $b$ cannot be any divisor of 10 or 15, as $\frac{11}{15} < \frac{8}{10}$ and $\frac{10}{15}< \frac{7}{10}$.

It remains to check a few values. As $\frac{7}{10}=0.70$ you are looking for a fraction which is 'slightly more' than $0.70$. $b=4$ does not work as $\frac{3}{4} > \frac{11}{15}$ and $b=6$ does not work either because we know $\frac{4}{6} < 0.70$ and $\frac{5}{6}$ is clearly too big. Now check $b=7$ and you find $\frac{5}{7}$ is in the required range, so $b=7$ is your answer.

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oh i see you edited your 7 as a 6... the problem is even easier now :) –  Niccolò Aug 25 '12 at 8:28

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