• My numerator AND denominator are both prime numbers less than 10. I am more than 1/2. I am less than 1. I am nit equivalent to 8/12 or 10/14. -What number am I?

  • If you double me, I'm more than 1. If you take half of me, I'm less than 1/2. I am less than 4/5, but more than 7/12. I am not equivalent to 2/3. The difference between my numerator and my denominator is 1. Both are whole numbers. -What number am I?



As we have only primes, the "I am not equivalent to..." fractions can be reduced to receive $$\frac{p}{q} \notin \left\{ \frac{2}{3}, \frac{5}{7} \right\}$$ $$\frac{1}{2} < \frac{p}{q} < 1 \Rightarrow p < q < 2p$$ $p=2$ would imply $q=3$ which is excluded, $p=3 \Rightarrow q=5$ is possible, $p=5 \Rightarrow q=7$ is also impossible.

$$x = \frac{3}{5}$$


Here we get $$\frac{2p}{q} > 1, \frac{p}{2q} < \frac{1}{2} \Rightarrow \frac{p}{q} < 1$$ So $2p > q > p$ as above and thus $q = p+1$. $$x = \frac{p}{p+1}, p > 1$$ The monotony gives us $$\frac{7}{12} < \frac{3}{4} \leq x < \frac{4}{5}$$ And the answer turns out to be $$x = \frac{3}{4}$$


For the first: There are only four prime numbers less than $10$, namely $2, 3, 5, 7$. The condition that the fraction is between $1/2$ and $1$ restricts it quite a bit, and there are very few possibilities to test against the last condition.

For the second: The number is either of the form $\frac n {n + 1}$ or $\frac{n + 1} n$. It's not the second (why?). Now since $\frac n {n + 1}$ gets close to $1$ as $n$ grows, there are only finitely many numbers to check.


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