Problem in valuation theory Find $\alpha\in \mathbb{Q}$, such that $v_2(\alpha-1/3)\ge 2$, $v_3(\alpha-1/2)\ge 3$ and $|\alpha-1|_\infty<1/2$, where $v_p$ is the $p$-adic exponential valuation and $|\cdot|_\infty$ is the usual absolute value.
Thanks in advance!
 A: The first two conditions are equivalent to $v_2(3\alpha-1)\ge2$ and $v_3(2\alpha-1)\ge3$. That is,
$$3\alpha-1\equiv 0 \mod 2^2\Bbb Z \qquad and \qquad 2\alpha-1 \equiv 0 \mod 3^3\Bbb Z.$$
Rewriting each side (note $2^{-1}\equiv 14\mod 27$), we obtain
$$\begin{cases} \alpha\equiv -1 \mod 4 \\ \alpha \equiv  14 \mod 27. \end{cases}$$
By CRT we have $\alpha=95 \mod 108$. We need $|p/q-1|_\infty <1/2$ while $pq^{-1} \equiv 95 ~(108)$. The two nontrivial factors of $95$ are $5$ and $19$; we have $5^{-1}\equiv 65$ and $19^{-1}\equiv 91 \mod 108$. The latter is close to $5+108=113$ in the $|\cdot|_\infty$ metric, so we check that $|113/91-1|_\infty<1/2$ indeed holds.
This gives $\alpha=113/91$ as one solution.
A: Hint: If $\alpha = \frac{a}{b}$ then
$\alpha-\frac{1}{2} = \frac{2a-b}{2b}$
$\alpha-\frac{1}{3} = \frac{3a-b}{3b}$
You need to make the first fraction divisible by a high power of 3, and the second divisible by a high power of 2. It's easiest to make $b$ relatively prime to 6 so you don't have to worry about the denominators. Pick such a $b$. Can you find an $a$ such that $2a-b$ is divisible by 27? Can you find an $a$ that makes $3a-b$ divisible by 4? Can you find an $a$ that does both? Finally, can you make $a$ close enough to $b$ so that $\frac{a}{b}$ is not too far from 1?
