Here is a *very concrete* question that can be explained by appealing to the $p$-adic continuity of addition and multiplication. We don't even need completions: the problem takes place entirely in the rational numbers and is not a fake problem in any sense.

We can form binomial coefficients $\binom{r}{n}$ when $r$ is not necessarily an integer, and this is important because they occur in the coefficients of the power series for $(1+x)^r$ in calculus. Their formula, for $n \geq 1$, is
$$
\binom{r}{n} = \frac{r(r-1)\cdots(r-n+1)}{n!}.
$$
If you look at the expansion for $\sqrt{1+x}$ and for $\sqrt[3]{1+x}$, corresponding to $r = 1/2$ and $r = 1/3$, the series start off as
$$
1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 + \cdots
$$
and
$$
1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 - \frac{10}{243}x^4 + \frac{22}{729}x^5 + \cdots
$$
The surprise is that the denominators are entirely powers of 2 in the first case and 3 in the second case. Think about that: $\binom{1/3}{5}$ involves division by $5!$, but the 2 and 5 factors cancel out. As a more extreme example, $\binom{-3/22}{7} = -\frac{1071892575}{39909726208}$ and $39909726208 = 2^{11}11^7$. Even though the definition of $\binom{-3/22}{7}$ involves division by $7!$, the primes that survive in the denominator seem to have nothing to do with $7!$ and everything to do with the denominator of $-3/22$.

Claim: For $n \geq 1$ and nonzero rational $r$, if a prime $p$ is in the denominator of $\binom{r}{n}$ then $p$ is in the denominator of $r$.

Proof: We show the contrapositive. If $p$ is not in the denominator of $r$ then $|r|_p \leq 1$, so the denominator of $r$ is invertible modulo any power of $p$, and therefore $r$ is a $p$-adic limit of positive integers, say $r = \lim_{k \rightarrow \infty} a_k$ with $a_k \in {\mathbf Z}^+$. That is a $p$-adic limit. By $p$-adic continuity of addition and multiplication (and division), we get $\binom{r}{n} = \lim_{k \rightarrow \infty} \binom{a_k}{n}$, another $p$-adic limit. By combinatorics we know $\binom{a_k}{n}$ is a positive integer, so
$|\binom{a_k}{n}|_p \leq 1$. The $p$-adic absolute value on ${\mathbf Q}$ is $p$-adically continuous, so $|\binom{r}{n}|_p = \lim_{k \rightarrow \infty} |\binom{a_k}{n}|_p \leq 1$. Thus $p$ is not in the denominator of $\binom{r}{n}$. QED

The special case $r = 1/2$ can be explained in terms of Catalan numbers: $\binom{1/2}{n} = (-1)^{n-1}C_{n-1}/2^{2n-1}$, where $C_{n-1}$ is the $(n-1)$th Catalan number (a positive integer). Therefore the denominator of $\binom{1/2}{n}$ is a power of 2. For the general case, I know of no argument that explains why primes in the denominator of $\binom{r}{n}$ must be primes in the denominator of $r$ in such a clean way as this $p$-adic method.

The converse of the claim is also true: for $n \geq 1$ and nonzero rational $r$, if a prime $p$ is in the denominator of $r$ then $p$ is in the denominator of $\binom{r}{n}$. That is, if $|r|_p > 1$ then $|\binom{r}{n}|_p > 1$. More precisely, if $|r|_p > 1$ then $|\binom{r}{n}|_p \geq |r|_p^n$, so in fact $|\binom{r}{n}|_p \rightarrow \infty$ as $n \rightarrow \infty$. Let's leave that as an exercise. (The data for the coefficients of $\sqrt{1+x}$ and $\sqrt[3]{1+x}$ suggest that perhaps the sequence $|\binom{r}{n}|_p$ is monotonically increasing if $|r|_p > 1$, and that too can be proved in general by looking at the $p$-adic absolute value of the ratio $\binom{r}{n+1}/\binom{r}{n}$.) In particular, for $n \geq 1$ the denominator of $\binom{1/2}{n}$ is a power of $2$ other than $1$.