$1 \cdot 27$ Find the smallest prime $p$ that cannot be expressed as $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers.
[Solution] It has been verified that 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 can all be written in the form $\left|3^{a}-2^{b}\right|$, where $a, b$ are non-negative integers:
$$
\begin{array}{l}
2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, \\
7=2^{3}-3^{0}, 11=3^{3}-2^{4}, 13=2^{4}-3^{1}, \\
17=3^{4}-2^{6}, 19=3^{3}-2^{3}, 23=3^{3}-2^{2}, \\
29=2^{5}-3^{1}, 31=2^{5}-3^{0}, 37=2^{6}-3^{3} .
\end{array}
$$

We conjecture that 41 is the smallest prime that cannot be expressed in this form. To verify this conjecture, we use proof by contradiction.
If the equation
$$
2^{u}-3^{v}=41
$$

has non-negative integer solutions $(u, v)$, then
$$
2^{u}>41 \text {, }
$$

which implies $u \geqslant 6$.
Therefore, $-3^{v} \equiv 1(\bmod 8)$
However, the possible remainders of $3^{v}$ modulo 8 are only 1 or 3, which contradicts the above equation. Hence, equation (1) has no non-negative integer solutions.
If the equation
$$
3^{x}-2^{y}=41
$$

has non-negative integer solutions $(x, y)$, then
$$
3^{x}>41 \text {, }
$$

which implies $x \geqslant 4$.
Therefore, $2^{y} \equiv 1 \quad(\bmod 3)$
Thus, $y$ must be even. Let $y=2 t$, then (2) becomes
$$
3^{x}-4^{t}=41,
$$

Thus, we have
$$
3^{x} \equiv 1 \quad(\bmod 4),
$$

which implies $x$ is also even. Let $x=2 s$, then (2) becomes
$$
\begin{array}{l}
3^{2 s}-2^{2 t}=41 \\
\left(3^{s}+2^{t}\right)\left(3^{s}-2^{t}\right)=41
\end{array}
$$

Since 41 is a prime, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
3^{s}+2^{t}=41 \\
3^{s}-2^{t}=1
\end{array}\right. \\
\left\{\begin{array}{l}
3^{s}=21 \\
2^{t}=20
\end{array}\right.
\end{array}
$$

This is impossible. Hence, equation (2) has no non-negative integer solutions.
In conclusion, 41 cannot be expressed in the form $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers.
Therefore, the smallest prime $p$ is 41.