11. (6 points) As shown in the figure, the front of each of the three cards has a number, and their backs are written with prime numbers $m, n, p$. If the sum of the numbers on both sides of each card is equal, then the minimum value of $m+n+p$ is $\qquad$ .
【Solution】According to the problem, we have:
$$
47+m=53+n=71+p,
$$

then $m=71+p-47=24+p, n=71+p-53=18+p$,
substituting into the expression $m+n+p$ we get,
$$
\begin{array}{l}
m+n+p \\
=71+p-47+71+p-53+p \\
=42+3 p \\
p=2, 3, 5, 7 \cdots
\end{array}
$$

The even prime number 2 does not fit the problem and is discarded;
When $p=3$, $n=18+p=18+3=21, 21$ is not a prime number, discard;
When $p=5$, $n=18+p=18+5=23, m=24+5=29, 23, 29$ are both prime numbers, which meet the problem's requirements;
Therefore, the minimum value of $m+n+p$ is:
$$
\begin{array}{l}
m+n+p \\
=42+3 p \\
=42+3 \times 5 \\
=42+15 \\
=57 .
\end{array}
$$

The answer is: 57.