Initially 35. 4 distinct prime numbers, none of whose last digit is 1, have a sum of squares equal to \(4pq\) (where \(p, q\) are primes, \(p \neq q\)). \(A=2^n pq\) (where \(n\) is an integer greater than 6) is a 6-digit number. The sum of the two 3-digit numbers formed by the first 3 digits and the last 3 digits of \(A\) equals \(10pq\), and the 3-digit number formed by the first 3 digits of \(A\) is of the form \(2r\) (where \(r\) is a prime). Find the product of these 4 prime numbers.
Let
$$
2^{\prime \prime} p q=10^{5} a+10^{4} b+10^{3} c+10^{2} d+10 e+f,
$$

where $a, b, \cdots, f$ are all integers, $1 \leqslant a \leqslant 9,0 \leqslant b, c, \cdots, f \leqslant$
9. Then (1) can be written as
$$
\begin{aligned}
2^{2} p q= & 199(1002+10 s+c)+100(a+d) \\
& +10(b+e)+c+f \\
= & 3^{3} \cdot 37 \cdot 2 r+10 p q .
\end{aligned}
$$

That is, $\left(2^{x-1}-5\right) p q=3^{3} \cdot 37 r$.
Given that $r>50$. Suppose $p6$ leads to a contradiction.

Therefore, (2) and (3) are both impossible.
Let $w, x, y, z$ be prime numbers, satisfying
$$
w^{2}+x^{2}+y^{2}+z^{2}=444 .
$$

Assume $w444, \therefore z444$, thus, only one of these 4 primes can be $\geqslant 13$. Therefore,
$$
\begin{array}{l}
w=3, x=5, y=7, \\
z^{2}=444-\left(3^{2}+5^{2}+7^{2}\right)=361, z=19 .
\end{array}
$$

Thus, the product of these 4 primes is $3 \times 5 \times 7 \times 19=1995$.
Note: It is easy to prove that when $p=3, q=37$, the $A, r, n$ that satisfy the conditions in the problem exist, i.e., $A=454656, r=227, n=12$.