5. There are 6 natural numbers that have the same remainder when divided by 12, and their product is 971425. Then the minimum value of the sum of these 6 natural numbers is $\qquad$ .
5. 150.

Since 971425 leaves a remainder of 1 when divided by 12, and
$$
971425=5 \times 5 \times 7 \times 7 \times 13 \times 61,
$$

among the prime factors, there are two each that leave a remainder of 5, 7, and 1 when divided by 12. Since the product of two numbers that leave a remainder of 5 (or 7) when divided by 12 also leaves a remainder of 1, and 971425 multiplied by several 1s is still 971425, and 1 leaves a remainder of 1 when divided by 12, it can only be that 971425 is the product of 6 numbers that leave a remainder of 1 when divided by 12. Calculation shows:
$$
971425=1 \times 1 \times 1 \times 1 \times 1 \times 971425,
$$

the sum of these 6 factors is
$$
\begin{array}{l}
1+1+1+1+1+971425=971430 ; \\
971425=1 \times 1 \times 1 \times 1 \times 13 \times 74725,
\end{array}
$$

the sum of these 6 factors is
$$
\begin{array}{l}
1+1+1+1+13+74725=74742 ; \\
971425=1 \times 1 \times 1 \times 13 \times 25 \times 2989,
\end{array}
$$

the sum of these 6 factors is
$$
1+1+1+13+25+2989=3030 \text {. }
$$

In fact, let $a$ and $b$ be natural numbers greater than 1 that leave a remainder of 1 when divided by 12, and $a \geqslant b$, then $a \geqslant b>2$. It is easy to see that
$$
a b>a \times 2=a+a>a+b .
$$

According to equation (1), we get
$$
\begin{array}{l}
971425=13 \times 74725>13+74725 \\
=13+25 \times 2989>13+25+2989 \\
=13+25+49 \times 61>13+25+49+61 .
\end{array}
$$

Since $971425=5^{2} \times 7^{2} \times 13 \times 61=1 \times 1 \times 13 \times 25$ $\times 49 \times 61$, 971425 can be expressed as the product of 6 natural numbers that leave a remainder of 1 when divided by 12, and the smallest sum of these numbers is
$$
1+1+13+25+49+61=150 \text {. }
$$