Problem (Algebra):
A lead box contains samples of two radioactive isotopes of iron. Isotope A decays so that after every 6 minutes, the number of atoms remaining is halved. Initially, there are twice as many atoms of isotope $\mathrm{A}$ as of isotope $\mathrm{B}$, and after 24 minutes there are the same number of atoms of each isotope. How long does it take the number of atoms of isotope B to halve?

Solution:
In 24 minutes, the number of atoms of isotope $\mathrm{A}$ has halved 4 times, so the initial number of atoms is $2^{4}=16$ times the number of atoms of isotope $\mathrm{A}$ at time 24 minutes.

But there were initially half as many atoms of isotope B as of isotope B, so there was 8 times the final number of atoms. Therefore, the number of atoms of isotope B halves 3 times in the 24 minutes, so it takes 8 minutes for the number of atoms of isotope B to halve.

Initially, there is twice as many atoms of isotope A as of isotope B, so let the original numbers of atoms of each be $2 x$ and $x$, respectively.

Considering isotope A, after 24 minutes, if it loses half of its atoms every 6 minutes, there will be $2 x\left(\frac{1}{2}\right)^{\frac{24}{6}}$ atoms remaining.

Similarly for isotope B, after 24 minutes, there will be $x\left(\frac{1}{2}\right)^{\frac{24}{T}}$ atoms remaining, where $T$ is the length of time (in minutes) that it takes for the number of atoms to halve.

From the given information,

$$
\begin{aligned}
2 x\left(\frac{1}{2}\right)^{\frac{24}{6}} & =x\left(\frac{1}{2}\right)^{\frac{24}{T}} \\
2\left(\frac{1}{2}\right)^{4} & =\left(\frac{1}{2}\right)^{\frac{24}{T}} \\
\left(\frac{1}{2}\right)^{3} & =\left(\frac{1}{2}\right)^{\frac{24}{T}} \\
\frac{24}{T} & =3 \\
T & =8
\end{aligned}
$$

Therefore, it takes 8 minutes for the number of atoms of isotope B to halve.

Final answer: 8 min