[tex]72 \; \text{days}[/tex]
It takes a half life [tex]t_{1/2} = 24 \; \text{days}[/tex] for the mass of thorium in a sample of to decrease by [tex]1/2[/tex], two half lives [tex]2 \; t_{1/2} = 48 \; \text{days}[/tex] to decrease by yet another [tex]1/2[/tex]- which leaves thorium atoms of mass [tex]1/4[/tex] that of the initial mass in the sample. Another half life [tex]t_{1/2} = 24 \; \text{days}[/tex] would bring this ratio down to [tex]1/2 \times 1/4 = 1/8[/tex]. It happens that
[tex]1.25/ 10 \\= 1 /8[/tex]
Meaning that it would take precisely
[tex]3 \; t_{1/2} = 72 \; \text{days}[/tex]
for the mass of thorium in the [tex]10 \; \text{gram}[/tex]-sample to reach [tex]1.25 \; \text{g}[/tex].
Note, that it might take a calculator to find the time required in case no integer powers of [tex]1/2 = 0.5[/tex] ([tex]3[/tex] in this case) matches the desired ratio or percentage. An exponential decay formula (which naturally works for this question as well) is given below:
[tex]t = t_{1/2} \cdot \log_{2}{(m_0/m)}[/tex]
