Decay Chains & Counting

A nucleus ${}^{228}_{88}\text{Ra}$ decays through a chain of 5 alpha and 4 beta-minus steps. How many protons and neutrons does the final nucleus have?

Answer: $A$ falls by 8, and only alpha changes $A$, so there are $8 \div 4 = 2$ alpha decays. After 2 alpha, $Z$ = $84 - 4 = 80$. The final $Z$ is 83, so $Z$ rose by 3, which needs 3 beta-minus decays. So 2 alpha and 3 beta-minus.