20. Box $\mathrm{P}$ has $p$ chocolates and box $\mathrm{Q}$ has $q$ chocolates, where $p$ and $q$ are both odd and $p>q$. What is the smallest number of chocolates which would have to be moved from box $\mathrm{P}$ to box $\mathrm{Q}$ so that box $\mathrm{Q}$ has more chocolates than box $\mathrm{P}$ ?
A $\frac{q-p+2}{2}$
B $\frac{p-q+2}{2}$
C $\frac{q+p-2}{2}$
D $\frac{p-q-2}{2}$
E $\frac{q+p+2}{2}$
Solution
B Because $p$ and $q$ are both odd, $p+q$ is even. If the chocolates were shared equally between the boxes there would be $\frac{1}{2}(p+q)$ chocolates in each box. So the least number of chocolates there must be in box $\mathrm{Q}$ if it is to have more chocolates than box $\mathrm{P}$ is $\frac{1}{2}(p+q)+1$. Because box $\mathrm{Q}$ starts with $q$ chocolates in it, to end up with $\frac{1}{2}(p+q)+1$ in box $\mathrm{Q}$ the number of chocolates we need to transfer from box $\mathrm{P}$ to box $\mathrm{Q}$ is
$$
\left(\frac{p+q}{2}+1\right)-q=\frac{p-q}{2}+1=\frac{p-q+2}{2} .
$$