The 16th Japanese Mathematical Olympiad - The First Round

It is clear that the smallest value exists. Let $B$ be a book with $n_i$ pages for the $i$-th chapter which attains the smallest value.

We first prove that none of $n_1,n_2,\dots ,n_{10}$ are equivalent to $3$ or $5$ modulo $6$. Assume that $n_1\equiv 3,5(\mathrm{mod}6)$. Checking the parity, we can assume without loss of generality that $n_2$ is odd. Consider book $B^{\prime }$ whose chapter one has 4 pages, chapter two has $n_1+n_2-4$ pages, and the remaining chapters have the same number of pages with corresponding chapters of $B$. Then easy computation shows that Akinori can read $B^{\prime }$ faster than $B$, and that Tomohiro can't read $B^{\prime }$ faster than $B$. Then the difference of the days needed to read $B^{\prime }$ is smaller than that of $B$, which contradicts minimality of $B$.

Now let $n_i=6m_i+k_i$ ($m_i$ nonnegative, $k_i$ nonnegative $<6$ and $k_i\ne 3,5$). Then it can be easily checked that Tomohiro reads $i$-th chapter $m_i$ days faster than Akinori. Since $$\sum _{i=1}^{10}{m}_i=\frac{1}{6}\left(\sum _{i=1}^{10}{n}_i-\sum _{i=1}^{10}{k}_i\right)=\frac{1}{6}\left(120-\sum _{i=1}^{10}{k}_i\right)$$ and $k_i\le 4$, it follows that $\sum m_i\ge 20-\frac{40}{6}$, hence $\sum m_i\ge 14$. This value is in fact attained if, for example, $n_1=n_2=\cdots =n_9=10$ and $n_{10}=30$. Therefore, the smallest value is 14.