Harvard-MIT Mathematics Tournament

Solution:
The number of zeroes in base $7$ is the total number of factors of $7$ in $1 \cdot 2 \cdots n$, which is
$$\left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{7^2} \right\rfloor + \left\lfloor \frac{n}{7^3} \right\rfloor + \cdots.$$
The number of zeroes in base $8$ is $\lfloor a \rfloor$, where
$$a = \left( \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{2^2} \right\rfloor + \left\lfloor \frac{n}{2^3} \right\rfloor + \cdots \right) / 3$$
is one-third the number of factors of $2$ in the product $n!$. Now $\left\lfloor \frac{n}{2^k} \right\rfloor / 3 \geq \left\lfloor \frac{n}{7^k} \right\rfloor$ for all $k$, since $\left( n / 2^k \right) / 3 \geq n / 7^k$. But $n$ can only be picante if the two sums differ by at most $2/3$, so in particular this requires $\left( \left\lfloor n / 2^2 \right\rfloor \right) / 3 \leq \left\lfloor n / 7^2 \right\rfloor + 2/3 \Leftrightarrow \left\lfloor n / 4 \right\rfloor \leq 3 \left\lfloor n / 49 \right\rfloor + 2$. This cannot happen for $n \geq 12$; checking the remaining few cases by hand, we find $n = 1, 2, 3, 7$ are picante, for a total of $4$ values.