What is the largest integer that can be placed in the box so that $\frac{\square}{11}<\frac{2}{3}$ ?

##
Since 33 is positive, $\frac{\square}{11}<\frac{2}{3}$ implies

$$
33\left(\frac{\square}{11}\right)<33\left(\frac{2}{3}\right)
$$

which simplifies to $3 \times \square<22$.

The largest multiple of 3 that is less than 22 is $3 \times 7$, so this means the number in the box cannot be larger than 7 .

Indeed, $\frac{7}{11}=\frac{21}{33}$ is less than $\frac{2}{3}=\frac{22}{33}$, so the answer is 7 .