Problem 248

For $\frac{3}{4}+\frac{4}{\square}=1$ to be true, we must have $\frac{4}{\square}=1-\frac{3}{4}=\frac{1}{4}$. Since $\frac{1}{4}=\frac{4}{16}$, we rewrite the right side using the same numerator to obtain $\frac{4}{\square}=\frac{4}{16}$. Therefore, $\square=16$ makes the equation true.