Answer
No, it is not possible to have 50 coins made up of pennies, dimes, and quarters that add up to 3 dollars. To show this, we try to find a positive integer solution for the equation $1p+10d+25q=300$ constrained by $p+d+q=50$, where $p$ is the number of pennies, $d$ the number of dimes, and $q$ the number of quarters. We can subtract the second equation from the first to get $9d+24q=250$. Solving for $d$, we get that $d=\frac{250-24q}{9}$. But this number cannot be an integer, since $250-24q$ is not divisible by $9$ for any positive integer value of $q$. We can readily verify this by the fact that $250-24q=9(\frac{250}{9}-\frac{8q}{9})$, but $\frac{250}{9}-\frac{8q}{9}$ has no integer solutions for $0\leq q\leq 12$. (We cannot have $q\gt12$, because we would then neeg negative pennies and dimes.) Therefore, it is impossible to have 50 coins made up of pennies, dimes, and quarters that add up to 3 dollars.
Work Step by Step
To elegantly prove that $9d+24q=250$ has no integer solutions, we would have to use the Euclidean algorithm, which is not introduced until section 4.8.