Problem 1449

Looking at the equation $\bmod 7$ gives $a \equiv 3(\bmod 7)$, so let $a=7a^{\prime}+3$. Then mod 4 gives $b \equiv 0(\bmod 4)$, so let $b=4b^{\prime}$. Finally, $\bmod 3$ gives $c \equiv 2(\bmod 3)$, so let $c=3c^{\prime}+2$. Now our equation yields $$84a^{\prime}+84b^{\prime}+84c^{\prime}+84d=2024-3 \cdot 12-2 \cdot 28=1932 \Longrightarrow a^{\prime}+b^{\prime}+c^{\prime}+d=23$$ Since $a, b, c, d$ are positive integers, we have $a^{\prime}$ and $c^{\prime}$ are nonnegative and $b^{\prime}$ and $d$ are positive. Thus, let $b^{\prime\prime}=b^{\prime}+1$ and $d^{\prime}=d+1$, so $a^{\prime}, b^{\prime\prime}, c^{\prime}, d^{\prime}$ are nonnegative integers summing to 21. By stars and bars, there are $\binom{24}{3}=2024$ such solutions.