How many triples $(a,b,c)$ of even positive integers satisfy $a^3 + b^2 + c \leq 50$?
This problem involves counting the number of solutions to an inequality involving three even positive integers. One way to approach this problem is to try different values of $a$ and see how they affect the possible values of $b$ and $c$. Since $a$ is even and positive, it can only take the values 2, 4, 6, 8, or 10. However, if $a \geq 6$, then $a^3 \geq 216$, which is already too big for the inequality to hold, since $b^2$ and $c$ are both positive. Therefore, $a$ can only be 2 or 4. If $a = 2$, then $a^3 = 8$ and the inequality becomes $b^2 + c \leq 42$. For $b$ to be even and positive, it can only be 2, 4, or 6, since $b^2 \geq 64$ for larger even values. If $b = 2$, then $b^2 = 4$ and the inequality becomes $c \leq 38$. The number of even positive integers less than or equal to 38 is 19, since they are the multiples of 2 from 2 to 38. Therefore, there are 19 triples $(a,b,c)$ with $a = 2$ and $b = 2$. If $b = 4$, then $b^2 = 16$ and the inequality becomes $c \leq 26$. The number of even positive integers less than or equal to 26 is 13, since they are the multiples of 2 from , to 26. Therefore, there are 13 triples $(a,b,c)$ with $a = 2$ and $b = 4$. If $b = 6$, then $b^2 = 36$ and the inequality becomes $c \leq 6$. The number of even positive integers less than or equal to 6 is 3, since they are 2, 4, and 6. Therefore, there are 3 triples $(a,b,c)$ with $a = 2$ and $b = 6$. In total, there are 19 + 13 + 3 = 35 triples $(a,b,c)$ with $a = 2$. If $a = 4$, then $a^3 = 64$ and the inequality becomes $b^2 + c \leq -14$. This is impossible, since $b^2$ and $c$ are both positive. Therefore, there are no triples $(a,b,c)$ with $a = 4$. Hence, the final answer is 35. 