Question:

We need to find positive integer solutions to $a+b+3c \leq 30$ where $a,b,c \geq 1$. Substituting $a=a'+1, b=b'+1, c=c'+1$ where $a',b',c' \geq 0$, we get $(a'+1)+(b'+1)+3(c'+1) \leq 30$, which simplifies to $a'+b'+3c' \leq 26$. For each fixed value of $c'$ from $0$ to $8$ (since $3c' \leq 26$), we count non-negative solutions to $a'+b' \leq 26-3c'$. The number of such solutions is $\sum_{i=0}^{26-3c'}(i+1) = \frac{(27-3c')(28-3c')}{2}$. Summing over $c'=0$ to $8$: $k = \sum_{c'=0}^{8}\frac{(27-3c')(28-3c')}{2} = \frac{1}{2}(27 \cdot 28 + 24 \cdot 25 + 21 \cdot 22 + 18 \cdot 19 + 15 \cdot 16 + 12 \cdot 13 + 9 \cdot 10 + 6 \cdot 7 + 3 \cdot 4) = \frac{1}{2}(756+600+462+342+240+156+90+42+12) = 1215$. Therefore $\frac{k}{5} = \frac{1215}{5} = 243$.