Twelve months in a year

Let non-negative integers $i$ , $j$ , and $k$ be such that $\begin{cases} i + j + k = 12 & ...(1) \\ 28i + 30j + 31k = 365 & ...(2) \end{cases}$

From $(2)$ :

$\begin{aligned} 28i + 30j + 31k & = 365 \\ 28{\color{#3D99F6}(i + j + k)} + 2j + 3k & = 365 & \small \color{#3D99F6} \text{Recall }(1): \ i+j+k= 12 \\ 28{\color{#3D99F6}(12)} + 2j + 3k & = 365 & \small \color{#3D99F6} \text{Rearranging} \\ \implies 2j + 3k & = 29 \end{aligned}$

$\implies k = \dfrac {29-2j}3$ .

For $k$ to be an integer, $29-2j$ must be a multiple of 3, then the acceptable $j = 1, 4, 7, 10, \cdots$ .

For $i+ j + k \le 12$ , we have $(i,j,k) = \begin{cases} (2, 1, 9) \\ (1, 4, 7) \\ (0, 5, 7) \end{cases}$ .

Therefore there are $\boxed 3$ solutions.