A number theory problem by أحمد الحلاق

Let $x+y+z=k$ , now $28k + 2y + 3z = 365$ .
If $k \leq 11$ , $2y + 3z \geq 57 \implies y > 11 \vee z > 11$ which is impossible since $k=x+y+z\leq11$ .
If $k \geq 13$ , $2y + 3z \leq 1 \implies y < 1 \vee z < 1$ which is impossible since $x,y,z > 0$ .

Hence, $\boxed{12}$ is the only possibility left and we indeed have:
$28$ days in February (except on leap years),
$30$ days in April, June, September and November,
$31$ days in January, March, May, Juli, August, October and December,
$12$ months in a year and
$365$ days in a year (except on leap years).

4 (31+30+28)=89 4=356. now 365-356=9. hence if we replace 4 28's with 4 30's the value increses by 8. hence now the expression looks like this 28 0+30 8+31 4=355. again we replace 1 30 with 1 31 .now the equation holds. i.e.28 0+30 7+31 5=356. hence x+y+z=0+7+5=12.