>>14364624I think I found the smart solution (meaning only imagination and arithmetic, as little algebra as possible, ideally none)
If it's assumed the most days without rain, you would get 5 days with no rain, leaving 1 morning with no rain. You would need to include that morning in one of the 7 days that rained, and on that day, consequently, it would have to have rained at the afternoon since if rains at morning it doesn't rain at afternoon and vice-versa.
But now there remains 6 days of rain, just one issue, you need to decide if it didn't rain at the morning or the afternoon for each day, but you already used all the 5 afternoons and 6 mornings where it didn't rain.
Then, you need to take 6 mornings or afternoons with no rain to complete the 6 remaining days where it rained.
Problem is, you can't just remove 5 mornings where it didn't rain and 1 afternoon where it didn't rain, and distribute it to the 6 remaining raining days, because the days with no rain where they were removed from would be missing half of the day. The only way to complete that now lacking half would be a raining morning/afternoon (contradiction as it is a day without rain) or completing it with the missing morning/afternoon with no rain (contradiction as that would go over the registered mornings and afternoons with no rain).
So the conclusion is that to fill the 6 remaining raining days you need to take a morning/afternoon pair from each non raining day. Since you need to fill 6 days, you need 3 pairs.
This, we're left with 7 raining days and 2 non raining days, N must be equal to 9.
