*** SPOILERS *** Solution below My score (M) = 10 Total table score (T) = (10 + G + R + B) G > R > B My score is more than 50% (or half) of the total score of the table. M > (T/2) 10 > (T/2) Multiplying both sides by 2 gives: 20 > T or T < 20 We now have two different definitions for the total table score (T) that we can compare: T = (10 + G + R + B) and T < 20 Therefore: (10 + G + R + B) < 20 Subtracting 10 from both sides gives: G + R + B < 10 The minimum score possible in Catan is 2, so: B >= 2 If we assume Blue has 2 points, we get: G + R < 8 So Green and Red total 7 points or fewer. Because G > R > B; If B = 2, the lowest R and G could be are 3 & 4 respectively which, together with Blue (2), total 9 points. Only this combination fits the rules. If Blue were more than 2, the score would be equal to or higher than your score of 10. So the final scorecard is: Black = 10 Green = 4 Red = 3 Blue = 2
This was a fun one!
More of a maths problem than a Catan problem but I did anyway 😅
lol this feels like it should belong in a 8th grade math class
*** SPOILERS ***
Solution below
My score (M) = 10
Total table score (T) = (10 + G + R + B)
G > R > B
My score is more than 50% (or half) of the total score of the table.
M > (T/2)
10 > (T/2)
Multiplying both sides by 2 gives:
20 > T or T < 20
We now have two different definitions for the total table score (T) that we can compare:
T = (10 + G + R + B)
and T < 20
Therefore: (10 + G + R + B) < 20
Subtracting 10 from both sides gives:
G + R + B < 10
The minimum score possible in Catan is 2, so:
B >= 2
If we assume Blue has 2 points, we get:
G + R < 8
So Green and Red total 7 points or fewer.
Because G > R > B;
If B = 2, the lowest R and G could be are 3 & 4 respectively which, together with Blue (2), total 9 points.
Only this combination fits the rules. If Blue were more than 2, the score would be equal to or higher than your score of 10.
So the final scorecard is:
Black = 10
Green = 4
Red = 3
Blue = 2