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At a carnival, there’s a balloon pop game with three different colored balloons: red, blue, and yellow. Popping a red balloon is worth 4 points, a blue balloon is worth 3 points, and a yellow balloon is worth 2 points. Players are allowed to throw up to 8 darts, and they start with a score of 0.
Suppose a player scores exactly 25 points after throwing 8 darts. Let r represent the number of red balloons popped, b represent the number of blue balloons popped, and y represent the number of yellow balloons popped.
Determine all possible combinations of (r, b, y) that result in a score of exactly 25 points.
We need to find all combinations of (r, b, y) that satisfy the equation:
4r + 3b + 2y = 25, where r + b + y ≤ 8
Let’s approach this systematically:
1. The maximum value for r is 6, as 7 × 4 = 28, which exceeds 25.
2. We’ll examine each possible value of r from 0 to 6.
For r = 6:
4(6) + 3b + 2y = 25
24 + 3b + 2y = 25
3b + 2y = 1
This is not possible with positive integers.
For r = 5:
4(5) + 3b + 2y = 25
20 + 3b + 2y = 25
3b + 2y = 5
The only solution is b = 1, y = 1
5 + 1 + 1 = 7 ≤ 8, so (5, 1, 1) is a valid solution.
For r = 4:
4(4) + 3b + 2y = 25
16 + 3b + 2y = 9
3b + 2y = 9
Possible solutions:
b = 3, y = 0 (4 + 3 + 0 = 7 ≤ 8)
b = 1, y = 3 (4 + 1 + 3 = 8 ≤ 8)
So (4, 3, 0) and (4, 1, 3) are valid solutions.
For r = 3:
4(3) + 3b + 2y = 25
12 + 3b + 2y = 13
3b + 2y = 13
Possible solution:
b = 3, y = 2 (3 + 3 + 2 = 8 ≤ 8)
So (3, 3, 2) is a valid solution.
For r = 2:
4(2) + 3b + 2y = 25
8 + 3b + 2y = 17
3b + 2y = 17
Possible solutions:
b = 5, y = 1 (2 + 5 + 1 = 8 ≤ 8)
So (2, 5, 1) is a valid solution.
For r = 1:
4(1) + 3b + 2y = 25
4 + 3b + 2y = 21
3b + 2y = 21
Possible solutions:
b = 7, y = 0 (1 + 7 + 0 = 8 ≤ 8)
So (1, 7, 0) is a valid solution.
For r = 0:
3b + 2y = 25
The only solution is b = 7, y = 2, but 7 + 2 = 9 > 8, so this is not valid.
Therefore, the valid combinations (r, b, y) are:
(5, 1, 1), (4, 3, 0), (4, 1, 3), (3, 3, 2), (2, 5, 1), (1, 7, 0)