To find the highest value that x may have, we need to consider the given conditions:
1) The product of x and y is less than 82.
2) y is a multiple of three.

Let's analyze the options one by one to determine the highest value for x.

Option A: 13
If x is 13, the possible values for y could be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. However, none of these combinations satisfy the condition that the product of x and y should be less than 82. So, option A is not the correct answer.

Option B: 42
If x is 42, the possible values for y could be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. Again, none of these combinations satisfy the given condition. So, option B is not the correct answer.

Option C: 27
If x is 27, the possible values for y could be 1, 2, 3, 4, 5, 6, 7, and 9. Let's check the product of x and y for each of these values:
- For y = 1, the product is 27*1 = 27, which is less than 82.
- For y = 2, the product is 27*2 = 54, which is less than 82.
- For y = 3, the product is 27*3 = 81, which is less than 82.
- For y = 4, the product is 27*4 = 108, which is not less than 82.
- For y = 5, the product is 27*5 = 135, which is not less than 82.
- For y = 6, the product is 27*6 = 162, which is not less than 82.
- For y = 7, the product is 27*7 = 189, which is not less than 82.
- For y = 9, the product is 27*9 = 243, which is not less than 82.

Therefore, the highest value that x may have is 27, which satisfies both conditions.

Option D: 30
If x is 30, the possible values for y could be 1, 2, 3, 4, 5, 6, 7, and 9. Let's check the product of x and y for each of these values:
- For y = 1, the product is 30*1 = 30, which is less than 82.
- For y = 2, the product is 30*2 = 60, which is less than 82.
- For y = 3, the product is 30*3 = 90, which is not less than 82.
- For y = 4, the product is 30*4 = 120, which is not less than 82.
- For y = 5, the product is 30*5 = 150, which is not less than 82.
- For y = 6,