a, b, c and d are four distinct positive integers that satisfy a + b + c + d = 60. Find the maximum possible value of (a – b)2 + (a – c)2 + (a – d)2 + (b – c)2 + (b – d)2 + (c – d)2.

To solve this problem, we need to understand that the expression (a – b)² + (a – c)² + (a – d)² + (b – c)² + (b – d)² + (c – d)² is maximized when the difference between the numbers is as large as possible.

Step 1: We know that a, b, c, and d are distinct positive integers that add up to 60.

Step 2: To maximize the differences between the numbers, we should choose one number to be as small as possible and the other three to be as large as possible.

Step 3: The smallest positive integer is 1, so we set a = 1.

Step 4: The remaining sum is 60 - 1 = 59, which we need to distribute among b, c, and d.

Step 5: To maximize the differences, we should set two of these numbers to be as close to each other as possible, and the third to be as large as possible.

Step 6: The largest number we can choose is 59, but this would leave no room for the other two numbers. So, we divide 59 by 3, which gives approximately 19.67.

Step 7: We can't have fractional numbers, so we round down to get 19. This means we set b = c = 19.

Step 8: The remaining sum is 59 - 19 - 19 = 21, so we set d = 21.

Step 9: Substitute a = 1, b = c = 19, and d = 21 into the expression to get the maximum value.

(a – b)² + (a – c)² + (a – d)² + (b – c)² + (b – d)² + (c – d)² = (1 - 19)² + (1 - 19)² + (1 - 21)² + (19 - 19)² + (19 - 21)² + (19 - 21)² = 324 + 324 + 400 + 0 + 4 + 4 = 1056.

So, the maximum possible value of the expression is 1056.