Final answer:
The maximum value for the largest integer factor of 180, when it is the product of four unequal positive integers all greater than one, is found to be 30.
Step-by-step explanation:
To find the maximum value of the largest integer when the number 180 is factored into four unequal positive integers all greater than one, one strategy is to first prime factorize the number 180. The prime factorization of 180 is 22 × 32 × 5. To maximize the largest factor, we want to combine the smaller factors as much as possible, considering they can't all be equal. First, we take one of each prime to make the smallest factor: 2 × 3 × 5 = 30. Next, since we are left with one 2 and one 3, taking them together makes 6. We now have two numbers: 30 and 6. We can't use a '1' because all integers must be greater than one, and we want three distinct factors besides 30. Therefore, the reasonable division of what we have left (2 × 3) would be to treat them as independent factors: 2 and 3. Thus, our factors are 2, 3, 30, and 6.
To further maximize the largest integer, we merge the smallest two integers (2 and 3) to form 6 so that the distribution of factors now is 6, 6, and 30. Now, if we calculate 180 divided by the product of our smallest factors (6 × 6), we get 5, which is the last factor we need to use. However, since 5 is equal to one of our existing factors and the factors need to be unique, we take the 5 into 30 to create a larger factor, thus creating 30 and 6 as 5 × 6 = 30. Because we've merged two factors into 30, our new factors become 6, 30, and 30. The largest factor we can have under these constraints is 180 / (6 × 30) which equals 1. Since 1 is not allowed (all integers must be greater than one), we cannot move forward with this grouping, and so we must revert to the previous arrangement of factors: 2, 3, 6, and 30. Following these constraints, the maximum value for the largest integer factor of 180 is thus 30.