The cost to equalize the colleagues is the sum of the reduce operations through all colleagues. Naturally you use as the baseline the minimum colleague and you are taking away from other colleagues until they are equal. However, given you can reduce by 1, 2, or 5, you naturally want to take bigger steps and use as many 5 reductions as possibly and then 2 and only after that reduce by 1.

Now imagine reducing the case 3 7 7. If you would be strictly equalizing them with the miminum colleague then you need to reduce them all to 3 in which case you cannot apply -5 because 7-5=2 is too low hence you need to apply -2 twice. In this case it is cheaper to adjust the minimum to 2. I.e. 377->327->322->222 is cheaper then 377->357->337->335->333.

Now thinking about it again, it is not necessary to calculate for 5 different baselines, 3 is enough: the minimum in the sequence M, M - 1, M - 2 (given M > 2).

In other words reducing by 1 or 2 is cheap (1 operation), by 3 or 4 is expensive (2 operations), 5 is cheap again. By cheaply reducing the basiline by 1 or 2 with a single operation reduction you can save multiple 2 operation reductions on other colleagues.