Question 332184: Four laborers arrived in town looking for work. They met with a buisness owner who said he could pay them for 200 hours of work. However, there was a catch. They could divide up the work anyway they wanted, but each laborer had to work the same number of days as he worked hours in one day. For example, if a laborer worked 8 hours a day, he had to work for 8 days. The other catch is that the laziest laborer did the negotiations with the buisness owner.
Answer by J2R2R(94)   (Show Source):
You can put this solution on YOUR website! The most obvious answer is where the sum of four whole number squares (hours times days - both the same) equals 200.
Just think of how 200 can be arrived at in this way. This is a bit of trial and error and playing with numbers where since they all work to some extent, none of the values can be 0 and the laziest is one on his own so there can only be one worker with the minimum value.
As 200 hours is in total we can only go up to 14 since 15 squared is greater than 200 and we cannot have squares that are negative (not with real numbers).
All groups of four numbers which when squared total 200 are as follows - we will eliminate what we don’t need later:
0, 0, 2, 14
0, 0, 10, 10
0, 6, 8, 10
2, 4, 6, 12
6, 6, 8, 8
Now we cannot have 0, so the first three are eliminated so we have:
2, 4, 6, 12
6, 6, 8, 8
Now the laziest one is on his own so we cannot use 6, 6, 8, 8 as two of them would be lazier doing 6 hours.
Therefore 2, 4, 6, 12 is the answer with the laziest one doing 2 hours.
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