Easy puzzle, they basically say the last number is the 3rd highest, just go from there...
Must be lower than the 2nd and the 5th number, and the 2nd is obv. bigger than the fifth as well, since the last one is only 1 lower than the 5th and the 2nd minus the 3rd is equal to the last. Even if you would only subtract the lowest number available (1) it wouldnt work, cause you`d then need 2 equal numbers, but all available ones are different. Therefore the 2nd number is very likely to be the highest without even thinking any further and wasting time...
x x x x x x
x x x x x 5 ergo
x x x x 6 5 ergo
x 7 2 x 6 5 ergo
1 7 2 4 6 5
Must be lower than the 2nd and the 5th number, and the 2nd is obv. bigger than the fifth as well, since the last one is only 1 lower than the 5th and the 2nd minus the 3rd is equal to the last. Even if you would only subtract the lowest number available (1) it wouldnt work, cause you`d then need 2 equal numbers, but all available ones are different. Therefore the 2nd number is very likely to be the highest without even thinking any further and wasting time...
x x x x x x
x x x x x 5 ergo
x x x x 6 5 ergo
x 7 2 x 6 5 ergo
1 7 2 4 6 5
