geeta719 —
Mechaniker: One line solutions do not necessarily always give the complete answer.My equation is valid for all values of m.
In this case, for eg.,
(1) your 1-liner assumes even values for m, but doesn’t say it.
Consider m = 15 ☟
According to my equation the clock face can be divided by
FLOOR( (m-1)/2 ) = 7 lines (horizontal lines in this case) giving 8 groups.
⠀⠀⠀⠀⠀⠀⠀15⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☜ See diagram to left.
⠀⠀⠀⠀14⠀⠀⠀⠀⠀1
⠀⠀13⠀⠀⠀⠀⠀⠀⠀⠀⠀2
⠀12⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀3
11⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀4
⠀10⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀5
⠀⠀ 9⠀⠀⠀⠀⠀⠀⠀⠀⠀6
⠀⠀⠀⠀ 8⠀⠀⠀⠀⠀7
Horizontal lines group the numbers in the fashion
{15}, {14,1}, {13,2}, {12,3}, {11,4}, {10,5}, {9,6}, {8,7}. The sum of the numbers in each group is 15.
(2) n has non-unique values. Your method gives just the max value for n, that sums 2 numbers (max and min of those left) at a time.It is a simple matter of getting other groupings. If every other separating line is removed, then the groups become {15,14,1}, {13,2,12,3}, {11,4,10,5}, {9,6,8,7}. Each group sums to 30.
Finally, if two more lines are removed, one has
{15,14,1,13,2,12,3} and {11,4,10,5,9,6,8,7}, both of which sum to 60.
The original problem did not ask for all sets of lines that divide the clock face into equal sums, but there it is. Frankly, I think the problem is trivial.
One line solutions do not necessarily always give the complete answer.They give the most aesthetically pleasing solution. That is why everyone in the universe knows of Einstein’s equation E = mc².