Given set of
(1 to 15)
Erase 3 numbers so that remaining integers can be arranged as a
3 by 4 array in which
the sum of the numbers in each row is a certain Sr and in each of the
columns a certain Sc.
Present the triplet you have chosen and one of the possible arrangements.
D4. Bonus question:
How many distinct solutions are there?
I'm guessing there are a great many solutions. Here's one I found pretty quickly:
14 1 8 5
3 13 2 10
4 7 11 6
The columns sum to 21 and the rows sum to 28.
All I did was note that the remaining numbers must sum to a multiple of 12 so as to be divisible by both 3 and 4.
Since the numbers 1 to 15 sum to a multiple of 12, the numbers removed must also sum to a multiple of 12.
I chose to remove 15, 12, 9 because they have a large sum are are spaced well apart (I'm not sure why, really)
The 4 columns then had to sum to 21. There are only a dozen partitions that are possible. This is the only quadruplet that doesn't repeat numbers. Then I arranged them to make the row sums all 28.
Posted by Jer
on 2015-06-16 13:27:52