An m*n rectangle is divided into mn unit squares by lines drawn parallel to its sides Let S=[k, k+1, k+2,... mn-1] be a set of mn consecutive integersthe object is to try and to write all the...
An m*n rectangle is divided into mn unit squares by lines drawn parallel to its sides
Let S=[k, k+1, k+2,... mn-1] be a set of mn consecutive integers
the object is to try and to write all the elements of S in the rectangle grid, one element of S per unit square, in such a way that the following two conditions are satisfied simultaneously:
in each row, the elements are congruent mod m;
in each column the elements are congruent mod n;
determine if the placement is possible when k=19 and
(i) m=3 and n=8
(ii) m=4 and n=6
So the answer to the first question is yes, and here is an example:
19 22 25 28 31 34 37 40
27 30 33 36 39 42 21 24
35 38 41 20 23 26 29 32
And the answer to the second question is no.
k actually doesn't matter
If you had a grid filled in, that worked, you could subtract j from each and every box. All of the modulo conditions would still match. But the numbers would be k-j, k-j+1, ..., k-j +mn -1
So for now, we will just assume k=0 (and then add the appropriate number later, if needed)
What matters is the relationship between m and n
If they are "relatively prime", you can fill out the grid. If not, you can't. "Relatively prime" means their gcd (greatest common denominator) is 1
If they are relatively prime, here is how to fill out the box:
0 m 2m 3m 4m 5m ... (n-1)m
n n+m n+2m ...
2n 2n+m ...
Then take all of the numbers modulo nm
You don't have any duplicate numbers.
In the case of 4 and 6, (and any time they aren't relatively prime), here is what goes wrong:
In some order, the numbers in one of the rows is:
0 4 8 12 16 20
but notice that, for example, 0 and 12 (or 4 and 16, or 8 and 20) are also the same modulo 6
So you need to put 12 in the same column as 0, even though it is already in the same row, and there is no way to fill out the grid.