Hi Ed,

I'm glad I didn't try to get all the answers by hand...

These are the 104 solutions to the problem propossed by Erich Friedman
that appears in your web page (unless something was wrong in my
program or the reasoning that lead me to it, which is quite possible).

Cheers,

Carlos Ungil

13 - (1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 27)

((1 2 15 27) (9 12 16 8) (20 18 3 4) (24 10 6 5))
((1 2 18 24) (8 15 10 12) (20 16 4 5) (27 9 6 3))
((1 2 18 24) (8 15 10 12) (20 16 6 3) (27 9 4 5))
((1 2 18 24) (10 12 8 15) (16 20 5 4) (27 9 6 3))
((1 2 18 24) (10 12 8 15) (16 20 6 3) (27 9 5 4))
((1 4 16 24) (8 10 15 12) (20 18 2 5) (27 6 9 3))
((1 8 12 24) (10 18 2 15) (16 5 20 4) (27 6 9 3))
((1 8 12 24) (10 18 2 15) (16 6 20 3) (27 5 9 4))
((1 9 15 20) (10 24 8 3) (16 5 18 6) (27 4 2 12))
((1 9 15 20) (10 8 24 3) (16 5 6 18) (27 12 2 4))
((1 9 15 20) (10 5 24 6) (16 8 3 18) (27 12 4 2))
((1 9 15 20) (10 5 18 12) (16 24 2 3) (27 4 8 6))
((1 9 15 20) (10 5 12 18) (16 24 3 2) (27 4 8 6))

22 - (1 2 3 4 6 7 8 9 12 14 16 18 21 24 27 28)

((1 3 18 28) (14 8 24 4) (16 21 7 6) (27 12 2 9))
((1 3 18 28) (14 21 6 9) (16 24 8 2) (27 4 7 12))
((1 3 18 28) (14 24 8 4) (16 7 21 6) (27 12 2 9))
((1 4 21 24) (8 12 16 14) (27 18 2 3) (28 7 9 6))
((1 4 21 24) (8 12 16 14) (27 18 3 2) (28 7 6 9))
((1 4 21 24) (8 14 16 12) (27 18 2 3) (28 6 9 7))
((1 4 21 24) (12 8 16 14) (18 27 2 3) (28 7 9 6))
((1 4 21 24) (12 8 16 14) (18 27 3 2) (28 7 6 9))
((1 6 16 27) (12 7 3 28) (21 18 9 2) (24 8 14 4))
((1 7 14 28) (12 3 8 27) (21 16 9 4) (24 18 6 2))
((1 7 14 28) (12 3 8 27) (21 18 9 2) (24 16 6 4))
((1 7 14 28) (12 8 3 27) (21 18 9 2) (24 6 16 4))
((1 7 14 28) (12 27 8 3) (21 2 9 18) (24 16 6 4))
((1 7 14 28) (12 27 8 3) (21 16 9 4) (24 2 6 18)) 
((1 7 14 28) (12 27 9 2) (21 8 3 18) (24 4 16 6))
((1 7 14 28) (16 3 27 4) (18 24 2 6) (21 12 8 9))
((1 7 14 28) (16 4 24 6) (18 27 2 3) (21 8 9 12))
((1 7 14 28) (16 4 27 3) (18 24 2 6) (21 9 8 12))
((1 7 18 24) (8 9 12 21) (27 6 14 3) (28 16 2 4))
((1 8 14 27) (12 7 3 28) (21 18 9 2) (24 6 16 4))
((1 9 12 28) (14 6 3 27) (18 7 21 4) (24 16 8 2))
((1 12 16 21) (8 4 14 24) (27 18 3 2) (28 7 9 6))

9 - (1 2 4 5 6 7 8 9 10 12 14 15 18 20 21 28)

((1 5 18 21) (12 9 4 20) (15 14 10 6) (28 8 7 2))
((1 7 9 28) (12 4 14 15) (20 18 5 2) (21 10 8 6))
((1 7 9 28) (12 8 10 15) (20 5 14 6) (21 18 4 2))
((1 7 9 28) (12 10 8 15) (20 18 5 2) (21 4 14 6))
((1 8 15 21) (9 18 12 6) (20 7 14 4) (28 5 2 10))
((1 8 15 21) (9 18 14 4) (20 7 12 6) (28 5 2 10))
((1 9 14 21) (12 8 5 20) (15 10 18 2) (28 7 4 6))
((1 10 14 20) (12 18 9 6) (15 4 5 21) (28 7 8 2))
((1 10 14 20) (12 21 5 7) (15 4 8 18) (28 6 9 2))

1 - (1 3 4 5 6 7 9 10 12 14 15 18 20 21 27 28)

((1 3 18 28) (14 12 15 9) (20 21 4 5) (27 10 7 6))

4 - (1 3 5 6 7 8 10 12 14 15 16 18 20 21 24 28)

((1 8 15 28) (20 14 6 12) (21 5 16 10) (24 18 7 3))
((1 10 20 21) (15 18 3 16) (24 8 14 6) (28 7 12 5))
((1 12 18 21) (15 7 14 16) (24 20 5 3) (28 6 8 10))
((1 12 18 21) (15 20 14 3) (24 7 5 16) (28 6 8 10))

22 - (2 3 4 5 7 8 9 10 14 15 16 18 20 21 24 28)

((2 3 18 28) (10 16 20 5) (21 14 7 9) (24 15 4 8))
((2 3 18 28) (10 16 20 5) (21 15 7 8) (24 14 4 9))
((2 3 18 28) (15 24 4 8) (16 10 20 5) (21 14 7 9))
((2 3 18 28) (15 24 4 8) (16 20 10 5) (21 7 14 9))
((2 3 18 28) (15 24 8 4) (16 20 5 10) (21 7 14 9))
((2 5 20 24) (15 8 18 10) (16 28 4 3) (21 9 7 14))
((2 5 20 24) (15 8 18 10) (21 9 7 14) (16 28 4 3))
((2 7 14 28) (10 8 15 18) (21 9 16 5) (24 20 3 4))
((2 7 14 28) (10 20 3 18) (21 9 16 5) (24 8 15 4))
((2 7 14 28) (10 20 16 5) (21 9 3 18) (24 8 15 4))
((2 7 14 28) (10 20 16 5) (21 18 3 9) (24 4 15 8))
((2 7 14 28) (10 20 16 5) (24 4 15 8) (21 18 3 9))
((2 7 14 28) (15 4 8 24) (16 20 10 5) (21 18 9 3))
((2 7 14 28) (15 4 24 8) (16 20 10 5) (21 18 3 9))
((2 7 14 28) (15 8 4 24) (16 10 20 5) (21 18 9 3))
((2 7 14 28) (15 8 4 24) (16 20 10 5) (21 9 18 3))
((2 7 14 28) (15 8 24 4) (16 20 10 5) (21 9 3 18))
((2 7 14 28) (15 24 4 8) (16 20 10 5) (21 3 18 9))
((2 7 14 28) (15 24 8 4) (16 20 5 10) (21 3 18 9))
((2 7 14 28) (15 24 8 4) (16 20 10 5) (21 3 9 18))
((2 14 15 20) (10 9 8 24) (18 5 21 7) (28 16 4 3))
((2 14 15 20) (10 9 24 8) (18 5 7 21) (28 16 4 3))

15 - (2 3 5 6 7 8 9 10 12 14 15 16 20 21 24 28)

((2 6 14 28) (10 20 5 15) (21 12 9 8) (24 7 16 3))
((2 8 16 24) (9 12 15 14) (20 21 6 3) (28 5 7 10))
((2 8 16 24) (9 21 6 14) (20 5 15 10) (28 12 7 3))
((2 8 16 24) (9 21 6 14) (20 12 15 3) (28 5 7 10))
((2 8 16 24) (9 21 14 6) (20 5 15 10) (28 12 3 7))
((2 8 16 24) (12 10 21 7) (15 9 6 20) (28 14 5 3))
((2 8 16 24) (12 14 21 3) (15 9 6 20) (28 10 5 7))
((2 8 16 24) (12 14 21 3) (15 10 5 20) (28 9 6 7))
((2 8 16 24) (12 21 14 3) (15 6 9 20) (28 10 5 7))
((2 8 16 24) (12 21 14 3) (15 10 5 20) (28 6 9 7))
((2 8 16 24) (12 28 3 7) (20 5 15 10) (21 9 14 6))
((2 8 16 24) (12 28 3 7) (20 9 15 6) (21 5 14 10))
((2 8 16 24) (12 28 7 3) (20 5 15 10) (21 9 6 14))
((2 9 15 24) (12 28 3 7) (20 8 16 6) (21 5 14 10))
((2 12 16 20) (10 28 3 9) (21 6 15 8) (24 5 14 7))

18 - (5 6 7 8 9 10 12 14 15 16 18 20 21 24 27 28)

((5 7 20 28) (12 27 6 15) (21 16 14 9) (24 10 18 8))
((5 7 20 28) (14 10 21 15) (16 24 8 12) (27 18 9 6))
((5 7 20 28) (14 10 21 15) (16 24 12 8) (27 18 6 9))
((5 7 20 28) (14 10 21 15) (18 27 6 9) (24 16 12 8))
((5 7 20 28) (14 10 21 15) (18 27 9 6) (24 16 8 12))
((5 9 18 28) (14 21 10 15) (16 8 24 12) (27 20 7 6))
((5 12 16 27) (9 10 21 20) (24 14 15 7) (28 18 6 8))
((5 12 16 27) (9 20 21 10) (24 7 15 14) (28 18 6 8))
((5 14 20 21) (8 12 24 16) (27 18 9 6) (28 10 7 15))
((5 14 20 21) (8 16 12 24) (27 9 18 6) (28 15 7 10))
((5 14 20 21) (8 24 12 16) (27 6 18 9) (28 15 7 10))
((5 14 20 21) (8 24 12 16) (27 9 18 6) (28 10 7 15))
((5 14 20 21) (9 18 27 6) (24 12 8 16) (28 10 7 15))
((5 14 20 21) (9 27 18 6) (24 8 12 16) (28 10 7 15))
((5 14 20 21) (12 8 24 16) (18 27 9 6) (28 10 7 15))
((5 14 20 21) (12 16 8 24) (18 9 27 6) (28 15 7 10))
((5 14 20 21) (12 24 8 16) (18 6 27 9) (28 15 7 10))
((5 14 20 21) (12 24 8 16) (18 9 27 6) (28 10 7 15))

-----------------------------------------------------------------

Here are the answers (27 total) I found for the problem.  I'm fairly 
sure that this is the complete list (excluding transpositions), but I
could've made a mistake.

I wrote a program to find all of the numbers that could be composed of
4 or more different sets of 4 numbers (with all 16 numbers being 
unique).  From there, I refined the program, to find arrangements of
the columns, such that the sums of the rows were equal.

product = 4320, sum = 45
1 8 20 27 2 15 16 9 24 12 3 5 18 10 6 4
1 9 20 24 2 12 18 10 15 16 3 6 27 8 4 5
1 10 16 27 15 24 6 2 20 3 18 4 9 8 5 12

product = 5040, sum = 45 
1 9 20 28 15 12 14 2 21 6 4 10 8 18 7 5
1 12 15 28 21 20 2 6 14 5 18 4 9 8 10 7
1 12 20 21 28 15 2 6 7 4 18 10 9 14 5 8

product = 6048, sum = 50
1 8 27 28 21 24 2 6 16 14 3 9 12 4 18 7
1 12 18 28 24 14 2 9 21 16 3 6 4 8 27 7
1 12 21 24 28 27 2 4 7 3 18 16 14 8 9 6
1 14 16 27 18 24 7 2 3 8 21 12 28 4 6 9
1 14 18 24 28 27 4 2 12 3 21 8 9 6 7 16
1 16 18 21 14 27 2 8 28 3 6 12 7 4 24 9

product = 7560, sum = 50
1 14 20 27 3 12 21 10 18 15 4 7 28 9 5 6

product = 10080, sum = 52
1 15 24 28 21 3 16 10 18 14 5 8 12 20 7 6
1 20 21 24 28 12 10 3 8 14 5 18 15 6 16 7

product = 10080, sum = 51
2 10 18 28 20 24 7 3 15 8 21 4 14 9 5 16
2 15 16 21 28 24 5 3 7 4 20 18 14 8 10 9

product = 10080, sum = 50
2 9 20 28 24 14 3 10 8 12 21 5 16 15 6 7
2 10 21 24 28 15 8 3 14 5 9 16 6 20 12 7
2 12 15 28 24 3 20 7 16 21 5 6 8 14 10 9
2 12 20 21 24 3 10 14 8 28 5 9 16 7 15 6

product = 30240, sum = 60
5 8 27 28 21 24 6 10 20 12 18 7 14 16 9 15
5 9 24 28 16 21 15 6 27 20 7 8 12 10 14 18
5 12 18 28 21 24 6 10 20 8 27 7 14 16 9 15
5 12 21 24 20 6 14 18 7 27 16 10 28 15 9 8
5 14 16 27 28 15 12 6 7 10 24 18 20 21 8 9
5 14 18 24 28 15 6 12 7 10 27 16 20 21 9 8

Nathan Stohler
------------------------------------------------------

3:
1,4,21,24
12,8,16,14
18,27,2,3
28,7,9,6

rows sum to 50,
columns multiply to 6048

L.T. Pebody

---------------------------------------------------------------------------
I've already sent Erich this sum-product matrix this month:

   1   2  15  27
   9  12  16   8
  20  18   3   4
  24  10   6   5
  
Joseph DeVincentis  

----------------------------------------------------------------------------
Ed,

I've found 2 distinct (albeit similar) solutions:

15 12 14  4
 6 21  8 10
 2 20  5 18
28  1  9  7

and

15 12  8 10
 6 21 14  4
 2 20  5 18
28  1  9  7

Row sums equal 45, column products equal 5040

So far, I investigated only about 50-75% of the solution search space. It's
possible that there are more solutions.
I might try to find them later.

To describe briefly my method:

1) Concentrate first on products, i.e. find four 4-number arrays that give
equal products and heir sum is divisible by 4.
2) Try to arrange sums later.

To find products, I eliminated "bad" numbers and finally came up with only
two 16-number sets that can possibly
give equal products:

a) 1 3 4 5 6 7 9 10 12 14 15 18 20 21 27 28
b) 1 2 4 5 6 7 8  9 10 12 14 15 18 20 21 28

The first set corresponds to the column product of 7560 (and the row sum
equals 50).
I analyzed it thoroughly and could get the only set of columns:
(4 5 14 27) (6 7 12 15) (1 18 20 21) (3 9 10 28).
However, they can't be arranged to give equal row sums.

So, seems like (b) is the only possible 16-number set. As I mentioned, I
explored only about a half of its search space
and came up with the 2 solutions above.

It's difficult to describe in more details how I analyzed the solutions -
different methods of reasoning were used.
I didn't use computer. Actually, I tried to, but the search space is too
big... Or, one can say, my heuristic algorithm was not good enough :) So,
I've got the result manually faster.

Mathematically Yours,

Igor Krivokon
------------------------------------------------------------------------------
Ed,

   One particular puzzle caught my attention, and I've attached my 
solution. It's the 4x4 matrix with row sums and column products all the 
same. My initial thought was that a computer could just search, but the 
search space is large and that's not really solving a puzzle is it... It 
seemed like it may be possilbe to solve by hand but not easily, so I 
decided to document my effort as I went along. People often give answers 
or proofs, but what you usually don't see is how they arrived at it. I 
fired up notepad and started typing (see attached). The highlight for me 
was discovering that there are at most 10 possible sets of 16 numbers in 
the solution. I suspect it can be solved with any of them, but that 
would be an job for a computer. One final solution was enough for me. A 
fine puzzle overall.

BTW, you have my permission to post my solution on mathpuzzle.com 
(reformatted as needed).

Thanks,
Paul H. Kahler

Problem: Find a 4x4 matrix whose entries are distinct positive integers less 
than 30, whose row sums are the same, and whose column products are the same.

Solution:

Definitions:
1) P = the product of the enties in any column
2) S = the sum of entries in any row

observations:
29 can not be one of the numbers, as P would contain 29 as a factor and there 
are no other numbers < 30 to provide a factor of 29 for the other columns.
By this reasoning, no prime > 30/4 can be used. This eliminates the following 
numbers from the game:  11,13,17,19,22,23,26,29
25 can not be used, as there are only 5 multiples of 5 less than 30 and using 
5^2=25 would require using 2 multiples of 5 in each of the 3 remaining columns.

This eliminates 9 numbers. Elimination of 4 more would identify the set of 16 
numbers.
There are 4 multiples of 7, one for each column, so they must either all be 
used or none at all.
The same can be said for the remaining multiples of 5 - all or none.

Factoring the remaining possibilities:
Number		2,3,5,7
-----------------------
1 = 1           0,0,0,0
2 = 2           1,0,0,0
3 = 3           0,1,0,0
4 = 2*2         2,0,0,0
5 = 5           0,0,1,0
6 = 2*3         1,1,0,0
7 = 7           0,0,0,1
8 = 2*2*2       3,0,0,0
9 = 3*3         0,2,0,0
10 = 2*5        1,0,1,0
12 = 2*2*3      2,1,0,0
14 = 2*7        1,0,0,1
15 = 3*5        0,1,1,0
16 = 2*2*2*2    4,0,0,0
18 = 2*3*3      1,2,0,0
20 = 2*2*5      2,0,1,0
21 = 3*7        0,1,0,1
24 = 2*2*2*3    3,1,0,0
27 = 3*3*3      0,3,0,0
28 = 2*2*7      2,0,0,1
------------------------
               23,13,4,4

Any primes in the set of 16 numbers must occur a multiple of 4 times.
Therefore, Some multiples of 2 and 3 must go. How many?
Elimination of the 7's would leave 16 numbers with 20 2's and 12 3's.
Elimination of the 5's would do the same. These are both valid sets.

The sum of the 16 numbers must be a multiple of 4 - the total is S*4.
The sum of these 20 remaining numbers is 250.
The sum of the numbers without multiples of 5 is 200. S = 50
The sum of the numbers without multiples of 7 is 180. S = 45

-------------------------------------------------------------
Can both the 5's and the 7's remain?
This would require removing 4 numbers from the set below while
reducing the number of 2's by 3,7,or 11. And the number of 3's
by 1,5,or 9. The sum of the removed numbers must = 2 mod 4.

Number          2,3,5,7
-----------------------
1 = 1           0,0,0,0
2 = 2           1,0,0,0
3 = 3           0,1,0,0
4 = 2*2         2,0,0,0
6 = 2*3         1,1,0,0
8 = 2*2*2       3,0,0,0
9 = 3*3         0,2,0,0
12 = 2*2*3      2,1,0,0
16 = 2*2*2*2    4,0,0,0
18 = 2*3*3      1,2,0,0
24 = 2*2*2*3    3,1,0,0
27 = 3*3*3      0,3,0,0

Some trial sets to remove:
1,2,3,4    Sum = 2 mod 4
1,3,8,16   Sum = 0 mod 4
2,3,4,16   Sum = 1 mod 4
Eliminating only 3 (from multiples of 3) requires removing 1 for parity.

1,4,6,16   Sum is odd. 6 as only multiple of 3 cannot be paired with 1.
2,4,6,8    Sum is 0 mod 4

2,8,16,24  Sum is 2 mod 4
It might be time to write a program to find all removable sets of 4.
There are probably not very many, which would leave few sets of 16.

If S is even, each row must contain an even number of odd numbers.
If S is odd, each row must contain an odd number of odd numbers, and odd
numbers are congruent to 1 mod 2, and 4*1 = 0 mod 2.
There must be an even number of odd numbers in the final set. Valid
subsets of the above numbers must have 0,2,or 4 odd numbers.
The odd numbers are: 1,3,9,27   There are only 6 possible pairs of these:
1,3             0,1,0,0
1,9             0,2,0,0
1,27            0,3,0,0
3,9             0,3,0,0
3,27            0,4,0,0
9,27            0,5,0,0

{1,3,9,27} can not be removed, because is contains 6 3's.
Obviously, removing odd numbers does not affect the number of 2's remaining.
The even numbers that contain 3's are: 6,12,18,24.
The even numbers with no 3's are: 2,4,8,16 which contain 1,2,3,4 2's.
This means for any odd pair, we can select 1 even number to get the right
number of 3's and then select a power of 2 to finnish the 2's.

The exception is if we pick 2 evens that contain 3's, this would have to
satisfy the number of 2's and 3's at the same time.

Remember the sum of the numbers removed must be 2 mod 4, and only 1 power
of 2 is non-zero mod 4.

No software needed, I'll find all sets of 16 tomorrow.... Good night.

Even Numbers    2,3,5,7
-----------------------
2 = 2           1,0,0,0
4 = 2*2         2,0,0,0
6 = 2*3         1,1,0,0
8 = 2*2*2       3,0,0,0
12 = 2*2*3      2,1,0,0
16 = 2*2*2*2    4,0,0,0
18 = 2*3*3      1,2,0,0
24 = 2*2*2*3    3,1,0,0

1,3,2,4         Sum = 2 mod 4
1,3,8,16        Sum = 0 mod 4
1,9,18,12       Sum = 0 mod 4
1,27,6,12       Sum = 2 mod 4
1,27,18,4       Sum = 2 mod 4
3,9,6,12        Sum = 2 mod 4
3,9,18,4        Sum = 2 mod 4
3,27,6,4        Sum = 0 mod 4
3,27,12,2       Sum = 0 mod 4
3,27,24,16      Sum = 2 mod 4
9,27,2,4        Sum = 2 mod 4
9,27,8,16       Sum = 0 mod 4

This completes the removable sets containing odd numbers.
There is still the possibility of removing 4 even numbers.
There are only a few ways to get 1 or 5 3's with the even numbers.
6,12,18,24 does not remove the correct number of 2's.
The only other way to get the right number of 3's is with one of {6,12,24} with 
three powers of 2.
2 = 2           1,0,0,0
4 = 2*2         2,0,0,0
8 = 2*2*2       3,0,0,0
16 = 2*2*2*2    4,0,0,0

{6,2,4,8,16} has 11 2's, and we need 3,7,or 11, so we must remove 16
6,2,4,8         Sum = 0 mod 4

{12,2,4,8,16} has 12 2's, so we throw out 2.
12,4,8,16       Sum = 0 mod 4

{24,2,4,8,16} has 13 2's, so we throw out 4
24,2,8,16       Sum = 2 mod 4

This last set is the only one containing no odd numbers that meets all the 
criteria.
It's time to list all the sets of 4 numbers that can be removed from the 
original set of 20.
The original set was:
{1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,21,24,27,28}

There are 10 sets of 4 that can be removed from this:
The number of primes removed (2,3,5,7) is to the right
{5,10,15,20}         3,1,4,0
{7,14,21,28}         3,1,0,4
{1,2,3,4}            3,1,0,0
{1,6,12,27}          3,5,0,0
{1,4,18,27}          3,5,0,0
{3,6,9,12}           3,5,0,0
{3,4,9,18}           3,5,0,0
{3,16,24,27}         7,5,0,0
{2,4,9,27}           3,5,0,0
{2,8,16,24}         11,1,0,0

This allows the following 10 sets of 16 numbers in the matrix:

{1,2,3,4,6,7,8,9,12,14,16,18,21,24,27,28}       20,12,0,4   P=6048   S=50
{1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,27}       20,12,4,0   P=4320   S=45
{5,6,7,8,9,10,12,14,15,16,18,20,21,24,27,28}    20,12,4,4   P=30240  S=58
{2,3,4,5,7,8,9,10,14,15,16,18,20,21,24,28}      20, 8,4,4   P=10080  S=51
{2,3,5,6,7,8,9,10,12,14,15,16,20,21,24,28}      20, 8,4,4   P=10080  S=50
{1,2,4,5,7,8,10,14,15,16,18,20,21,24,27,28}     20, 8,4,4   P=10080  S=55
{1,2,5,6,7,8,10,12,14,15,16,20,21,24,27,28}     20, 8,4,4   P=10080  S=54
{1,2,4,5,6,7,8,9,10,12,14,15,18,20,21,28}       16, 8,4,4   P=5040   S=45
{1,3,5,6,7,8,10,12,14,15,16,18,20,21,24,28}     20, 8,4,4   P=10080  S=52
{1,3,4,5,6,7,9,10,12,14,15,18,20,21,27,28}      12,12,4,4   P=7560   S=50

These are the only sets of 16 positive numbers that satisfy these criteria:
1) all less than 30.
2) All prime factors occur in multiples of 4 (needed for the products)
3) sum of all numbers is a multiple of 4 (needed for sums)
4) there are an even number of odd numbers in the set (needed for criteria 3)

A final solution can have any permutations of it's rows or columns.
That allows 576 permutations of any valid solution to be valid as well.

New day: Today I will try to find a solution by hand.
The approach will be to divide a set of numbers into 2 sets of 8 whose product 
is P^2.
Then each of those sets can be divided into 2 sets whose product is P.
A similar approach can be used for the sums.

The 5th set looked like a good start:
{2,3,5,6,7,8,9,10,12,14,15,16,20,21,24,28}      20, 8,4,4   P=10080  S=50
But I found it difficult to work with the number 16 which had too many 2's in 
it.

The 8th set does not have 16 or 27, so the high counts are reduced.
{1,2,4,5,6,7,8,9,10,12,14,15,18,20,21,28}       16, 8,4,4   P=5040   S=45

5 = 5           0,0,1,0
15 = 3*5        0,1,1,0
14 = 2*7        1,0,0,1
28 = 2*2*7      2,0,0,1

1 = 1           0,0,0,0
4 = 2*2         2,0,0,0
12 = 2*2*3      2,1,0,0
18 = 2*3*3      1,2,0,0
-----------------------
                8,4,2,2

2 = 2           1,0,0,0
6 = 2*3         1,1,0,0
8 = 2*2*2       3,0,0,0
9 = 3*3         0,2,0,0

10 = 2*5        1,0,1,0
20 = 2*2*5      2,0,1,0
21 = 3*7        0,1,0,1
7 = 7           0,0,0,1
-----------------------
                8,4,2,2

Using just the top half:
5 = 5           0,0,1,0
14 = 2*7        1,0,0,1
4 = 2*2         2,0,0,0
18 = 2*3*3      1,2,0,0
-----------------------
                4,2,1,1

1 = 1           0,0,0,0
12 = 2*2*3      2,1,0,0
15 = 3*5        0,1,1,0
28 = 2*2*7      2,0,0,1
-----------------------
                4,2,1,1

Now for the bottom half:
20 = 2*2*5      2,0,1,0
21 = 3*7        0,1,0,1
2 = 2           1,0,0,0
6 = 2*3         1,1,0,0
-----------------------
                4,2,1,1

8 = 2*2*2       3,0,0,0
9 = 3*3         0,2,0,0
10 = 2*5        1,0,1,0
7 = 7           0,0,0,1
-----------------------
                4,2,1,1

===========================================================
The set has been split into 4 groups with the same product.
These groups can be used as columns in the matrix.

Columns       Row Sum
----------------------
  1  2  4  7    14
 12  6  5  8    31
 15 20 14  9    58
 28 21 18 10    77

Any numbers in a column can be exchanged without affecting the products, so 
arrange to get sums = 45
This set contains 6 odd numbers, so one row will need to contain 3 of them, 
with 1 in each remaining row.

Columns       Row Sum
----------------------
  1 21 14  9    45
 12  6  4  8    30
 15 20  5  7    47
 28  2 18 10    58

The top row contains 3 odd numbers, and each remaining row will need to have 1 
odd number to reach 45.

Columns       Row Sum
----------------------
  1 21 14  9    45
 15  2 18 10    45
 12 20  5  8    45
 28  6  4  7    45

Column products are all 5040 and row sums are all 45.
A challenging problem, but one not requiring a computer to solve.

             {  1 21 14  9 }
             { 15  2 18 10 }
             { 12 20  5  8 }
             { 28  6  4  7 }

Solution by Paul Kahler November 17th 2002.
The only tools used were the standard "notepad" and "calculator" that come with 
MS Windows.
The only portion deleted was my first attempt to partition a set of 16 into 
columns.
Rows and column entries were swapped in this file like a movable puzzle, so my 
effort to
solve these sub-puzzles is not documented here.

Copyright 2002 Paul H. Kahler