Place the numbers 1 through 9 in the grid below:

   1 2 3  
A |_|_|_|  
B |_|_|_|  
C |_|_|_|  
 

1. The product of the numbers in row C is twice the sum of the numbers in row C.
2. The product of the numbers in column 3 is twice the sum of the numbers in column 3.
3. If you ignore column 1, all the columns have even products.
4. If you take the six-digit number C3 B3 B1 C1 A3 B2 and multiply it by 3, you get the six-digit number B3 B1 C1 A3 B2 C3.

Try to figure it out without using a computer program.

1*4*5 = 20 = 10*2 = (1+4+5)*2
1*3*8 = 23 = 12*2 = (1+3+8)*2

These, I think, are the only three digit combinations that satisfy the equations for statements 1 and 2. Therefore, the digit in common, 1, must be C3.

Now, using statement 4, some number multiplied by 3 results in some number ending with a 1. So the ones digit in the original number must be a 7. Now, some number ending in a 7 multiplied by 3 results in some number ending with 71. So the original number must end in 57. We can repeat this process to figure out that the equation referenced in statement 4 is 142857 * 3 = 428571. So we know all the positions except A1 and A2, which must consist of the 6 and the 9. Since B2 and C2 are odd but we know from statement 3 that the second column's product must be even, we conclude that A2 must be the 6. Hence we have:

965
274
831

Posted by Avin on 2005-05-17 18:47:04