
Cross Sums Example
The Contest Center 59 DeGarmo Hills Road Wappingers Falls, NY 12590 

Let's work an example of a Cross Sum puzzle. To make it easier
to explain the process, the boxes have been labeled A through I. You may
also find this useful when you work the puzzles yourself.
On the top row we have A÷B×C equal to 20. So A÷B and C are two numbers that multiply to make 20. The only numbers that can be multiplied to make 20 are 1×20, 2×10, 4×5, 5×4, 10×2 and 20×1. Since the highest possible value for A÷B is 9÷1, which is 9, we can eliminate 10×2 and 20×1. Since the highest possible value for C is also 9 we can eliminate 1×20 and 2×10. So either A÷B is 4 and C is 5, or A÷B is 5 and C is 4. There are only a few ways that A÷B could be 4 or 5, namely 4÷1, 8÷2, or 5÷1. This means that B can only be 1 or 2, and that C can only be 4 or 5. Let's put these all together, and look at the possibilities.
Next let's look at the right column. We have C×FI equal to 39. Here, C must be 4 or 5. We can immediately eliminate 4 because in order for 4×FI to be 39, F would have to be at least 10. This is impossible since all of the numbers must be from 1 to 9. So C must be 5. Now C×FI must be either 5×81 or 5×96. We can eliminate 5×81 because of the middle row, D×E×F which equals 189. If F were 8, then D×E×F would be even, but 189 is an odd number, so F cannot be 8. This means C×FI has to be 5×96. The possibilities are now
In each case there is only one possiblity for the left column A×DG to equal 20. These are 4×78 and 8×34. This gives us
Now there is only one blank space left in each row, and we can just fill those in by sight to make the rows work out.
In the left grid above, the middle column is 1+3+2 which does not equal 10, while the right grid has 2+7+1 which is 10. So the answer is

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