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

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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. Example     ÷     ×     20 ×   +   ×     ×   ×   189 -   +   -     ÷   +   10 20   10   39   Labels  A  ÷  B  ×  C  20 ×   +   ×   D × E × F 189 -   +   -   G ÷ H + I 10 20   10   39          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.  4  ÷  1  ×  5  20 ×   +   ×   D × E × F 189 -   +   -   G ÷ H + I 10 20   10   39    8  ÷  2  ×  5  20 ×   +   ×   D × E × F 189 -   +   -   G ÷ H + I 10 20   10   39    5  ÷  1  ×  4  20 ×   +   ×   D × E × F 189 -   +   -   G ÷ H + I 10 20   10   39          Next let's look at the right column. We have C×F-I equal to 39. Here, C must be 4 or 5. We can immediately eliminate 4 because in order for 4×F-I 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×F-I must be either 5×8-1 or 5×9-6.        We can eliminate 5×8-1 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×F-I has to be 5×9-6. The possibilities are now  4  ÷  1  ×  5  20 ×   +   ×   D × E × 9 189 -   +   -   G ÷ H + 6 10 20   10   39    8  ÷  2  ×  5  20 ×   +   ×   D × E × 9 189 -   +   -   G ÷ H + 6 10 20   10   39          In each case there is only one possiblity for the left column A×D-G to equal 20. These are 4×7-8 and 8×3-4. This gives us  4  ÷  1  ×  5  20 ×   +   ×   7 × E × 9 189 -   +   -   8 ÷ H + 6 10 20   10   39    8  ÷  2  ×  5  20 ×   +   ×   3 × E × 9 189 -   +   -   4 ÷ H + 6 10 20   10   39          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.  4  ÷  1  ×  5  20 ×   +   ×   7 × 3 × 9 189 -   +   -   8 ÷ 2 + 6 10 20   10   39    8  ÷  2  ×  5  20 ×   +   ×   3 × 7 × 9 189 -   +   -   4 ÷ 1 + 6 10 20   10   39          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  8  ÷  2  ×  5  20 ×   +   ×   3 × 7 × 9 189 -   +   -   4 ÷ 1 + 6 10 20   10   39

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