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1/28/2013 5:01:36 PM

wa4lrm
wa4lrm
Posts: 54
The March 2013 on Logic Lover's Math & Logic Problems has the worst kind on Sum Logic on page 47. Now they use same numbers total of three or four times on both two rows and two columns. It makes it impossible to solve this problem starting with three different assumption and guessing which one to use. No matter which rows or column you use, you cannot even get this puzzle started. I had erased the wrong guess so many times that it rip the paper and couldn't do Cross Sum 18 on next page. If this is the future design, I won't do them again. I like the way it was on past issues which you can get it started.upset

2/4/2013 9:51:29 AM

admin
admin
Administrator
Posts: 108
wa4lrm wrote:
The March 2013 on Logic Lover's Math & Logic Problems has the worst kind on Sum Logic on page 47. Now they use same numbers total of three or four times on both two rows and two columns. It makes it impossible to solve this problem starting with three different assumption and guessing which one to use. No matter which rows or column you use, you cannot even get this puzzle started. I had erased the wrong guess so many times that it rip the paper and couldn't do Cross Sum 18 on next page. If this is the future design, I won't do them again. I like the way it was on past issues which you can get it started.upset


We’re sorry this puzzle caused you frustration! We checked with our editors and they suggested the following strategy (because there are 2 puzzles on the page you mention, we’ve provided strategies for both):

For the first Sum Logic:
Take a look at the second row of numbers: you’re given 26, 15, and 7, which total 48. The goal number is 63, so for this row you need 15 more. The only two possible numbers that total 15 are the pairs 14 and 1, or 11 and 4. Since the 15 is already entered in the fifth box in that row, this eliminates the 4-11 possibility, leaving the two numbers that go in that row to be 14 and 1. Now there are two 14s and two 1s available, but again the 15 is in the fifth box, eliminating one of the 14s, and therefore leaving the other 14 to be placed in the second box of that row.

For the second Sum Logic:
This puzzle is a little trickier, since it uses negative numbers. Examining the first column of numbers, it can be seen that it consists of 7, 4, and 5, which total 16. Since the goal number is 13, this means you need to subtract 3. The two available numbers that total –3 are –14 and 11. Since there are 3 –14s and only one 11, that single 11 can immediately be placed in the diagram in the fourth box down the first column.

Whew, we hope this helps!

Best,
Admin

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