I'm thinking not, but........

This is what I'm working with--> I have 4 boxes, A,B,C,D....................

(A+B)-C=7

C+D=4

C Equal to or less than 3.

3 posts
• Page **1** of **1**

I'm thinking not, but........

This is what I'm working with--> I have 4 boxes, A,B,C,D....................

(A+B)-C=7

C+D=4

C Equal to or less than 3.

This is what I'm working with--> I have 4 boxes, A,B,C,D....................

(A+B)-C=7

C+D=4

C Equal to or less than 3.

- KakuroFool
**Posts:**2**Joined:**12 July 2014

From your conditions I assume you have a 3-digit horizontal (A,B,C) intersecting a 2-digit vertical (C,D), or vice versa.

There are many ways to satisfy this set of equations.

First, note that your third condition is redundant, because by the second condition we must have either C=1, D=3 or C=3, D=1.

If C=3, plugging that into the first condition yields A+B=10. This can be solved with A,B being either 1,9 or 2,8 or 3,7 or 4,6 or the reverse of any of these. 3,7 doesn't work because we already have the 3. That still leaves three (six, including reversals) possible solutions.

If C=1, then we have A+B=8. That, too, has several solutions: 1,7 and 2,6 and 3,5 (and their reversals). 1,7 is no good because we already have the 1.

So, from just what you state, there are ten solutions (involving both C=3 and C=1, and all reversals).

In general you cannot expect to solve a puzzle based on only 4 cells. You need to think more globally.

By the way, where does (A+B)-C=7 come from? Normally you are given the sums, not the differences, of the cells in a word. If your condition instead is that A+B+C=7, then the solution is easy, because A,B,C must then be 1,2,4 (in some order). Since C must be 1 or 3, it follows that it must be 1. Then D=3, and A,B are 2,4 (in some order).

Bill Smythe

There are many ways to satisfy this set of equations.

First, note that your third condition is redundant, because by the second condition we must have either C=1, D=3 or C=3, D=1.

If C=3, plugging that into the first condition yields A+B=10. This can be solved with A,B being either 1,9 or 2,8 or 3,7 or 4,6 or the reverse of any of these. 3,7 doesn't work because we already have the 3. That still leaves three (six, including reversals) possible solutions.

If C=1, then we have A+B=8. That, too, has several solutions: 1,7 and 2,6 and 3,5 (and their reversals). 1,7 is no good because we already have the 1.

So, from just what you state, there are ten solutions (involving both C=3 and C=1, and all reversals).

In general you cannot expect to solve a puzzle based on only 4 cells. You need to think more globally.

By the way, where does (A+B)-C=7 come from? Normally you are given the sums, not the differences, of the cells in a word. If your condition instead is that A+B+C=7, then the solution is easy, because A,B,C must then be 1,2,4 (in some order). Since C must be 1 or 3, it follows that it must be 1. Then D=3, and A,B are 2,4 (in some order).

Bill Smythe

- Smythe Dakota
**Posts:**536**Joined:**11 February 2006

You came up with what I came up with, lol. I was hoping my algebra was just weak..............

A+B was paired together going down. C was in a corner, and wrapped through 15 or so squares away around a couple other corners was D, putting A+B in the middle. I solved it with some discover info, thank you.

And I didn't post more, because I wanted as little help as possible. I wanted to make sure that I was understanding the problem correctly.

A+B was paired together going down. C was in a corner, and wrapped through 15 or so squares away around a couple other corners was D, putting A+B in the middle. I solved it with some discover info, thank you.

And I didn't post more, because I wanted as little help as possible. I wanted to make sure that I was understanding the problem correctly.

- KakuroFool
**Posts:**2**Joined:**12 July 2014

3 posts
• Page **1** of **1**