We’ll solve for A. First we will find the possibilities for A by the row and then A by the column.
A by the Row: Unique combination
According to A’s column: A+B= 4
The only possible combination for A and B are 3 and 1 (in some order).
Possible values for A by the row:
| By: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Row | NO | NO | NO | NO | NO | NO | NO |
A by the Column: Lowest Possible – 2 Cells
According to A’s Column: A+D = 12
If A was 2, then D would have to be 10 and that is not possible. So A is 3 or greater. The LOWEST possible value for A or D is 3.
Possible Values for A by the column are:
| By: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Column | NO | NO |
Solve for A: Combine Row and Column Possibilities
Combine column possibilities (A = 1 or 3) and row possibilities (A >= 3). The only value that would make both of these possible is A=3
| By: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Row |
NO |
|
NO | NO | NO | NO | NO | NO | |
| Column | NO |
NO |
|
||||||
| Both | NO |
NO |
YES |
NO | NO | NO | NO | NO | NO |
So A=3
Solve B,C,D,E: Complete Sums
Solve B: A+B = 4; 3+B=4; B=4-3; B=1
Solve D: A+D=12; 3+D=12; D=12-3; D=9
Solve C: B+C=9; 1+C=9; C=9-1; C=8
Solve E: C+D+E=19; 8+9+E=19; E=19-8-9; E=2


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