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Boxing Day (Posted on 2005-09-28) Difficulty: 3 of 5
Fill in each of the boxes with the numbers 1 to 9 so that the three horizontal equations are true.

Each number is used once; no repetition is allowed.

The calculations are done from left to right (i.e., 7-2x3=15, not 1) and may involve only positive numbers (so 5-3-7=-5 is not allowed).

                               _     _     _     _
 _     _     _     _     _    | | - | | x | | = | |
| | x | | x | | = | | + | | + |_| + |_| + | | + |_|
|_| - |_| x |_| = |_| - |_|             - |_|

See The Solution Submitted by Lisa    
Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 7
Here is the answer ordered how they appear left to right

8 4 1 9 2 7 5 3 6

I started with the middle equation since it offered the greatest contraints on the solutions.

Any 6 digits from the set given will add to a minimum of 21 and a maximum of 39.

Then I wrote out the sets of 3 digits I could multiply together from the set that whose product fit the contraints above.  Order is not important yet.

Since the sum of 1 through 9 is 45, I wrote the sum of the 3 digits next to the product of these 3 digit, and found which product and sum added to 45.  This gave me the set of numbers on either side of the middle equation.  Only one set met the contraint.

This made the other 2 equations easy to solve.





  Posted by Robert on 2005-09-28 17:00:43
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