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Arithmetic Grid II (Posted on 2014-06-12) Difficulty: 3 of 5
In the following grid:

(1) The numbers in each column form an increasing arithmetic sequence
(2) The numbers in each row form an increasing arithmetic sequence
(3) The X's in each spot indicate how many digits are in the number
(4) A is at least 1200
(5) B is at least 2900

+---+---+----+----+----+----+----+----+
|  0| XX|  XX|  XX|  XX|  XX|  XX| XXX|
+---+---+----+----+----+----+----+----+
| XX|XXX| XXX| XXX| XXX| XXX| XXX| XXX|
+---+---+----+----+----+----+----+----+
|XXX|XXX| XXX| XXX| XXX| XXX| XXX| XXX|
+---+---+----+----+----+----+----+----+
|XXX|XXX| XXX| XXX| XXX| XXX|XXXX|XXXX|
+---+---+----+----+----+----+----+----+
|XXX|XXX| XXX| XXX|XXXX|XXXX|XXXX|XXXX|
+---+---+----+----+----+----+----+----+
|XXX|XXX| XXX|XXXX|XXXX|XXXX|XXXX|XXXX|
+---+---+----+----+----+----+----+----+
|XXX|XXX|XXXX|XXXX|XXXX|XXXX|XXXX|XXXX|
+---+---+----+----+----+----+----+----+
|XXX|XXX|   A|XXXX|XXXX|XXXX|XXXX|   B|
+---+---+----+----+----+----+----+----+
Reconstruct the grid.
An analytic solution is preferred.

No Solution Yet Submitted by Jer    
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Solution Analytical solution | Comment 1 of 2

Let r1 be the common difference among terms in the first row, r2 be the common difference between terms in the second row, etc. 

This solution relies on the insight that r1, r2, ..., r7 must themselves form an arithmetic progression, with common difference d.*  Similarly, if c1, c2, ..., c7 are the common differences between terms in the columns, respectively, then these are also an arithmetic progression and share the same common difference d.

We can then compute the value of any individual cell as a combination of r1, c1, d, and its location in the grid (given as the ith row and jth column, starting with 0), as follows:

i,j = i*c1 + j*r1 + i*j*d

So in particular we are told that:

B = 7c1 + 7r1 + 49d >= 2900

A = 7c1 + 2r1 + 14d >= 1200

Also note that cell (3,5) is only three digits, and thus 3c1 + 5r1 + 15d < 1000.

The first row contains 6 two-digit numbers and then becomes a three-digit number in the final column.  This is only possible if r1 = 15 or r1 = 16. 

Let r1 = 15.  Plugging this in and doing a little simplfying to the above inequalities gives us the following:

[1]  c1 + 7d >= 400

[2]  c1 + 2d >= 168

[3]  c1 + 5d <= 308

Considering [1] and [3], we can see that 2d must be at least 400 - 308 = 92, so d >= 46.

Considering [2] and [3], we can see that 3d can be no more than 308 - 168 = 140, so d <= 46.67.

Therefore, d = 46.  Plugging this into the above inequalities shows us that 77.3 <= c1 <= 78.3, so c1 = 78.

With r1, c1, and d we can populate the entire grid and confirm that it conforms to all of the conditions of the puzzle:

0 15 30 45 60 75 90 105
78 139 200 261 322 383 444 505
156 263 370 477 584 691 798 905
234 387 540 693 846 999 1152 1305
312 511 710 909 1108 1307 1506 1705
390 635 880 1125 1370 1615 1860 2105
468 759 1050 1341 1632 1923 2214 2505
546 883 1220 1557 1894 2231 2568 2905

 


* - I briefly thought I had a concise way to show why this is the case but as I worked through the solution the insight I had has escaped me.  I'll leave it to someone less scatterbrained than I am to complete this solution by proving that the differences themselves form an arithmetic progression.

 

Edited to add: When r1 = 16, we also find that d = 46 but then we get something like 76.2 < c1 < 76.7, so the solution presented above is the only one.

 

Edited on June 12, 2014, 5:45 pm
  Posted by tomarken on 2014-06-12 17:25:01

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