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Fill the 4x4 grid with numbers to make eight arithmetic progressions Question
In the 4x4 grid below, fill each empty square with a number such that the numbers in each row form an arithmetic progression when read from left to right. Similarly, the numbers in each column should form an arithmetic progression when read from top to bottom.
(An arithmetic progression is a sequence in which each term after the first is obtained from the previous term by adding the same constant which might be negative, zero or positive.
For example, 3, 5, 7, 9 is a four term arithmetic progression.)
Inspiration: 2016 Gauss Contest
2 answers
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In some Mathematical puzzles, there are several unknown values. In this particular puzzle, there are 12 unknown numbers that are to be placed in the 12 empty squares.
We could attempt to solve this puzzle by using 12 variables, one for each unknown number, plus a number of equations relating these 12 variables. But this would be a tedious way to determine the numbers for the empty squares.
With a bit of ingenuity, we can solve this puzzle using only 1 variable.
Solution
Step 1:
Let the common difference between consecutive numbers in the top row be d.
So the grid is:
27 27+d 27+2d 27+3d
55
45
25
Step 2:
In the rightmost column the common difference between consecutive numbers is:
55-(27+3d) = 28-3d.
So the number below the 55 is:
55+(28-3d) = 83-3d.
So now the grid is:
27 27+d 27+2d 27+3d
55
45 83-3d
25
Step 3:
In the third row the common difference between consecutive numbers is:
(83-3d)-45 = 38-3d.
So the number to the left of the 45 is:
45-(38-3d) = 7+3d.
So now the grid is:
27 27+d 27+2d 27+3d
55
7+3d 45 83-3d
25
Step 4:
In the second column, the first and third numbers are:
27+d and 7+3d.
Twice the common difference between consecutive numbers in the second column is:
(7+3d) - (27+d) = -20+2d.
So the common difference between consecutive numbers in the second column is:
(-20+2d)/2 = -10+d.
So the second number in the second column is:
(27+d) + (-10+d) = 17+2d.
So now the grid is:
27 27+d 27+2d 27+3d
17+2d 55
7+3d 45 83-3d
25
Step 5:
Now all our hard work is about to yield a great result.
Using the common difference between consecutive numbers in the second column, the number at the bottom of the second column (25) is also equal to:
(7+3d) + (-10+d) = -3+4d.
Therefore, 25 = -3+4d.
Solving we get d = 7.
Using this value of d, we can calculate some of the grid numbers.
So now the grid is:
27 34 41 48
31 55
28 45 62
25
Step 6:
Finally, we can now fill in the rightmost column and then the rows to get:
27 34 41 48
19 31 43 55
11 28 45 62
3 25 47 69
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Solution
27 34 41 48
19 31 43 55
11 28 45 62
3 25 47 69
Reasoning
Filling in the first column as an arithmetic progression and then the rows as arithmetic progressions we get27 27+b 27+2b 27+3b
27+a 27+a+c 27+a+2c 27+a+3c
27+2a 27+2a+d 27+2a+2d 27+2a+3d
27+3a 27+3a+e 27+3a+2e 27+3a+3e
We have five unknowns. Taking any of the other columns as an arithmetic progression we get $c-2d+e = 0$ and $b-2c+d = 0$; together with the three other given values we have five linear equations in five unknowns and we can apply Gaussian elimination.
The stated inspiration is a hint that this was the intended solution process. It could also have been done with twelve variables, one for each empty square, but it would have been less straightforward to identify the redundant equations, and they would also have been more fiddly to manipulate.
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