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Calendar Capers

Stage: 3 Challenge Level: Challenge Level:1

Adding the numbers in the corners

Sophie from Burford School in the UK sent us some examples and noticed that:

The four numbers in the corners have to be either ALL even or ALL odd. If ALL four numbers are odd, when they are added together, the result will be even! If ALL four numbers are even, then the result will also be even.

Luisa, Polina and Irene from King's College Alicante in Spain wrote the equation:

$(p+k+l+h)/4=x$
where
$p$=top right number
$k$=top left number
$l$=bottom right number
$h$=bottom left number
$x$=the middle number
 
Furthermore, they claimed that
It works for every odd number square - eg: 3 x 3, 5 x 5...
We also have worked out the equation for the even number square like 4 x 4,
6 x 6...
It's basically the same but instead of finding the centre number (which does not exist), we had to find the mean of the centre four numbers. That mean is called $x,$ and then you can
apply it to the equation.
Later on we discovered there are more shapes that can apply to this equation. It works as long as the corner numbers have equal gaps.

Patrick from Bangkok Patana School in Thailand did a similar thing in words, and explained why it works:
If the totals are divided by four it makes the number in the centre of the square. This is because by dividing by four it makes the average. It is the number in the centre because:
Diagonally from the northeast corner to the southwest corner, the distance is 6 because it is one week minus one day. The plus 6 and the minus 6 cancel each other out.
Diagonally from the northwest corner to the southeast corner, the distance is 8 because it is one week plus one day. The plus 8 and the minus 8 cancel each other out.

Aline, also from King's School Alicante, noticed that
to find the numbers at the top left corner and bottom right of the 3 by 3 square, you would have to use $x-8$ to find the top left, $x+8$ to find the bottom right.

I used $-8$ because it would be up a line (relative to $x$) because it's in weeks ($7$ days) then $-1$ ($8$ days) and the same with $+8$ but adding.

Siddhant from Indus International School in India used this idea to show why the formula works algebraically.
Suppose we take the centre number as $x$. This will give us a variable to compare all the numbers.

To make it easier, we will take the sum of the 4 corners to be $y$. The top left corner will always be $x - 8$, and the top right would be $x-6$. The bottom left would be $x + 6$, and the bottom right would be $x + 8$. So if you add them all together, you would get $(x -8) +(x - 6) + (x + 6) + (x + 8)$. When you open the brackets and group the numbers and the variables, you will get $4x
+ 0 = y$. Therefore, the sum of the four corners will always be 4 times the middle number.


Adding the numbers in the rows, columns and diagonals that pass through the centre
Patrick noticed that
adding the numbers in each row, column and diagonal that pass through the centre number results in the same total.

The column is equal to the diagonal because in the diagonal northeast to southwest the corner northeast is 1 more than the northern number and the southwest corner is 1 less than the southern number. This then equalizes to make the same result; it is the same for the opposite diagonal.

The column is equal to the horizontal because the northern number is 6 less than the eastern number, and the southern number is 6 more than the western number which then equalizes. The difference is six because inside the columns of a calendar the numbers grow by 7, which corresponds to the number of days in a week. Inside the rows of a calendar the numbers grow by 1, which corresponds to 1 day at a time. So it is 6 because 7-1=6.

When the total of the diagonals, horizontals or the verticals is divided by three it makes the centre number. This is because the sum divided by three is the average. The average is the number in the centre because the distance from the centre is the same on both sides.


Adding the numbers in the top and bottom rows and left and right columns
Sophie noticed that 

In the row sums, there are jumps of +3 going right, and in the column sums, there are jumps of +21 going down.

Patrick and Siddhant both used algebra. This is Patrick's work:
The sum of the {left and right} columns and the sum of the {top and bottom} rows are equal. The sum of the columns or the sum of the rows divided by 6 is equal to the centre number. This is because divided by 6 is finding the average.

Let $x$ denote the number at the centre. The top row is $x-8,$ $x-7,$ $x-6$ and the bottom row is $x+6,$ $x+7,$ $x+8$ which then equalizes.

The left column is $x-8,$ $x-1,$ $x+6$ and the right column is $x-6,$ $x+1,$ $x+8$ which then equalizes.
Siddhant also noted that the expressions in both cases add up to $6x.$


Circling and putting lines through numbers
Ben from Sheldon School in England used an example to explain why this process always gives the same result, regardless of which numbers in the block are chosen:

They go to the same number because, if the first number was say 6 in the top left hand corner the next number will be 14 or 15. If you chose 15 the next number will be 21, and if you chose 14 the number will be one higher because 14 is one lower than 15, which means it must be 22, which adds up to 42, the same as 6+15+21.

Patrick and Siddhant both explained this in a more general case. This is Siddhant's work:
When you look at the columns in a calender, you will see that the left column consists of the lowest number, the middle column will have the numbers in between, and the right column will have the highest number.
Hence, no matter which numbers you will choose, the numbers will always be one low number, one middle number, and one high number. You will then notice that these numbers always cut each other, and you will invariably be left with $3x.$
For the given example, they have chosen the low number as $x - 8,$ the high number as $x + 1$ and the middle number as $x + 7.$ The top left, the middle right and middle bottom.
If you do the maths, you will be left with $3x + 0.$
Therefore, the 3 numbers will always add up to 3 times the middle number.