BLUEBIRD MATH CIRCLE �
13 April 2023
Mark Saul, Ph.D.
“Think deeply of simple things.”
-- Arnold Ross
MAGIC SQUARES
There are six identical glasses in a row. The first three are full of water, and the last three are empty.
How can you arrange it so that the full and empty glasses alternate—touching only one glass?
WARM UP 1
3 | 9 | 2 |
5 | x | 7 |
y | 3 | 8 |
Find numbers x and y so that the three rows of the table above have the same sum.
Rows go left to right (side-to-side)
WARM UP 1
3 | 9 | 2 |
5 | x | 7 |
y | 3 | 8 |
Find numbers x and y so that the three rows of the table above have the same sum.
Solution: The sum of the first row is 14.
For the sum of the second row to be 14, x must be 2.
For the sum of the third row to be 14, y must be 3.
WARM UP 2
4 | 8 | 7 |
6 | 10 | 4 |
1 | 9 | 8 |
In the table above, the sums of each of the three columns is different.
But if you switch the positions of two of the nine numbers, the sums of the columns will be the same. Can you find the two numbers to switch?
4 8 7
6 10 4
+ 1 + 9 +8
11 27 19
Columns go top to bottom (up-and-down)
WARM UP 2
4 | 8 | 7 |
6 | 10 | 4 |
1 | 9 | 8 |
In the table above, the sums of each of the three columns is different.
But if you switch the positions of two of the nine numbers, the sums of the columns will be the same. Can you find the two numbers to switch?
4 8 7
6 10 4
+ 1 + 9 +8
11 27 19
Hint: If the three column-sums are equal, what must each one be?
WARM UP 2
4 | 8 | 7 |
6 | 10 | 4 |
1 | 9 | 8 |
In the table above, the sums of each of the three columns are different.
But if you switch the positions of two of the nine numbers, the sums of the columns will be the same. Can you find the two numbers to switch?
4 8 7
6 10 4
+ 1 + 9 +8
11 27 19
11+27+ 19 = 57
= 3x19
SOLUTION: We follow the hint. The sum of all the numbers (see above left) is 57. Since 57/3 = 19, each of the equal column sums must be 19.
Hint: If the three column-sums are equal, what must each one be?
WARM UP 2
4 | 8 | 7 |
6 | 10 | 4 |
1 | 9 | 8 |
In the table, the sums of each of the three columns is different.
But if you switch the positions of two of the nine numbers, the sums of the columns will be the same. Can you find the two numbers to switch?
4 8 7
6 10 4
+ 1 + 9 +8
11 27 19
11+27+ 19 = 57
= 3x19
SOLUTION (cont’d): The third column-sum is already 19, so we will try not to touch it.
The first column-sum is 11 (too small by 8) and the second is 27 (too big, also by 8). So we need to look for two numbers to switch which differ by 8.
WARM UP 2
4 | 8 | 7 |
6 | 10 | 4 |
1 | 9 | 8 |
In the table, the sums of each of the three columns is different.
But if you switch the positions of two of the nine numbers, the sums of the columns will be the same. Can you find the two numbers to switch?
4 8 7
6 10 4
+ 1 + 9 +8
11 27 19
11+27+ 19 = 57
= 3x19
SOLUTION (cont’d): In fact, if we just switch the 1 and the 9, we will have what we need:
4 | 8 | 7 |
6 | 10 | 4 |
9 | 1 | 8 |
WARM UP 3
3 | 17 | 15 |
13 | 9 | 5 |
11 | 1 | 7 |
Each of the three columns in the table above adds up to 27. (Check this!)
But each of the rows has a different sum. Can you exchange two numbers so that the column-sums remain equal, and the row-sums also become equal?
WARM UP 3
3 | 17 | 15 |
13 | 9 | 5 |
11 | 1 | 7 |
Each of the three columns in the table adds up to 27. (Check this!)
But each of the rows has a different sum.
Can you exchange two numbers so that the column-sums remain equal, and the row-sums also become equal?
Solution: The trick will be to exchange two numbers in the same column. That way the column-sums will not change, but the row-sums will.
It’s easy to check that each column sum is 27. So this must be the new sum for each row.
Sum: 27 27 27
WARM UP 3
3 | 17 | 15 |
13 | 9 | 5 |
11 | 1 | 7 |
Each of the three columns in the table adds up to 27. (Check this!)
But each of the rows has a different sum.
Can you exchange two numbers so that the column-sums remain equal, and the row-sums also become equal?
Solution: For the row-sums, we now have
3 + 17 + 15 = 35, which is 8 too big.
13 + 9 + 5 = 27, which is just right
11 + 1 + 7 = 19, which is 8 too small.
Column Sum: 27 27 27
Row Sum
35
27
19
WARM UP 3
3 | 17 | 15 |
13 | 9 | 5 |
11 | 1 | 7 |
Each of the three columns in the table adds up to 27. (Check this!)
But each of the rows has a different sum.
Can you exchange two numbers so that the column-sums remain equal, and the row-sums also become equal?
Solution: We now have 3 + 17 + 15 = 35, which is 8 too big.
13 + 9 + 5 = 27, which is just right
11 + 1 + 7 = 19, which is 8 too small.
So we must decrease the bottom row by 8 and increase the top row by 8. Since 15 – 7 = 8, we can just exchange these two numbers.
3 | 17 | 7 |
13 | 9 | 5 |
11 | 1 | 15 |
Breakout Rooms
2. Using your answer to (1), find a way to arrange the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 in three rows of three so that the row-sums and column-sums are all the same.
3. Can you now make the diagonals also have the same sum?
1 | 2 | 3 |
4 | 5 | 6 |
7 | 8 | 9 |
BREAKOUT
Breakout Rooms
*3. Can you accomplish this task in more than one way?
1 | 2 | 3 |
4 | 5 | 6 |
7 | 8 | 9 |
BACK FROM BREAKOUT
DISCUSSION
Here are some solutions that some groups came up with:
2 | 9 | 4 |
7 | 5 | 3 |
6 | 1 | 8 |
8 | 1 | 6 |
3 | 5 | 7 |
4 | 9 | 2 |
8 | 3 | 4 |
1 | 5 | 9 |
6 | 7 | 2 |
How are they different?
How are they the same?
First two: each row has the same numbers, maybe in different orders.
And the rows are in different orders
Second is the same as the first rotated by 180 Degrees.
The five always stays in the middle.
All the even numbers are in the corners.
2 | 7 | 6 |
9 | 5 | 1 |
4 | 3 | 8 |
Discussion notes
The Magic Constant Must be 15
Why? Well, if we add up the sums of the three rows, we must get:
1+2+3+4+5+6+7+8+9 = 45 (except the addends will be in a different order) .
And 45/3 = 15.
a | b | c |
d | e | f |
g | h | i |
If the magic constant (sum of each row) is S,
then:
a+b+c = S
d+e+f = S
g+h+I = S
So S+S+s = 3S = a+b+c+d+e+f+g+h+I,
which is 1+2+3+4+5+6+7+8+9, = 45
(with the addends in some order.
Thus 3S = 45, and S = 15.
The middle number must be 5
Why?
P | | |
| 6 | |
| | q |
Well, try to put some other number in the middle, say 6.
Then the numbers in the two opposite corners, p and q must add up to 15-6 = 9
The middle number must be 5
Why?
P | w | u |
r | 6 | s |
t | v | q |
Well, try to put some other number in the middle, say 6.
Then the numbers in the two opposite corners, p and q must add up to 15-6 = 9
But so must r + s and t+u and v+w. Each pair must add up to 9.
That means that all the nine numbers will add up to 9+9+9+9+6 = 42.
But they don’t! They add up to 45. So 6 in the middle must be wrong.
The middle number must be 5
Why?
The sums along the arrows below are all 15, and the arrows cover each number once EXCEPT the middle number, which is counted FOUR times (once for each arrow).
| | |
| | |
| | |
So the sum of the numbers along all four arrows together is 4x15 = 60.
This sum contains one copy of each of the numbers 1 through 9, plus three EXTRA copies of the middle number.
The sum of the numbers 1 through 9 is 45, so the three extra copies of the middle number add up to 60-45=15.
So the middle number is 15/3 = 5.
The numbers in the corners must all be even
Why?
(This is not so simple)
| | |
| 5 | |
| | |
The middle number is 5: we’ve figured that out.
The sum along any row, column, or diagonal must be 15
To get a sum of 15, we can have three odd numbers, or one odd number and two even numbers in a line.. We cannot have just one even number, and we cannot have three even numbers. (Otherwise, the sum will be even, and 15 is not even.)
We will show that a ‘middle’ square cannot contain an even number.
The numbers in the corners must all be even
Why?
(This is not so simple)
| | |
e | 5 | e |
| | |
We will show that a ‘middle’ square cannot contain an even number.
Indeed: Suppose we put an even number to the left of the 5.
Then the number to the right must also be even (because we need two even numbers in that middle row.
The numbers in the corners must all be even
Why?
(This is not so simple)
e | | |
e | 5 | e |
| | |
We will show that a ‘middle’ square cannot contain an even number.
Indeed: Suppose we put an even number to the left of the 5.
Then the number to the right must also be even (because we need two even numbers in that middle row.
Now look at the first column. It must contain just one other even number (because it cannot contain just one or all three even numbers). Suppose we put an even number on top.
(We will see that it doesn’t matter if we put an even number on the bottom of the first row: we will reach the same conclusion. )
The numbers in the corners must all be even
Why?
(This is not so simple)
e | | |
e | 5 | e |
| | e |
We will show that a ‘middle’ square cannot contain an even number.
Indeed: Suppose we put an even number to the left of the 5.
Then the number to the right must also be even (because we need two even numbers in that middle row.
Now look at the first column. It must contain just one other even number (because it cannot contain just one or all three even numbers). Suppose we put an even number on top.
(We will see that it doesn’t matter if we put an even number on the bottom of the first row: we will reach the same conclusion. )
Now look at the diagonal. As we saw, it must have another even entry, which must be the lower right hand corner.
The numbers in the corners must all be even
Why?
(This is not so simple)
e | o | o |
e | 5 | e |
o | o | e |
We will show that a ‘middle’ square cannot contain an even number.
Now look at the diagonal. As we saw, it must have another even entry, which must be the lower right hand corner.
But we have now used up all four even numbers we have (2,4,6,8), and the other numbers must be odd.
This is trouble: Look at the top row. Its sum will be even (odd+odd+even = even , and so cannot be 15.
Thus we cannot place an even number to the left of the 5.
The numbers in the corners must all be even
Why?
(This is not so simple)
e | o | o |
e | 5 | e |
o | o | e |
We will show that a ‘middle’ square cannot contain an even number.
Now look at the diagonal. As we saw, it must have another even entry, which must be the lower right hand corner.
But we have now used up all four even numbers we have (2,4,6,8), and the other numbers must be odd.
This is trouble: Look at the top row. Its sum will be even, and so cannot be 15.
Thus we cannot place an even number to the left of the 5.
The same reasoning will hold for any of the other squares in the middle of a row or column.
The numbers in the corners must all be even
Why?
(This is not so simple)
o | o | e |
e | 5 | e |
e | o | o |
The same reasoning will hold for any of the other squares in the middle of a row or column.
And you can check that if we had placed the second even number of the first column on bottom, instead of on top, we would get into the same trouble. We would ‘use up’ the even numbers, and one line would have to have two odd numbers and one even number.
The numbers in the corners must all be even
Why?
(This is not so simple)
e | | e |
| 5 | |
e | | e |
We have shown that no even number can be in the middle square of a row or column.
Where can they go? Only in the corners. There are four corners, and four even numbers, so it all works out.
The corners MUST be even.
The numbers in the corners must all be even.
| | 2 |
| | |
| | |
So how many magic squares are there with the numbers 1,2,3,4,5,6,7,8,9?
Put a 2 in the upper left corner. How many ways can you arrange the other numbers to get a magic square?
Then put a 2 in each of the other corners, and count the possibilities.
Keep a copy of each of the magic squares you create.
BREAKOUT
The numbers in the corners must all be even
2 | | |
| | |
| | |
So how many magic squares are there with the numbers 1,2,3,4,5,6,7,8,9?
Put a 2 in the upper left corner. How many ways can you arrange the other numbers to get a magic square?
Then put a 2 in each of the other corners, and count the possibilities.
Keep a copy of each of the magic squares you create.
BACK FROM BREAKOUT
How many magic squares?
How many magic squares?
Eight:
2 | 9 | 4 |
7 | 5 | 3 |
6 | 1 | 8 |
2 | 7 | 6 |
9 | 5 | 1 |
4 | 3 | 8 |
4 | 3 | 8 |
9 | 5 | 1 |
2 | 7 | 6 |
4 | 9 | 2 |
3 | 5 | 7 |
8 | 1 | 6 |
6 | 7 | 2 |
1 | 5 | 9 |
8 | 3 | 4 |
6 | 1 | 8 |
7 | 5 | 3 |
2 | 9 | 4 |
8 | 3 | 4 |
1 | 5 | 9 |
6 | 7 | 2 |
8 | 1 | 6 |
3 | 5 | 7 |
4 | 9 | 2 |