Std 6 Mathematics · Chapter 19

19. Magical Mathematics - 2

Complete solutions and explanations for the Ramanujan Square, the Brahma Tower (Tower of Hanoi), the pan balance puzzle and the sum of areas of squares. Maharashtra State Board (Balbharati), Grade Six Mathematics.

About the chapter

This chapter has some interesting puzzles. We study the magic square made by Srinivas Ramanujan with his date of birth. We play the Brahma Tower game and find a formula for it. We find a light object with a pan balance. We also join two squares to make one new square.

Key points

In the Ramanujan square, every row, column and diagonal has the same sum, 139.
In the Brahma Tower, only one disc is moved at a time. A big disc is never placed on a small disc.
Each time a disc is added, the number of moves doubles and one more move is added.
A pan balance can give three results. So we divide the objects into three groups.
When two squares are cut and joined into one square, the areas add up. The perimeters do not.

Formulas and terms

Magic square – a square of numbers in which every row, column and diagonal has the same sum.
Minimum moves in Brahma Tower: t = 2ⁿ − 1 (n = number of discs)
2ⁿ – 2 multiplied by itself n times. 2³ = 2 × 2 × 2 = 8
Area of a square = side × side

1. Sir Srinivas Ramanujan – the Ramanujan Square

✦Add up the group of four numbers in the square formed by his date of birth. Does the sum of horizontal, vertical and diagonal lines add up the same?

The Ramanujan square built from 22-12-1887.

The first row is Ramanujan's date of birth, 22-12-1887. It is written as 22, 12, 18 and 87.

Row 1 : 22 + 12 + 18 + 87 = 139
Row 2 : 88 + 17 + 9 + 25 = 139
Row 3 : 10 + 24 + 89 + 16 = 139
Row 4 : 19 + 86 + 23 + 11 = 139
Column 1 : 22 + 88 + 10 + 19 = 139
Column 2 : 12 + 17 + 24 + 86 = 139
Column 3 : 18 + 9 + 89 + 23 = 139
Column 4 : 87 + 25 + 16 + 11 = 139
Diagonal : 22 + 17 + 89 + 11 = 139
Diagonal : 87 + 9 + 24 + 19 = 139

Yes. The sum of every horizontal, vertical and diagonal line is 139.

The four corner numbers also add up to 139. 22 + 87 + 19 + 11 = 139

✦Can you make a square like this with your date of birth?

Yes. Write your date of birth as four numbers: day (D), month (M), first two digits of the year (Y1) and last two digits of the year (Y2). Then fill the square in this way:

D	M	Y1	Y2
Y2 + 1	Y1 − 1	M − 3	D + 3
M − 2	D + 2	Y2 + 2	Y1 − 2
Y1 + 1	Y2 − 1	D + 1	M − 1

Every row, column and diagonal will add up to D + M + Y1 + Y2.

Check with 22-12-1887
D = 22, M = 12, Y1 = 18, Y2 = 87
Row 2 : 87 + 1 = 88, 18 − 1 = 17, 12 − 3 = 9, 22 + 3 = 25
Row 4 : 18 + 1 = 19, 87 − 1 = 86, 22 + 1 = 23, 12 − 1 = 11
This gives the same square as Ramanujan's.

2. Brahma Tower (Tower of Hanoi)

1Play the game with three discs. Count the total number of times you move discs to move all three discs to stick 3.

The three discs must end up on stick 3 in the same order.

Smallest disc: stick 1 → stick 3
Middle disc: stick 1 → stick 2
Smallest disc: stick 3 → stick 2
Largest disc: stick 1 → stick 3
Smallest disc: stick 2 → stick 1
Middle disc: stick 2 → stick 3
Smallest disc: stick 1 → stick 3

The three discs are moved to stick 3 in 7 moves.

2Increase the number of discs one by one and write the number of moves taken in a table. Find the minimum number of moves required each time.

Number of discs (n)	Minimum number of moves (t)
1	1
2	3
3	7
4	15
5	31
6	63

Each time one disc is added, the number of moves is doubled and 1 is added. For example, 2 × 7 + 1 = 15.

3Can a formula be prepared showing the relationship between the number of disks 'n' and the moves 't'?

Yes. Each number of moves is one less than a power of 2.

1 = 2 − 1 = 2¹ − 1
3 = 4 − 1 = 2² − 1
7 = 8 − 1 = 2³ − 1
15 = 16 − 1 = 2⁴ − 1
31 = 32 − 1 = 2⁵ − 1

So, the formula is t = 2ⁿ − 1.

3. Identify the lightweight object using a pan balance

✦You have nine identical looking objects. One of them is light in weight. How many times will you use the pan balance to identify this lightweight object?

The pan that rises holds the lighter group.

First time: Put 4 objects in each pan. Keep 1 aside. The pan that goes up has the light object. If the pans balance, the object kept aside is the light one.

Second time: Put 2 of those 4 objects in each pan. The pan that goes up has the light object.

Third time: Put 1 of those 2 objects in each pan. The pan that goes up has the light object.

By this method, the pan balance is used 3 times.

✦If the pan balance is to be used twice, how will you find the lightweight object? Find the solution and share it with the class.

First time: Make three groups of 3 objects each. Put one group in each pan. Keep the third group aside.
If one pan goes up, the light object is in that group.
If the pans balance, the light object is in the group kept aside.

Second time: Take the 3 objects of that group. Put one object in each pan. Keep the third aside.
If one pan goes up, that object is the light one.
If the pans balance, the object kept aside is the light one.

So, the light object is found by using the pan balance only twice.

✦When can a lightweight object be found using the pan balance only once?

When there are only 2 or 3 objects.

With 2 objects, put one in each pan. The pan that goes up has the light object.
With 3 objects, put one in each pan and keep one aside. If the pans balance, the object kept aside is the light one.

4. Sum of areas of squares

1Find the relationship between the area and perimeter of two squares and the newly formed square.

(i) The two squares with the cuts TP and TD.

(ii) The five pieces rearranged into one new square.

Areas
Area of square PQRS = 6 × 6 = 36 sq. units
Area of square ABCD = 3 × 3 = 9 sq. units
Sum of the two areas = 36 + 9 = 45 sq. units
Area of the new square = 45 sq. units

The area of the new square is equal to the sum of the areas of the two squares.

Perimeters
Perimeter of PQRS = 4 × 6 = 24 units
Perimeter of ABCD = 4 × 3 = 12 units
Sum of the two perimeters = 36 units
Side of the new square is about 6.7 units.
Perimeter of the new square = 4 × 6.7 = about 26.8 units

The perimeter of the new square is less than the sum of the perimeters of the two squares.

2Side BC is half of side QR. If it is not half, will a new square be formed?

Yes, a new square can still be formed. But then T should not be the midpoint of QR. T should be taken so that l(QT) = l(BC).

Example: QR = 6 units and BC = 2 units
Take T so that QT = 2 units.
(TP)² = 2² + 6² = 4 + 36 = 40
(TD)² = 6² + 2² = 36 + 4 = 40
Both cuts are equal. The new square has area 40 = 36 + 4.

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