Experts have designed these Class 7 Maths Notes and Chapter 6 Number Play Class 7 Notes for effective learning.
Class 7 Maths Chapter 6 Notes Number Play
Class 7 Maths Notes Chapter 6 – Class 7 Number Play Notes
→ A number that is exactly divisible by 2 is called an even number.
→ A number that is not exactly divisible by 2 is called an odd number.
→ Parity of a number tells us whether the number is even or odd.
→ The parity of the expression an + b, where a and b are fixed numbers, depends upon a and b, and the value of n.
→ A square grid is called a magic square if the sum of numbers in small squares along any row, any column, or any diagonal is the same. This sum is called the magic sum of the magic square under consideration.
→ In a 3 × 3 magic square with numbers 1 to 9 without repetition:
- The magic sum is always 15.
- 5 can be placed only at the centre position.
- 1 and 9 can be placed only in the middle positions.
→ 1, 2, 3, 5, 8, 13, 21, 34,… are called Virahanka numbers.
→ The sequence 1, 2, 3, 5, 8, 13, 21, 34, ….. is called the Virahanka sequence.
→ Digits in disguise means a math game or activity in which letters are assigned values 1, 2, 3, …, and players try to find which letter represents which digit based on given equations or clues. This type of problem is also called Cryptarithms or Alphametics.
Numbers Tell Us Things Class 7 Notes
Consider the following figure of 7 children standing in a row.
Let us find the number of children taller than each child on his left in the above figure.
There is no child taller than child 1 on his left.
Child 1 is taller than Child 2 on his left.
∴ There is 1 child taller than child 2 on his left.
Children 1 and 2 are taller than child 3 on his left.
∴ There are 2 children taller than child 3 on his left.
Only child 1 is taller than child 4 on his left.
∴ There is 1 child taller than child 4 on his left.
Children 1,2,3, and 4 are taller than child 5 on his left.
∴ There are 4 children taller than child 5 on his left.
Children 1, 2, and 4 are taller than child 6 on his left.
∴ There are 3 children taller than child 6 on his left.
There is no child taller than child 7 on his left.
Now let us discuss the number of children shorter than each child on his left in the above figure.
There is no child shorter than child 1 on his left.
There is no child shorter than child 2 on his left.
There is no child shorter than child 3 on his left.
Children 2 and 3 are shorter than child 4 on his left.
∴ There are 2 children shorter than child 4 on his left.
There is no child shorter than child 5 on his left.
Children 3 and 5 are shorter than child 6 on his left.
∴ There are 2 children shorter than child 6 on his left.
Children 1 to 6 are all shorter than child 7 on his left.
∴ There are 6 children shorter than child 7 on his left.
Question 1.
7 students are standing in a row in the given figure. Explain the number written above each student.
Solution:
Let us check the number of taller students on the left of each student.
No student is taller than student 1 on his left.
∴ ‘0’ written above student 1 is justified.
Only one student (No. 1) is taller than student 2 on his left.
∴ ‘1’ written above student 2 is justified.
No student is taller than student 3 on his left.
∴ ‘0’ written above student 3 is justified.
Two students (No. 1 and 3) are taller than student 4 on his left.
∴ ‘2’ written above student 4 is justified.
Two students (No. 1 and 3) are taller than student 5 on his left.
∴ ‘2’ written above student 5 is justified.
Two students (No. 1 and 3) are taller than student 6 on his left.
∴ ‘2’ written above student 6 is justified.
Only one student (No. 3) is taller than student 7 on his left.
∴ ‘1’ written above student 7 is justified.
Picking Parity Class 7 Notes
1. Parity of a Number
Parity of a number tells us whether the number is even or odd.
2. Even Numbers
A number that is exactly divisible by 2 is called an even number.
For example, 2, 6, 8, 14, 24 are even numbers.
Even numbers always end up with the last digit as 0, 2, 4, 6, or 8.
Any even number can be arranged in pairs without any leftover.
The following are some demonstrations of even numbers.
We consider the sum of even numbers 4 and 8 by using dots.
We see that adding any two even numbers results in a number that can still be arranged in pairs without any leftover. Thus, we find that the sum of two even numbers is again an even number.
Similarly, we can show that the sum of any number of even numbers is an even number. The sum of any number of even numbers is an even number.
∴ The sum of any number of even numbers is an even number.
3. Odd Numbers
A number that is not exactly divisible by 2 is called an odd number.
For example, 1, 3, 7, 17, 45 are odd numbers.
Odd numbers always end up with the last digit as 1, 3, 5, 7, or 9.
Odd numbers cannot be arranged in pairs. They are one less than a pair.
The following are some demonstrations of odd numbers.
We consider the sum of odd numbers 7 and 9 by using dots.
We see that adding any two odd numbers results in a number that can still be arranged in pairs without any leftover. Thus, we find that the sum of two odd numbers is an even number.
Now let us consider the sum of an even number and an odd number. 6 is an even number, and 9 is an odd number. We consider their sum by using dots.
We see that the sum 15 is an odd number. Thus, the sum of an even number and an odd number is always an odd number.
Now, we shall study the parity of the sum of some odd numbers. We know that the sum of two odd numbers is an even number, whereas the sum of an even number and an odd number is an odd number.
Suppose we are to add ‘n’ odd numbers. The number ‘n’ is either even or odd.
Let n be an even number.
∴ n = 2k, where k is a counting number.
∴ Sum of n odd numbers = Sum of k pairs of odd numbers
= Sum of k even numbers.
= an even number.
Let n be an odd number.
∴ n = 2k + 1, where k is a counting number.
∴ Sum of n odd numbers = Sum of k pairs of odd numbers + one odd number
= Sum of k even numbers + one odd number
= an even number + one odd number
= an odd number.
∴ The sum of an even number of odd numbers is an even number, and the sum of an odd number of odd numbers is an odd number.
Question 1.
What about adding 3 odd numbers? Can the resulting sum be arranged in pairs?
Solution:
Sum of 3 odd numbers = Sum of 2 odd numbers + one odd number
= One even number + one odd number
= One odd number
∴ The sum of 3 odd numbers cannot be arranged in pairs.
Question 2.
Explore what happens to the sum of:
(a) 4 odd numbers
(b) 5 odd numbers
(c) 6 odd numbers
Solution:
(a) The sum of 4 odd numbers is an even number, because the sum of an even number of odd numbers is an even number.
(b) The sum of 5 odd numbers is one odd number, because the sum of an odd number of odd numbers is an odd number.
(c) The sum of 6 odd numbers is an even number, because the sum of an even number of odd numbers is an even number.
Question 3.
Kishor has some number cards and is working on a puzzle: There are 5 boxes, and each box should contain exactly 1 odd-numbered card. The sum of all the numbers in the boxes must equal 30. Can you help him find a way to do it?
Solution:
On the left side, there are five boxes, and one odd-numbered card is to be placed in each box.
5 is an odd number.
We know that the sum of an odd number of odd numbers is an odd number.
∴ Left side, being the sum of 5 odd numbers, must be an odd number.
i.e., it can never be equal to the even number 30.
∴ The given puzzle cannot be solved.
Question 4.
Two siblings, Martin and Maria, were born exactly one year apart. Today they are celebrating their birthday. Maria exclaims that the sum of their ages is 112. Is this possible? Why or why not?
Solution:
Since Martin and Maria were bom exactly one year apart, their ages must be two consecutive numbers.
We know that two consecutive numbers can neither be both even nor be both odd.
∴ Their ages are one even number and one odd number.
∴ The sum of their ages is an odd number.
Since 112 is an even number, the sum of the ages of Martin and Maria cannot be 112.
Small Squares in Grids
If there are m rows and n columns of small squares in a grid, then the total number of small squares in the grid is equal to m × n.
For example, let there be 3 rows and 4 columns in the grid given in the figure.
The number of small squares in this grid is equal to 3 × 4, i.e., 12.
Let there be a grid having m rows and n columns of small squares.
In this grid, we have m × n small squares.
If m and n are both odd numbers, then the parity of m × n is an odd number.
If one of m and n is an even number, then the parity of m × n is even.
If m and n are both even numbers, then the parity of m × n is even.
Parity of Expressions
Let a and b be fixed numbers. Consider the expression an + b. The parity of the expression an + b would depend upon a, b, and the value of n.
Question 1.
Consider the parity of the expression 3n + 4 for n equal to 3, 8, 10, 15.
Solution:
The given expression is 3n + 4.
n = 3
Here 3n + 4 is equal to 3(3) + 4.
The parity of 3(3) is an odd number, because 3 is an odd number.
.’. The parity of 3(3) + 4 is an odd number, because ‘odd + even is odd’.
n = 8
Here 3n + 4 is equal to 3(8) + 4.
The parity of 3(8) is an even number, because 8 is an even number.
∴ The parity of 3(8) + 4 is an even number, because ‘even + even is even’.
n = 10
Here 3n + 4 is equal to 3(10) + 4.
The parity of 3(10) is an even number, because 10 is an even number.
∴ The parity of 3(10) + 4 is an even number, because ‘even + even is even’.
n = 15
Here 3n + 4 is equal to 3(15) + 4.
The parity of 3(15) is an odd number, because 3 and 15 are both odd numbers.
∴ The parity of 3(15) + 4 is an odd number, because ‘odd + even is odd”.
Question 2.
Discuss the parity of the following expressions for different values of n:
(i) 100n
(ii) 48n – 2
(iii) 4n + 1
(iv) 2n + 3
(v) 6n + 2
Solution:
(i) The given expression is 100n.
Since 100 is an even number, the parity of the expression 100n is always even for every value of n.
(ii) The given expression is 48n – 2.
Since 48 is an even number, the parity of the expression 48n is always even.
∴ The parity of the expression 48n – 2 is an even number, because ‘even-even is even’.
(iii) The given expression is 4n + 1.
Since 4 is an even number, the parity of the expression 4n is always even.
∴ The parity of the expression 4n + 1 is an odd number, because ‘even + odd is odd’.
(iv) The given expression is 2n + 3.
Since 2 is an even number, the parity of the expression 2n is always even.
∴ The parity of 2n + 3 is an odd number, because ‘even + odd is odd’.
(v) Given expression is 6n + 2.
Since 6 is an even number, the parity of the expression 6n is always even.
∴ The parity of 6n + 2 is an even number, because ‘even + even is even’.
Question 3.
Give an example of an expression that always has:
(i) even parity
(ii) odd parity
Solution:
(i) Consider the expression 4n, where n is any number.
Since 4 is an even number, the parity of the expression 4n is always even.
(ii) Consider the expression 4n + 1, where n is any number.
Since 4 is an even number, the parity of the expression 4n is always even.
∴ The parity of the expression 4n + 1 is an odd number, because ‘even + odd is odd’.
Question 4.
Give an example of an expression like 3n + 4, which could have either odd or even parity.
Solution:
Consider the expression 5n + 8, where n is any number.
Care I – n is Even: Since 5 is odd and n is even, the parity of the number 5n is even.
∴ The parity of the expression, 5n + 8, is an even number, because ‘even + even is even’.
Care II – n is odd: Since 5 and n are both odd, the parity of the number 5n is odd.
∴ The parity of the expression, 5n + 8, is an odd number, because ‘odd + even is odd’.
Question 5.
Are there expressions that we can use to list all the even numbers?
Solution:
We consider the expression 2n, where n is any number.
Since 2 is an even number, the parity of the expression 2n is always even.
For n = 1, we have 2n = 2(1) = 2
For n = 2, we have 2n = 2(2) = 4
For n = 3, we have 2n = 2(3) = 6
For n = 4, we have 2n = 2(4) = 8
∴ Using the expression 2n, we get all even numbers 2, 4, 6, 8,……….
Question 6.
Are there expressions that we can use to list all odd numbers?
Solution:
We consider the expression 2n – 1, where n is any number.
Since 2 is an even number, the parity of the expression 2n is always even.
∴ The parity of the expression 2n – 1 is an odd number, because ‘even-odd is odd’.
For n = 1, we have 2n – 1 = 2(1) – 1 = 2 – 1 = 1
For n = 2, we have 2n – 1 = 2(2) – 1 = 4 – 1 = 3
For n = 3, we have 2n – 1 = 2(3) – 1 = 6 – 1 = 5
For n = 4, we have 2n – 1 = 2(4) – 1 = 8 – 1 = 7
∴ Using the expression 2n – 1, we get all odd numbers 1, 3, 5, 7,…….
Question 7.
Write a formula to find the nth:
(i) even number.
(ii) odd number.
Solution:
(i) Consider the expression 2n, where n is any number.
For n = 1, 2, 3, 4,…, the values of 2n are 2, 4, 6, 8,…, and these are all even numbers.
∴ The formula for even numbers is 2n, where n is any number.
Also, the nth even number is equal to 2n.
(ii) Consider the expression 2n – 1, where n is any number.
For n = 1, 2, 3, 4, …, the values of 2n – 1 are 1, 3, 5, 7,…, and these are all odd numbers.
∴ The formula for odd numbers is 2n – 1, where n is any number.
Also, the nth odd number is equal to 2n – 1.
Question 8.
What is the 100th?
(a) even number
(b) odd number
Solution:
(a) We know that the nth even number is 2n.
∴ 100th even number = 2(100) = 200.
(b) We know that the nth odd number is 2n – 1.
∴ 100th odd number = 2(100) – 1 = 200 – 1 = 199.
Some Explorations in Grids Class 7 Notes
Consider the following 3 × 3 grid. This grid has 9 small squares. This grid is filled with any 9 numbers. The sum of numbers in each row and column is given in circles outside the grid. We observe that the sum of numbers in circles horizontally is 27 + 32 + 11 = 70, and the sum of numbers in circles vertically is 24 + 17 + 29 = 70.
Let us consider one more 3 × 3 grid in which 9 numbers 1 to 9 are filled without repetition.
Here, we observe that the sum of numbers in circles horizontally is 9 + 18 + 18 = 45, and the sum of numbers in circles vertically is 13 + 12 + 20 = 45.
We notice that each sum is equal to 45.
This has happened because 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.
Question 1.
Fill the following grids using only numbers from 1 to 9 without repetition:
(i)
Solution:
(i) We observe that:
13 + 14 + 18 = 45 and 24 + 9 + 12 = 45.
Also, 1 + 2 + 3 + …… + 8 + 9 = 45
∴ We shall be able to complete the grid by using 9 numbers from 1 to 9 without repetition.
The following is the required filled grid.
(ii) We observe that:
24 + 15 + 6 = 45 and 12 + 16 + 17 = 45.
Also, 1 + 2 + 3 + ……… + 8 + 9 = 45
∴ We shall be able to complete the grid by using 9 numbers from 1 to 9 without repetition.
The following is the required filled grid.
Question 2.
Can the following 3 × 3 grid be filled using numbers from 1 to 9 without repetition?
Solution:
In the given grid, 9 is already filled.
So, we have to use only 1, 2, 3,…….., 8 to fill the grid.
In the third column, the sum is to be 26, and 9 is already filled.
∴ The sum of the other two numbers in small squares must be 26 – 9 = 17.
From numbers 1, 2, 3,……., 8, we cannot find two numbers whose sum is 17, because the largest sum in the third column can be 9 + 8 + 7 = 24.
Also, in the first row, the sum cannot be 5, because the smallest sum can be 1 + 2 + 3 = 6.
∴ It is impossible to fill the given grid.
1. Positions of Small Squares in a 3 × 3 Grid
In a 3 × 3 grid, there are 9 small squares.
The positions of these small squares are classified as Comer position, Middle position, and Centre position.
The following figures show the different positions of small squares in a 3 × 3 grid.
2. Magic Square
A square grid is called a magic square if the sum of numbers in small squares along any row, any column, or any diagonal isthe same. This sum is called the magic sum of the magic square under consideration.
In our further discussion, we shall be considering 3 × 3 magic squares using only the numbers 1, 2, 3,…, 9 without repetition. The following are the important observations for a 3 × 3 magic square with numbers from 1 to 9 without repetition.
Observation I:
We know that: 1 + 2 + 3 + ….. + 9 = 45.
∴ The sum of the totals along three rows must be 45.
Since, in a magic square, the sums along any row are the same.
∴ Sum of numbers along any row = 45 ÷ 3 = 15
The grid is a magic square, so the sum of numbers along any column or any diagonal is also 15.
∴ The magic sum is 15.
Thus, in a 3 × 3 magic square with numbers 1 to 9 without repetition, the magic sum is always equal to 15.
Observation II:
Here, we shall consider the possibility of numbers at the centre position.
Let 9 be placed at the centre position. The number 8 will be at some other position.
Let the numbers 9 and 8 be filled as follows:
The sum of numbers along the marked diagonal must be 15.
∴ 8 + 9 + other number = 15
∴ 17 + other number = 15
This is impossible.
∴ 9 cannot be placed at the centre.
Similarly, 8 cannot be placed at the centre.
Let 7 be placed at the centre.
The number 9 will be at some other position.
Let the numbers 7 and 9 be filled as follows:
The sum of numbers along the marked column must be 15.
∴ 9 + 7 + other number = 15
This is impossible.
∴ 7 cannot be placed at the centre.
Similarly, 6 cannot be placed at the centre.
Let 1 be placed at the centre.
The number 2 will be at some other position.
Let the numbers 1 and 2 be filled as follows:
The sum of numbers along the marked row must be 15.
∴ 2 + 1 + other number = 15
∴ Other number = 15 – 3 = 12
This is impossible.
∴ 1 cannot be placed at the centre.
Similarly, 2 cannot be placed at the centre.
Let 3 be placed at the centre.
The number 1 will be placed at some other position.
Let the numbers 1 and 3 be filled as follows:
The sum of numbers along the marked diagonal must be 15.
∴ 1 + 3 + other number = 15
∴ Other number = 15 – 4 = 11
This is impossible.
∴ 3 cannot be placed at the centre; similarly, 4 cannot be placed at the centre.
∴ Only 5 can be placed at the centre.
Thus, in a 3 × 3 magic square with numbers 1 to 9 without repetition, only 5 can be placed at the centre position.
Observation III:
Here, we shall study the positions of 1 and 9. We know that 1 and 9 cannot be placed at the centre. We shall show that 1 and 9 cannot be placed at comer positions.
Let 1 be placed at a corner position in a 3 × 3 grid.
The sum of numbers along the diagonals is always 15.
∴ The third number in the diagonal of 1 and 5 must be 9.
In the first row, we will have 6 and 8 so as to make a total of 15.
In the first column, the sum of the remaining two numbers must be 15 – 1 = 14.
The remaining unused numbers are 2, 3, 4, 7, and the sum of any two of these cannot be 14.
∴ 1 cannot be placed at a corner position.
Let 9 be placed at a corner position in a 3 × 3 grid.
The sum of numbers along the diagonals is always 15.
∴ The third number in the diagonal of 9 and 5 must be 1.
In the first column, we will have 2 and 4 to make a total of 15.
In the third row, the sum of the two numbers must be 15 – 9 = 6.
The remaining unused numbers are 3, 6, 7, 8, and the sum of any two of these cannot be 6.
∴ 9 cannot be placed at a corner position.
Thus, in a 3 × 3 magic square with numbers 1 to 9 without repetition, the numbers 1 and 9 can be placed at only the middle positions.
We conclude that a 3 × 3 magic square with numbers I to 9 without repetition can be one of the following 4 forms:
3. Generalising a 3 × 3 Magic Square
Till now, we have been discussing 3 × 3 magic squares using numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 without repetition.
Now, we shall consider a general method of forming a 3 × 3 magic square with a given magic sum, which is a multiple of 3.
Step 1: Let the magic sum of the required magic square be 3n.
Step 2: Consider the 9 consecutive numbers: n – 4, n – 3, n – 2, n – 1, n, n + 1, n + 2, n + 3, and n + 4.
We shall use these 9 numbers in our magic square.
Step 3: Starting with n-4, fill these 9 numbers in 9 squares along the arrows as shown below:
Step 4: In the above arrangement, there are 9 numbers in 3 rows and 3 columns, and the sum of numbers in any row or any column or along any diagonal is the same and it is equal to 3n.
Step 5: The above grid is the required 3 × 3 magic square with given magic sum 3n.
Remark: In forming a magic square with magic sum 3n, a multiple of 3, we have used 9 consecutive numbers: n – 4, n – 3, n – 2, n – 1, n, n + 1, n + 2, n + 3, and n + 4.
Activity 1
Construct a 3 × 3 magic square with a magic sum of 141, a multiple of 3.
Solution:
Step 1: The Magic sum is 141.
Step 2: \(\frac{141}{3}\) = 47
Consider the 9 consecutive numbers:
47 – 4, 47 – 3, 47 – 2, 47 – 1, 47, 47 + 1, 47 + 2, 47 + 3, 47 + 4,
i.e., 43, 44, 45, 46, 47, 48, 49, 50, 51.
Step 3: Starting with 43, fill these 9 numbers in 9 squares along the arrows as shown below:
Step 4: In the above arrangement, there are 9 numbers in 3 rows and 3 columns, and the sum of numbers in any row or any column or along any diagonal is the same and it is equal to the given magic sum 141.
Step 5: The above grid is the required 3 × 3 magic square with given magic sum 141.
Activity 2
If m is the letter-number of the number in the centre of a 3 × 3 magic square, express how other numbers are related to m, how much more or less than m.
Solution:
Step 1: Given number in the centre of a 3 × 3 magic square is m.
Step 2: Consider the 9 consecutive numbers, m – 4, m – 3, m – 2, m – 1, m, m + 1, m + 2, m + 3, and m + 4.
Step 3: Starting with m-4, fill these 9 numbers in 9 squares along the arrows as shown below:
Step 4: In the above arrangement, there are 9 numbers in 3 rows and 3 columns, and the sum of numbers in any row or any column or along any diagonal is the same and it is equal to 3m.
Step 5: The above magic square expresses how other numbers in the magic square are related to the number in the centre, i.e., m.
4. The First-ever 4 × 4 Magic Square
The first-ever recorded 4 × 4 magic square is called Chautlsa Yantra. This magic square was found in the 10th century. The following is the Chautisa Yantra.
The above is a magic square with a magic sum of 34 (Chautis in Hindi).
Another magic square with a magic sum of 34 can be obtained by interchanging rows and columns in the above Chautisa Yantra. This is given below:
5. Magic Squares in History and Culture
(i) Lo Shu Square: The Lo Shu Square is said to have appeared on the shell of a turtle emerging from the Lo River, a tributary of the Yellow River in ancient China.
Lo Shu Square is a magic square of numbers 1 to 9 without repetition.
(ii) The great India Mathematician Narayana Pandita described the method of writing a magic square containing any odd number of rows in a work named Ganita Kaumudi (c. 1356 CE).
(iii) The great Indian mathematician Srinivasa Ramanujan (1887 – 1920) also formulated many results in the field of magic squares.
(iv) A magic square of 3 rows is engraved on a pillar in Dhemipureeswarar Temple in Tamil Nadu. This temple dates back to the 8th century.
(v) Navagraha Yantras: The Navagraha Yantras are also in the form of magic squares of three rows.
(vi) Kubera Yantra: The Kubera Yantra is also in the form of a magic square of three rows.
Nature’s Favourite Sequence: The Virahanka-Fibonacci Numbers! Class 7 Notes
1. Discovery of Virahanka Numbers:
In the poetry of many Indian languages, each syllable is classified as either short or long. A long syllable is pronounced exactly twice as a short syllable. When singing a poem, a short syllable lasts one beat of time, and a long syllable lasts two beats of time. This leads to many mathematical questions, which the ancient poets considered very deeply.
To take a simple example, let us consider the number of ways of making rhythms using 4 beats consisting of short and long syllables. We have the following possibilities:
- Short Short Short Short
- Short Short Long
- Short Long Short
- Long Short Short
- Long Long
So, in all, we have 5 different ways of using 4 beats for making rhythms.
To deal with such problems mathematically, we represent a short syllable by ‘1’ and a long syllable by ‘2’. Using ‘1’ and ‘2’, we can express the above problem as:
- 4 = 1 + 1 + 1 + 1
- 4 = 1 + 1 + 2
- 4 = 1 + 2 + 1
- 4 = 2 + 1 + 1
- 4 = 2 + 2
∴ 4 can be expressed in 5 ways as sums of 1’s and 2’s.
Let us find the number of ways of writing numbers 1 to 5 as sums of 1’s and 2’s.
We can also write the ways of expressing 5 as sum of 1’s and 2’s as follows:
- 2 + (1 + 1 + 1)
- 2 + (1 + 2)
- 2 + (2 + 1)
- 1 + (1 + 1 + 1 + 1 + 1)
- 1 + (1 + 1 + 2)
- 1 + (1 + 2 + 1)
- 1 + (2 + 1 + 1)
- 1 + (2 + 2)
Here, we have added 2 to the number of ways for 3 and 1 to the number of ways for 4.
∴ The number of 5-beat rhythms is the sum of 3-beat rhythms and 4-beat rhythms.
Similarly, we can find the number of rhythms for 6 beats, 7 beats, 8 beats, and so on.
We have 5- beat rhythms = 3-beat rhythms + 4-beat rhythms
= 3 + 5
= 8
6-beat rhythms = 4-beat rhythms + 5-beat rhythms
= 5 + 8
= 13
7-beat rhythms = 5-beat rhythms + 6-beat rhythms
= 8 + 13
= 21
8-beat rhythms = 6-beat rhythms + 7-beat rhythms
= 13 + 21
= 34
This method of counting the number of rhythms of short syllables and long syllables for any given number of beats was given by the great scholar Virahanka around the year 700 CE.
The numbers 1, 2, 3, 5, 8, 13, 21, 34, 55,……… are called Virahanka numbers.
Question 1.
Use the systematic method to write down all 6-beat rhythms, i.e., write 6 as the sum of 1’s and 2’s in all possible ways. Did you get 13 ways?
Solution:
The following are the ways of writing 6 as the sum of 1’s and 2’s.
- 6 = 1 + 1 + 1 + 1 + 1 + 1
- 6 = 1 + 1 + 1 + 1 + 2
- 6 = 1 + 1 + 1 + 2 + 1
- 6 = 1 + 1 + 2 + 1 + 1
- 6 = 1 + 2 + 1 + 1 + 1
- 6 = 2 + 1 + 1 + 1 + 1
- 6 = 1 + 1 + 2 + 2
- 6 = 1 + 2 + 1 + 2
- 6 = 1 + 2 + 2 + 1
- 6 = 2 + 1 + 1 + 2
- 6 = 2 + 1 + 2 + 1
- 6 = 2 + 2 + 1 + 1
- 6 = 2 + 2 + 2
∴ We get 13 ways in total.
2. Virahanka Sequence
The sequence 1, 2, 3, 5, 8, 13, 21, 34,……. is called the Virahanka sequence. This sequence contains all Virahanka numbers. Much later than Virahanka, an Italian mathematician, Fibonacci developed these numbers in the year 1202 CE.
For this reason, the numbers 1, 2, 3, 5, 8, 13, 21, 34,… are also referred to as Virahanka-Fibonacci numbers.
For example, the number of petals in a daisy flower is generally a Virahanka number.
Question 2.
Write the next three numbers in the sequence:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, _____, _____, _____, _____, _____
Solution:
The given sequence is 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,……..
This is the Virahanka sequence.
Let Tn denote its nth term.
We are to find T11, T12, and T13.
T11 = T9 + T10 = 55 + 89 = 144
T12 = T10 + T11 = 89 + 144 = 233
T13 = T11 + T12 = 144 + 233 = 377
∴ The required terms are 144, 233, and 377.
Question 3.
What is the parity of each number in the Virahanka sequence? Do you notice any pattern in the sequence of parities?
Solution:
The Virahanka sequence is 1, 2, 3, 5, 8, 13, 21, 34, 55,….
Let Tn denote the nth term of this sequence.
From this sequence, we observe that:
T3n and T3n+1 are odd, where n = 0, 1, 2, 3,….
T3n+2 is even, where n = 0, 1, 2, 3,….
For the Virahanka sequence, the sequence of parities is O, E, O, O, E, O, O, E, O, O, E,…….
Digits in Disguise Class 7 Notes
Digits in disguise means a math game or activity in which letters are assigned values 1, 2, 3,…, and players try to find which letter represents which digit based on given equations or clues. This type of problem is also called cryptarithms or alphametics.
Question 1.
Is there a one-digit number that, when added to itself twice more, gives a two-digit sum? The unit digit of the sum is the same as the digit of the one-digit number. What could this number be? Can it be 2? Can it be 3?
Solution:
Let the given one-digit number be T.
The given problem can be expressed as:
where U is the digit at the tens’ place of the sum.
Let T = 2.
We have
∴ Digit at unit’s place of sum is not T, i.e., 2.
∴ T cannot be 2.
Let T = 3
We have
∴ Digit at unit’s place of sum is not T, i.e., 3.
∴ T cannot be 3.
Let us take T = 5
Here, the digit at the unit’s place of sum is also T, i.e., 5.
∴ The required number is 5.
Question 2.
Consider the following puzzle:
Here, K2 is a 2-digit number with 2 at the ones’ place. Find the digit represented by M? What about H? Can it be 2? Can it be 3?
Solution:
Given puzzle is:
We take K as the digit 7.
∴ Puzzle becomes:
∴ Digit M is 4, and digit H is 1.
The value of H cannot be 2 or 3, because the maximum value of K is 9, and 9 + 9 = 18.
So, the value of H is either 0 or 1.
∴ The value of H is neither 2 nor 3.
Question 3.
Solve the following puzzles.
Solution:
(i) The given puzzle is
We choose Y = 9 and Z = 1.
∴ Puzzle becomes
We take Z = 1 and O = 0.
∴ The solution is Y = 9, Z = 1, O = 0.
(ii) The given puzzle is
We choose B = 7 and D = 0.
∴ Puzzle becomes
We take E = 1.
∴ The solution is B = 7, D = 0, and E = 1.
(iii) The given puzzle is
We choose K = 6 and P = 1.
∴ Puzzle becomes
We take R = 2
∴ The solution is K = 6, P = 1, and R = 2.
(iv) The given puzzle is
We choose C = 9.
∴ Puzzle becomes
We take F = 0.
∴ The solution is C = 9 and F = 0.
Question 4.
Who is where? How many are there?
Some children are standing in increasing order of their height. Following the clues given, find out how many children there are.
(a) Each child has either an odd number of children on both sides or an even number of children on both sides.
(b) No one else is taller than E.
(c) All but one are taller than A.
(d) H and A are T’s neighbours, and
(e) L, H, and S are together in some order.
H has more children taller than him than shorter. L has more children shorter than him than taller.
Solution:
Consider the following diagram of 9 children standing in increasing order of their heights.
As per the given conditions, the positions of the children E, A, H, T, L, and S are marked in the figure.
In all, there are 9 children.
(a) It can be verified that each child has either an odd number of children on both sides or an even number of children on both sides.
(b) No one is taller than E.
(c) All but one child S is taller than A.
(d) H and A are neighbours of T.
(e) L, H, and S are together in decreasing order.
H has 5 children taller than him and 3 children shorter than him. L has 5 children shorter than him and 3 children taller than him.