NCERT Solutions Class 7 Maths Chapter 6 Number Play

NCERT Solutions Class 7 Maths Chapter 6 Number Play introduces students to interesting patterns and tricks with numbers. In this chapter, students explore concepts like divisibility rules, prime and composite numbers, and various number puzzles that make learning maths enjoyable.

These NCERT Solutions follow the official NCERT guidelines and provide clear, step-by-step answers for each question. The solutions are explained in a way that is easy to understand and helps students solve the problems without confusion. Practicing from NCERT Solutions for this chapter builds strong number sense and improves a student’s ability to work with different types of numbers confidently.

1.0NCERT Solutions Class 7 Maths Chapter 6 Number Play – Download PDF

This ALLEN‑prepared PDF features easy-to-follow answers to every question of NCERT Solutions Class 7 Maths Chapter 6 , aligned with the updated syllabus and optimized for clarity, exam readiness, and revision.

2.0Key Concepts in Chapter 6: Number Play

1. Factors and Multiples

• A factor divides a number evenly; e.g., 6 is a factor of 18.
• A multiple of a number is obtained by multiplying it by an integer; e.g., 18 is a multiple of 6.
• Students learn to list factors, multiples, and understand their significance in solving problems.

2. Common Factors and Common Multiples

• Common factors are numbers dividing two or more numbers without a remainder.
• Common multiples are numbers divisible by each of a set.
• Learning about common factors/multiples prepares students for GCD and LCM concepts and their applications.

3. Highest Common Factor (HCF) & Lowest Common Multiple (LCM)

• HCF (or GCD) is the greatest number that divides two given numbers.
• LCM is the smallest number that is a multiple of both.
• NCERT introduces prime factorization, a fundamental method to compute HCF and LCM.
• Practical examples: synchronizing schedules or comparing fractions with different denominators.

4. Prime and Composite Numbers

• A prime number has exactly two factors: 1 and itself.
• A composite number has more than two factors.
• Recognizing primes helps in factorization and understanding number structure.

5. Tests of Divisibility

• Divisibility rules (for 2, 3, 4, 5, 6, 9, 10) allow quick mental testing of numbers.
• These rules are explained with logic to help in fast problem solving and number pattern detection.

6. Relationships between HCF and LCM

• NCERT emphasizes the product rule: HCF × LCM = product of two numbers, given they share only those factors.
• This powerful property simplifies many computational tasks.

7. Playing with Numbers – Magic Squares & Patterns

• The chapter includes puzzles like magic squares, where rows, columns, and diagonals sum to the same value.
• Such exercises develop logical reasoning, pattern recognition, and creativity.

8. Word Problems Involving Numbers

• Real-life scenarios are framed as word problems that require HCF/LCM, divisibility, or arranging numbers in patterns.
• Through guided practice, students learn to model such problems mathematically.

3.0NCERT Class 7 Maths Chapter 6 Number Play: Detailed Solutions

FIGURE IT OUT-01

• Arrange the stick figure cutouts given at the end of the book or draw a height arrangement such that the sequence reads: (a) $0,1,1,2,4,1,5$ (b) $0,0,0,0,0,0,0$ (c) $0,1,2,3,4,5,6$ (d) $0,1,0,1,0,1,0$ (e) $0,1,1,1,1,1,1$ (f) $0,0,0,3,3,3,3$ Sol. (a) $0,1,1,2,4,1,5$ $0,1,1,2,4,1,5$ The rule says that the given numbers are equal to the number of children in front of each stick figure who are taller than them. That means here 0 means no taller child in front of the stick figure. $1=1$ child in front of him who is taller than him. $1=1$ children in front of him who is taller than him. $2=2$ children in front of him who is taller than him. $4=4$ children in front of him who is taller than him. $1=1$ children in front of him who is taller than him. $5=5$ children in front of him who is taller than him. Hence the arrangement will be
  
  (b) $0,0,0,0,0,0,0$
  
  (c) $0,1,2,3,4,5,6$
  
  (d) $0,1,0,1,0,1,0$
  
  (e) $0,1,1,1,1,1,1$
  
  

FIGURE IT OUT-02

• Using your understanding of the pictorial representation of odd and even numbers, find out the parity of the following sums: (a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even + odd + odd) (b) Sum of 2 odd numbers and 3 even numbers (c) Sum of 5 even numbers (d) Sum of 8 odd numbers Sol. (a) Even + Even $=$ Even and Odd + Odd = Even Adding the two results, we get Even + Even $=$ Even . The parity of the result is even. Example: $4+6+3+5=10+8=18$ (Even) (b) Odd + Odd = Even and Even + Even + Even $=$ Even Adding the two results, we get Even + Even $=$ Even The parity of the result is even. Example: $5+7+4+6+8=12+18=30$ (Even) (c) Adding any 5 even numbers always gives an even number. The parity of the result is even. Example: $2+4+6+8+10=30$ (Even) (d) Odd + Odd = Even (4 such pairs). Adding the 4 such results, we get Even + Even + Even + Even $=$ Even The parity of the result is even. Example: $1+3+5+7+9+11+13+15=64$ (Even)
• Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an even number of ₹10 coins in his piggy bank. He calculated the total and got ₹205. Did he make a mistake? If he did, explain why. If he didn't, how many coins of each type could he have? Sol. Lakpa has an odd number of ₹ 1 coins, so, their total value is odd. He also has an odd number of ₹ 5 coins, so their total is also odd. The ₹ 10 coins are even in numbers, so their total is even. Now, adding two odd sums, we get Odd (from ₹ 1) + Odd (from ₹ 5) = Even Adding the ₹ 10 coins' total (even sum) to this even sum, Even + Even $=$ Even Since 205 is an odd number, the total of ₹ 205 is not possible with an odd number of ₹ 1 and ₹ 5 coins and an even number of ₹ 10 coins. Therefore, Lakpa made a mistake in his calculation.
• We know that: (a) even + even $=$ even (b) odd + odd = even (c) even + odd = odd Similarly, find out the parity for the scenarios below: (d) even - even = $\_\_\_\_$ (e) odd - odd = $\_\_\_\_$ (f) even - odd = $\_\_\_\_$ (g) odd - even = $\_\_\_\_$ Sol. (d) even - even $=$ $\_\_\_\_$ $8-4=4\to$ even $10-4=6\to$ even Parity of result = even $\therefore$ even - even $=$ even (e) odd - odd = $\_\_\_\_$ $5-3=2\to$ even $9-3=6\to$ even Parity of result = even $\therefore$ odd - odd $=$ even (f) even - odd = $\_\_\_\_$ $6-3=3$ $10-5=5$ Parity of result = odd $\therefore$ even - odd $=$ odd (g) odd - even = $\_\_\_\_$ $5-2=3$ $11-6=5$ Parity of result = odd $\therefore$ odd - even $=$ odd

FIGURE IT OUT-03

• How many different magic squares can be made using the numbers 1-9? Sol. There are 8 different magic square that can be made using numbers 1-9.
  
  Sol. The numbers 2-10 are 9 consecutive numbers, just like 1-9, but increased by 1 . We know that the sum of the numbers from 2 to 10 is $2+3+4+5+6+7+8+9+10=54$ Magic sum $=\frac{54}{3}=18$

• Take a magic square, and (a) increase each number by 1 (b) double each number In each case, is the resulting grid also a magic square? How do the magic sums change in each case? Sol.
  

Original (magic sum $=15$ ) (a) increase each number by 1

New magic sum=18 (b) double each number

New magic sum = 30 In case (a), adding a constant to every number, $\to$ magic sum increases by $3\times$ that constant. In case (b), multiplying all by a constant $\to$ magic sum multiplied by that constant.

• What other operations can be performed on a magic square to yield another magic square? Sol. A magic square is a square grid where the sums of numbers in each row, column, and both diagonals are all the same that's called the magic constant. There are several operations we can do on a magic square that still preserve its magic properties. Here are some common ones: (i) Rotation We can rotate the square by: $90^{\circ }$ clockwise, $180^{\circ }$, $270^{\circ }$ clockwise Each rotation gives us a new magic square. (ii) Reflection (Flipping) We can reflect (flip) the square:

• Horizontally (left $\leftrightarrow$ right)
• Vertically (top $\leftrightarrow$ bottom)
• Diagonally (across either main diagonal) Each reflection also gives us a new magic square.Adding or multiplying every number
• Add the same number to every cell (e.g., +5 to each) $\to$ Still a magic square, but with a new magic constant.
• Multiply every number by the same nonzero value (e.g., $\times 2$ ) $\to$ Also gives a new magic square. Note: This doesn't preserve the original numbers, but the magic structure remains.

• Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2 - 10, $3-11$, 9-17, etc.).

Sol.

Original for number 1-9 After adding 1 in each number we get The magic square for number 2-10

Here, magic sum = 18 The magic square for number $3-11$ is formed by adding 2 to each no. of the original magic square using no. 1 to 9 .

Here, magic sum $=21$ The magic square for number $9-17$ is formed by adding 8 to each no. of the original magic square.

Here, magic sum = 39

FIGURE IT OUT-04

• Using the generalised form, find a magic square if the centre number is 25 . Sol. A magic square with the center value 25 , where other numbers in the grid are expressed in relation to 25 .

| $25+3$ | $25-4$ | $25+1$ | | $25-2$ | 25 | $25+2$ | | $25-1$ | $25+4$ | $25-3$ |

$\Rightarrow$| 28 | 21 | 26 | | :---: | :---: | :---: | | 23 | 25 | 27 | | 24 | 29 | 22 |

• What is the expression obtained by adding the 3 terms of any row, column or diagonal? Sol. Row sum $(1^{\text{st }}$ row $)=28+21+26=75$ Column sum $(1^{\text{st }}$ column $)=28+23+24=75$ Diagonal sum $(1^{\text{st }}$ column $)=28+25+22=75$ The expression obtained $=3\times \mathrm{m}$ where $m$ is the letter-number representing the number in the centre.
• Write the result obtained by- (a) adding 1 to every term in the generalised form. (b) doubling every term in the generalised form Sol.

original (a)

(b)

• Create a magic square whose magic sum is 60 . Sol. A $3\times 3$ magic square's sum is $3\times$ the middle element. So, for a sum of 60 , the middle element should be $\frac{60}{3}=20$ To get a magic sum of 60 , we will multiply the original magic square by 4 , i.e.,

• Is it possible to get a magic square by filling nine non-consecutive numbers? Sol. Yes, it is possible. Let us consider the two magic squares with a magic sum 45. 9 consecutive numbers:

9 non consecutive numbers:

FIGURE IT OUT-05

• A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why? Sol. Dorjee toggles the switch 77 times. Now, let's break it down: Starting from the initial ON state of the bulb: $\begin{array}{ll}{1}^{\text{st }}& \text{ toggle }\to \text{ OFF }\\ 2^{\text{nd }}& \text{ toggle }\to \text{ ON }\\ 3^{\text{rd }}& \text{ toggle }\to \text{ OFF }\end{array}$ ...and so on. Each time you toggle the switch, the state changes. So, the key is determining whether the number of toggles is odd or even:

• If the number of toggles is odd, the light bulb will end up in the opposite state from where it started (which was ON ). If the number of toggles is even, the light bulb will end up in the same state (ON). Since 77 is odd, after 77 toggles, the light bulb will be in the opposite state to its starting position. So, it will be $\mathbf{OFF}$. Answer: The light bulb will be OFF after 77 toggles.

• Liswini has a large old encyclopaedia. When she opened it, several loose pages fell out of it. She counted 50 sheets in total, each printed on both sides. Can the sum of the page numbers of the loose sheets be 6000 ? Why or why not? Sol. Total sheets $=50$ Each sheet has one even and one odd page number. Thus, the total of 50 sheets consists of 50 even numbers and 50 odd numbers. The sum of the even numbers is even. (because adding any number of even numbers is even). The sum of the odd numbers is even. (because adding an even number of odd numbers is even). Thus, the total sum is even + even $=$ even. Since 6000 is an even number, it is possible for the sum of the page numbers of the loose sheets to be 6000 .
• Here is a $2\times 3$ grid. For each row and column, the parity of the sum is written in the circle; ' $e$ ' for even and 'o' for odd. Fill the 6 boxes with 3 odd numbers (' $o$ ') and 3 even numbers (' $e$ ') to satisfy the parity of the row and column sums.
  
  Sol.
  
• Make a $3\times 3$ magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed. Sol. One Solution to this problem could be: $\left[\begin{array}{ccc}-1& 4& -3\\ -2& 0& 2\\ 3& -4& 1\end{array}\right]$ Let's check the row, column, and diagonal sums to ensure everything adds up to 0 : (i) Row sums: $\begin{array}{rl}& -1+4+(-3)=0\\ & -2+0+2=0\\ & 1+3+(-4)=0\end{array}$$(ii)Columnsums:\begin{array}{rlrl}& (-1)+(-2)+3=0& 4+0+(-4)=0& 3+(-4)+1=0\end{array}$$(iii)Diagonalsums:$$\begin{array}{rlr}& (-1)+0+1=0& 3+0+(-3)=0\end{array}$Conclusion: This$3\times 3\text{magic square works, and the sum of the numbers in each row, column, and diagonal is 0 .}$
• Fill in the following blanks with 'odd' or 'even': (a) Sum of an odd number of even numbers is $\_\_\_\_$ (b) Sum of an even number of odd numbers is $\_\_\_\_$ (c) Sum of an even number of even numbers is $\_\_\_\_$ (d) Sum of an odd number of odd numbers is $\_\_\_\_$ Sol. (a) Even (b) Even (c) Even (d) Odd
• What is the parity of the sum of the numbers from 1 to 100 ? Sol. There are 100 numbers from 1 to 100. Parity: odd, even, odd, even...There are exactly 50 odd numbers and 50 even numbers. Adding 50 odd numbers gives an even sum (because an even number of odd numbers sums to even). Adding 50 even numbers gives an even sum. Total sum: even + even $=$ even. Therefore, the parity of the sum of numbers from 1 to 100 is even.
• Two consecutive numbers in the Virahāṅka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence? Sol. In the Virahanka sequence, the next number is obtained by adding the two previous numbers. Given that two consecutive numbers in sequence are 987 and 1597. The next two numbers are: $987+1597=2584$. $1597+2584=4181$. The previous two numbers are: $1597-987=610$. $987-610=377$. the sequence is .....,377, 610, 987, 1597, 2584, 4181,.....
• Angaan wants to climb an 8 -step staircase. His playful rule is that he can take either 1 step or 2 steps at a time. For example, one of his paths is $1,2,2,1,2$. In how many different ways can he reach the top? Sol. Different ways by which Angaan can climb the 8-step staircase by taking either 1 step or 2 steps at a time are:

So, the total ways in which Angaan reaches the top $=1+7+15+10+1=34$ ways

• What is the parity of the 20th term of the Virahāṅka sequence? Sol. Virahāṅka sequence: $1,2,3,5,8,13,21,34,55,89,\dots$ 1(odd), 2(even), 3(odd), 5(odd), 8(even), 13(odd), 21(odd), 34(even), 55(odd), 89(odd),... Parity cycle: odd, even, odd This pattern repeats every 3 terms. $20÷3=6$ remainder 2 . This means the $20^{\text{th }}$ term corresponds to the second position in the parity cycle, which is: odd, even, odd Thus, the parity of the $20^{\text{th }}$ term is even.
• Identify the statements that are true. (a) The expression $4m-1$ always gives odd numbers. (b) All even numbers can be expressed as $6j-4$. (c) Both expressions $2p+1$ and $2q-1$ describe all odd numbers. (d) The expression $2f+3$ gives both even and odd numbers. Sol. (a) Substituting $\mathrm{m}=1$ in $4\mathrm{m}-1$ $\Rightarrow 4(1)-1=4-1=3($ odd $)$. Substituting $\mathrm{m}=3$ in $4\mathrm{m}-1$ $\Rightarrow 4(3)-1=12-1=11$ (odd). 4 m is always even, so $4\mathrm{m}-1$ is always odd. Thus, the statement is True.
  

4.0Key Features of NCERT Solutions Class 7 Maths Chapter 6 : Number Play

• Clear Explanation of General Forms: The solutions provide thorough explanations of how to represent two-digit and three-digit numbers in their general forms (10a+b, 100a+10b+c), which is crucial for proving divisibility rules and solving advanced number problems.
• Comprehensive Divisibility Rule Coverage: All divisibility rules (for 2, 3, 4, 5, 6, 8, 9, 10) are explained with illustrative examples, helping students master these essential shortcuts.
• Logical Reasoning Development: Solutions to letter puzzles emphasize the logical deduction process, guiding students through the thought process required to solve such problems, thereby enhancing their critical thinking skills.
• Adherence to NCERT Syllabus: The content strictly follows the updated NCERT Ganita Prakash curriculum for Class 7, ensuring that students are well-prepared for their examinations and aligned with academic standards.
• Downloadable PDF Format: Conveniently available as a PDF, these NCERT Solutions allow for easy access and offline study, enabling flexible and effective learning for all students.