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Playing with Numbers – CBSE NCERT Study Resources

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8th

8th - Mathematics

Playing with Numbers

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Overview

This chapter, "Playing with Numbers," introduces students to various concepts related to numbers, including divisibility rules, factors, multiples, and basic number patterns. The chapter aims to strengthen foundational arithmetic skills and logical thinking through engaging exercises and real-life applications.

Divisibility Rules

A number is divisible by another if it can be divided exactly without leaving a remainder.

Key divisibility rules covered in this chapter include:

  • Divisible by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).
  • Divisible by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
  • Divisible by 5: A number is divisible by 5 if its last digit is 0 or 5.
  • Divisible by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.
  • Divisible by 10: A number is divisible by 10 if its last digit is 0.

Factors and Multiples

A factor of a number is an integer that divides the number exactly, while a multiple is the product of the number and an integer.

Key concepts include:

  • Finding factors of a given number.
  • Understanding common factors and highest common factor (HCF).
  • Identifying multiples and least common multiple (LCM).

Prime and Composite Numbers

A prime number has exactly two distinct factors: 1 and itself. A composite number has more than two factors.

Students learn to:

  • Identify prime and composite numbers.
  • Understand the Sieve of Eratosthenes method to find primes.
  • Express composite numbers as a product of primes (prime factorization).

Number Patterns

This section explores sequences and patterns in numbers, such as:

  • Arithmetic sequences (constant difference between terms).
  • Geometric sequences (constant ratio between terms).
  • Triangular and square numbers.

Applications

The chapter concludes with real-life applications of these concepts, such as:

  • Using divisibility rules to simplify calculations.
  • Applying LCM and HCF in problem-solving (e.g., scheduling, measurements).
  • Recognizing patterns in nature and daily life.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
What is the sum of the first five prime numbers?
Answer:
28
Question 2:
Find the HCF of 24 and 36.
Answer:
12
Question 3:
Is 1 a prime number?
Answer:

No, 1 is not prime.

Question 4:
What is the LCM of 8 and 12?
Answer:
24
Question 5:
Write the divisibility rule for 3.
Answer:

Sum of digits divisible by 3.

Question 6:
Find the smallest odd composite number.
Answer:
9
Question 7:
What is the product of the first three even numbers?
Answer:
48
Question 8:
Check if 57 is divisible by 3.
Answer:

Yes, 5+7=12 is divisible.

Question 9:
Find the difference between the largest 2-digit and smallest 2-digit prime numbers.
Answer:
86
Question 10:
What is the square of the smallest prime number?
Answer:
4
Question 11:
Express 36 as the sum of two odd primes.
Answer:

17 + 19 = 36

Question 12:
What is the divisibility rule for 3?
Answer:

A number is divisible by 3 if the sum of its digits is divisible by 3.
Example: 123 → 1 + 2 + 3 = 6, which is divisible by 3.

Question 13:
Find the HCF of 18 and 24 using the prime factorization method.
Answer:

Step 1: Prime factors of 18 = 2 × 3 × 3
Step 2: Prime factors of 24 = 2 × 2 × 2 × 3
Step 3: Common factors = 2 × 3
HCF = 6

Question 14:
What is the smallest 4-digit number divisible by 6?
Answer:

The smallest 4-digit number is 1000.
Divide 1000 by 6 → remainder = 4.
Required number = 1000 + (6 - 4) = 1002.

Question 15:
If a number is divisible by both 2 and 3, what other number is it divisible by?
Answer:

If a number is divisible by both 2 and 3, it must also be divisible by 6 (since 6 = 2 × 3).

Question 16:
Write the general form of a 3-digit number abc in expanded form.
Answer:

The expanded form of a 3-digit number abc is:
100a + 10b + c, where a is from 1 to 9 and b, c are from 0 to 9.

Question 17:
Check if 1452 is divisible by 11 using the divisibility rule.
Answer:

Step 1: Alternating sum = (1 + 5) - (4 + 2) = 6 - 6 = 0.
Step 2: Since 0 is divisible by 11, 1452 is divisible by 11.

Question 18:
What is the LCM of 12 and 15?
Answer:

Step 1: Prime factors of 12 = 2 × 2 × 3
Step 2: Prime factors of 15 = 3 × 5
Step 3: LCM = 2 × 2 × 3 × 5 = 60.

Question 19:
Express 36 as the sum of two odd primes.
Answer:

Possible pairs:
5 + 31 = 36
7 + 29 = 36
13 + 23 = 36
17 + 19 = 36.

Question 20:
Find the missing digit in 3_4 to make it divisible by 9.
Answer:

Step 1: Sum of known digits = 3 + 4 = 7.
Step 2: Next multiple of 9 after 7 is 9.
Step 3: Missing digit = 9 - 7 = 2 (so, 324).

Question 21:
What is the HCF of two consecutive even numbers?
Answer:

The HCF of two consecutive even numbers is 2.
Example: HCF of 8 and 10 is 2.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
What is the divisibility rule for 9?
Answer:

A number is divisible by 9 if the sum of its digits is divisible by 9.
For example, 729: 7 + 2 + 9 = 18, which is divisible by 9.

Question 2:
Find the HCF of 36 and 48 using the prime factorization method.
Answer:

Step 1: Prime factors of 36 = 2 × 2 × 3 × 3
Step 2: Prime factors of 48 = 2 × 2 × 2 × 2 × 3
Step 3: Common prime factors = 2 × 2 × 3
HCF = 12

Question 3:
What is the smallest 4-digit number divisible by 12, 15, and 20?
Answer:

Step 1: Find LCM of 12, 15, and 20 = 60
Step 2: Smallest 4-digit number = 1000
Step 3: Divide 1000 by 60 → Remainder = 40
Step 4: Required number = 1000 + (60 - 40) = 1020

Question 4:
Check if 456 is divisible by 6 using the divisibility rules.
Answer:

Rule for 6: A number must be divisible by both 2 and 3.
Step 1: 456 is even → divisible by 2
Step 2: Sum of digits = 4 + 5 + 6 = 15 → divisible by 3
Thus, 456 is divisible by 6.

Question 5:
Express 150 as a product of its prime factors.
Answer:

Step 1: 150 ÷ 2 = 75
Step 2: 75 ÷ 3 = 25
Step 3: 25 ÷ 5 = 5
Step 4: 5 ÷ 5 = 1
Prime factors: 2 × 3 × 5 × 5

Question 6:
Find the LCM of 8 and 12 using the division method.
Answer:

Step 1: Divide by common prime factors (2): 8 ÷ 2 = 4, 12 ÷ 2 = 6
Step 2: Divide again by 2: 4 ÷ 2 = 2, 6 ÷ 2 = 3
Step 3: Multiply divisors and remainders: 2 × 2 × 2 × 3 = 24
LCM = 24

Question 7:
What is the largest 3-digit number divisible by both 5 and 10?
Answer:

Condition: A number divisible by both 5 and 10 must end with 0.
The largest 3-digit number ending with 0 is 990.

Question 8:
If a number is divisible by both 4 and 9, by which other number will it always be divisible?
Answer:

Since 4 and 9 are co-prime, any number divisible by both will also be divisible by their product.
Thus, the number will always be divisible by 36 (4 × 9).

Question 9:
Find the HCF of 24 and 32 using the Euclidean algorithm.
Answer:

Step 1: 32 ÷ 24 → Remainder = 8
Step 2: 24 ÷ 8 → Remainder = 0
HCF = 8 (last non-zero remainder)

Question 10:
Write the general form of a 2-digit number and its reverse.
Answer:

Let the number be 10a + b, where a is the tens digit and b is the units digit.
Its reverse is 10b + a.

Question 11:
What is the smallest 4-digit number divisible by 6, 8, and 12?
Answer:

Step 1: Find LCM of 6, 8, 12 = 24
Step 2: Smallest 4-digit number = 1000
Step 3: Divide 1000 by 24 → Remainder = 16
Step 4: Required number = 1000 + (24 - 16) = 1008

Question 12:
Check if 1452 is divisible by both 3 and 11.
Answer:

Divisible by 3: Sum of digits = 1 + 4 + 5 + 2 = 12 (divisible by 3)
Divisible by 11: (1 + 5) - (4 + 2) = 0 (divisible by 11)
∴ 1452 is divisible by both.

Question 13:
Write the general form of a 3-digit number abc expanded using place values.
Answer:

The general form is: 100a + 10b + c
Where a is hundreds digit, b is tens digit, and c is units digit.

Question 14:
Find all factors of 36 using the multiplication method.
Answer:

Factors of 36:
1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6
∴ 1, 2, 3, 4, 6, 9, 12, 18, 36

Question 15:
What is the difference between LCM and HCF of 15 and 20?
Answer:

LCM: 15 = 3 × 5, 20 = 2 × 2 × 5 → LCM = 60
HCF: Common factor = 5
Difference: 60 - 5 = 55

Question 16:
Express 180 as a product of its prime factors using exponents.
Answer:

180 = 2 × 2 × 3 × 3 × 5
In exponential form: 2² × 3² × 5¹

Question 17:
If a number is divisible by both 5 and 12, by which other number will it always be divisible?
Answer:

Since 5 and 12 are co-prime, the number will always be divisible by their product.
∴ 60 (5 × 12 = 60)

Question 18:
Find the missing digit x in 37_x such that the number is divisible by 9.
Answer:

Sum of digits = 3 + 7 + x = 10 + x
For divisibility by 9: 10 + x must be divisible by 9
∴ x = 8 (since 18 is divisible by 9)

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Explain the divisibility rule for 3 with an example.
Answer:

A number is divisible by 3 if the sum of its digits is divisible by 3.


Example: Take the number 246.
Sum of digits = 2 + 4 + 6 = 12.
Since 12 is divisible by 3, 246 is also divisible by 3.
Question 2:
What is a perfect number? Give an example.
Answer:

A perfect number is a positive integer that is equal to the sum of its proper divisors (excluding itself).


Example: 6.
Proper divisors of 6: 1, 2, 3.
Sum = 1 + 2 + 3 = 6.
Thus, 6 is a perfect number.
Question 3:
Check if 1452 is divisible by both 3 and 11 using their divisibility rules.
Answer:

For 3: Sum of digits = 1 + 4 + 5 + 2 = 12.
Since 12 is divisible by 3, 1452 is divisible by 3.


For 11:
Alternate sum = (1 + 5) - (4 + 2) = 6 - 6 = 0.
Since 0 is divisible by 11, 1452 is divisible by 11.


Thus, 1452 is divisible by both 3 and 11.
Question 4:
Write the general form of a 3-digit number abc and express it in expanded form.
Answer:

The general form of a 3-digit number abc is:


100 × a + 10 × b + c.

Example: For the number 357,
a = 3, b = 5, c = 7.
Expanded form = 100 × 3 + 10 × 5 + 7 = 300 + 50 + 7 = 357.

Question 5:
Explain the divisibility rule for 9 with an example.
Answer:

The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is divisible by 9.

For example, take the number 729:
Step 1: Sum the digits → 7 + 2 + 9 = 18.
Step 2: Check if 18 is divisible by 9 → 18 ÷ 9 = 2 (no remainder).
Since the sum (18) is divisible by 9, the number 729 is divisible by 9.

Question 6:
What is a perfect number? Give one example.
Answer:

A perfect number is a positive integer that is equal to the sum of its proper divisors (excluding itself).

Example: 6.
Proper divisors of 6: 1, 2, 3.
Sum: 1 + 2 + 3 = 6.
Since the sum equals the number itself, 6 is a perfect number.

Question 7:
Check if 1452 is divisible by 6 using the divisibility rules.
Answer:

For a number to be divisible by 6, it must satisfy both divisibility rules for 2 and 3.

Step 1: Check divisibility by 2 → Last digit is 2 (even), so 1452 is divisible by 2.
Step 2: Check divisibility by 3 → Sum of digits = 1 + 4 + 5 + 2 = 12. Since 12 is divisible by 3, 1452 is divisible by 3.
Conclusion: 1452 is divisible by 6 as it meets both conditions.

Question 8:
Find the LCM of 15 and 20 using the division method.
Answer:

To find the LCM using the division method:

Step 1: Write the numbers (15, 20) and divide by the smallest prime (2).
20 is divisible by 2 → 10, but 15 is not. Write 15 as is.
Step 2: Divide again by 2 → 10 becomes 5, 15 remains.
Step 3: Now divide by 3 → 15 becomes 5, 5 remains.
Step 4: Divide by 5 → Both become 1.
Multiply all divisors: 2 × 2 × 3 × 5 = 60.
Thus, the LCM of 15 and 20 is 60.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain the divisibility rule for 9 with an example. How is it useful in checking calculations?
Answer:
Introduction

We studied that a number is divisible by 9 if the sum of its digits is divisible by 9. This rule helps verify calculations quickly.


Argument 1
  • Example: 729 → 7+2+9=18, which is divisible by 9.
  • Our textbook shows this rule simplifies checking large numbers.

Argument 2

In real life, it ensures accuracy in billing or inventory counts. For instance, verifying 9-item bundles.


Conclusion

This rule saves time and reduces errors, making it practical for daily use.

Question 2:
Find the general form of a 3-digit number abc and prove it is divisible by 11 if (a-b+c) is divisible by 11.
Answer:
Introduction

A 3-digit number abc can be written as 100a + 10b + c. We will prove its divisibility by 11.


Argument 1
  • Rewrite as 99a + a + 11b - b + c = (99a + 11b) + (a - b + c).
  • Since 99a + 11b is always divisible by 11, the term (a - b + c) must also be divisible by 11.

Argument 2

Example: 132 → (1-3+2)=0, which is divisible by 11. Hence, 132 ÷ 11 = 12.


Conclusion

This rule helps quickly identify multiples of 11, useful in algebra and number puzzles.

Question 3:
A number is divisible by both 5 and 12. By which other numbers will it always be divisible? Justify with LCM and examples.
Answer:
Introduction

If a number is divisible by 5 and 12, it must be divisible by their LCM. We will find other divisors.


Argument 1
  • LCM of 5 and 12 is 60. So, the number is a multiple of 60.
  • Example: 60 is divisible by 1, 2, 3, 4, 6, 10, 15, 20, 30.

Argument 2

Our textbook shows that common multiples inherit divisors of their factors. Real-life applications include scheduling events.


Conclusion

Understanding LCM helps identify all possible divisors, simplifying problem-solving.

Question 4:
Find the HCF of 36 and 48 using prime factorization. Why is HCF important in real life?
Answer:
Introduction

HCF (Highest Common Factor) is the largest number dividing two given numbers. We find it using prime factors.


Argument 1
  • 36 = 2×2×3×3, 48 = 2×2×2×2×3.
  • Common factors: 2×2×3 = 12 (HCF).

Argument 2

HCF is used to divide things equally, like distributing 36 pencils and 48 erasers among students.


Conclusion

Prime factorization makes HCF easy to calculate, helping in fair distribution.

Question 5:
A number is divisible by both 5 and 12. What is the smallest such number? Explain using LCM.
Answer:
Introduction

To find the smallest number divisible by both 5 and 12, we use LCM (Least Common Multiple).


Argument 1
  • 5 = 5, 12 = 2×2×3.
  • LCM = 2×2×3×5 = 60.

Argument 2

Our textbook shows LCM applications in scheduling, like buses arriving every 5 and 12 minutes.


Conclusion

60 is the smallest number divisible by both, proving LCM's practical use.

Question 6:
Find the HCF of 36 and 60 using the prime factorization method. Why is HCF important in real-life scenarios?
Answer:
Introduction

HCF is the highest common factor of numbers. We find it using prime factors.


Argument 1
  • 36 = 2×2×3×3, 60 = 2×2×3×5.
  • Common factors: 2×2×3 = 12 (HCF).

Argument 2

HCF helps divide things equally, like distributing 36 pencils and 60 erasers in packets.


Conclusion

Prime factorization makes HCF easy to calculate for practical uses.

Question 7:
A number is divisible by both 5 and 12. What is the smallest such number? Explain using the LCM concept.
Answer:
Introduction

LCM gives the smallest number divisible by two or more numbers. Here, we find LCM of 5 and 12.


Argument 1
  • 5 = 5, 12 = 2×2×3.
  • LCM = 5×2×2×3 = 60.

Argument 2

Our textbook shows LCM is used in scheduling, like buses arriving every 5 and 12 minutes.


Conclusion

60 is the smallest number divisible by both 5 and 12, proven by LCM.

Question 8:
Explain the divisibility rule for 9 with examples. How can this rule help us check if 783 is divisible by 9?
Answer:
Introduction

We studied that a number is divisible by 9 if the sum of its digits is divisible by 9. This rule simplifies calculations.


Argument 1
  • Example: For 27, 2 + 7 = 9, which is divisible by 9.
  • Our textbook shows similar problems like 99 (9 + 9 = 18, divisible by 9).

Argument 2

For 783, sum of digits: 7 + 8 + 3 = 18. Since 18 is divisible by 9, 783 is divisible by 9.


Conclusion

This rule helps quickly verify divisibility without full division.

Question 9:
A number is reversed and added to itself (e.g., 23 + 32 = 55). Prove that the resulting number is always divisible by 11 using a two-digit example.
Answer:
Introduction

We explored how reversing digits and adding creates numbers divisible by 11. Let's test this with a two-digit number.


Argument 1
  • Let the number be 34. Reversed: 43. Sum: 34 + 43 = 77.
  • 77 ÷ 11 = 7, proving divisibility.

Argument 2

Algebraically, a two-digit number 10a + b, when reversed (10b + a), sums to 11(a + b), always divisible by 11.


Conclusion

This pattern holds for all two-digit numbers, as shown in NCERT examples.

Question 10:
Describe how to find the HCF of 36 and 60 using the prime factorization method. Why is HCF useful in real life?
Answer:
Introduction

HCF is the highest common factor of numbers. We use prime factorization to find it efficiently.


Argument 1
  • 36 = 2 × 2 × 3 × 3, 60 = 2 × 2 × 3 × 5.
  • Common factors: 2 × 2 × 3 = 12. So, HCF is 12.

Argument 2

HCF helps in real life, like dividing 36 pencils and 60 erasers equally among students.


Conclusion

Prime factorization simplifies HCF calculation, as taught in our textbook.

Question 11:
Explain the divisibility rule for 11 with examples. Also, verify the rule using the number 9185.
Answer:

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of its digits in the odd positions and the sum of its digits in the even positions is either 0 or a multiple of 11 (including negative multiples).


Steps to apply the rule:

  • Label the digits from right to left as position 1, 2, 3, etc.
  • Sum the digits in odd positions (1st, 3rd, 5th, etc.).
  • Sum the digits in even positions (2nd, 4th, 6th, etc.).
  • Find the difference between the two sums.
  • If the difference is 0 or divisible by 11, the original number is divisible by 11.

Example with 9185:


Digits: 9 (position 4), 1 (position 3), 8 (position 2), 5 (position 1).
Sum of odd positions (1st and 3rd): 5 + 1 = 6.
Sum of even positions (2nd and 4th): 8 + 9 = 17.
Difference: 17 - 6 = 11.
Since 11 is divisible by 11, the number 9185 is divisible by 11.

Verification: 9185 ÷ 11 = 835, which confirms the rule.

Question 12:
A number is divisible by both 5 and 12. By which other numbers will that number always be divisible? Justify your answer with an example.
Answer:

If a number is divisible by both 5 and 12, it must also be divisible by the Least Common Multiple (LCM) of 5 and 12. This is because the LCM is the smallest number that is a multiple of both.


Steps to find LCM of 5 and 12:

  • Prime factorization of 5: 51.
  • Prime factorization of 12: 22 × 31.
  • LCM = product of the highest powers of all primes: 22 × 31 × 51 = 60.

Thus, any number divisible by both 5 and 12 must be divisible by 60. Consequently, it will also be divisible by all the factors of 60, such as 1, 2, 3, 4, 6, 10, 15, 20, 30, and 60.


Example: Take 120 (divisible by 5 and 12).
120 ÷ 5 = 24, 120 ÷ 12 = 10.
120 is also divisible by 60 (120 ÷ 60 = 2) and all factors of 60.

Question 13:
Explain the divisibility rule for 11 with examples. Also, verify whether the number 90827 is divisible by 11 using this rule.
Answer:

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of its digits in the odd positions (from the right) and the sum of its digits in the even positions is either 0 or a multiple of 11.


Steps to apply the rule:

  • Start from the rightmost digit (units place) and label positions as 1 (odd), 2 (even), 3 (odd), and so on.
  • Add the digits in the odd positions.
  • Add the digits in the even positions.
  • Find the difference between the two sums.
  • If the difference is 0 or divisible by 11, the number is divisible by 11.

Example: Check if 90827 is divisible by 11.


Step 1: Identify positions and digits:
Position 1 (odd): 7
Position 2 (even): 2
Position 3 (odd): 8
Position 4 (even): 0
Position 5 (odd): 9


Step 2: Sum of odd-positioned digits = 7 + 8 + 9 = 24
Sum of even-positioned digits = 2 + 0 = 2


Step 3: Difference = 24 - 2 = 22


Step 4: Since 22 is divisible by 11 (11 × 2 = 22), 90827 is divisible by 11.

Question 14:
A number is divisible by both 5 and 12. By which other numbers will it always be divisible? Justify your answer with reasoning and examples.
Answer:

If a number is divisible by both 5 and 12, it must be divisible by the Least Common Multiple (LCM) of 5 and 12. This is because the LCM represents the smallest number that is a multiple of both.


Step 1: Find the LCM of 5 and 12.
Prime factors of 5 = 5
Prime factors of 12 = 2 × 2 × 3
LCM = 2 × 2 × 3 × 5 = 60


Step 2: Since the number is divisible by 60, it will also be divisible by all the factors of 60. These include:

  • 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.


Example: Let the number be 120 (divisible by both 5 and 12).
120 ÷ 5 = 24
120 ÷ 12 = 10
120 ÷ 60 = 2
120 is also divisible by other factors of 60, such as 15 (120 ÷ 15 = 8).


Conclusion: Any number divisible by both 5 and 12 will always be divisible by the LCM (60) and its factors.

Question 15:
Explain the divisibility rule for 11 with examples. Also, verify the divisibility of the number 90827 by 11 using this rule.
Answer:

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of its digits in the odd positions (from the right) and the sum of its digits in the even positions is either 0 or a multiple of 11.


Example 1: Check if 121 is divisible by 11.
Sum of odd-positioned digits (1st and 3rd): 1 + 1 = 2
Sum of even-positioned digits (2nd): 2
Difference: 2 - 2 = 0 → Divisible by 11.


Example 2: Check if 1331 is divisible by 11.
Sum of odd-positioned digits: 1 + 3 = 4
Sum of even-positioned digits: 3 + 1 = 4
Difference: 4 - 4 = 0 → Divisible by 11.


Verification for 90827:
Digits from right: 7 (1st), 2 (2nd), 8 (3rd), 0 (4th), 9 (5th).
Sum of odd-positioned digits (1st, 3rd, 5th): 7 + 8 + 9 = 24
Sum of even-positioned digits (2nd, 4th): 2 + 0 = 2
Difference: 24 - 2 = 22
Since 22 is a multiple of 11 (11 × 2 = 22), 90827 is divisible by 11.

Question 16:
A number is divisible by both 5 and 12. By which other numbers will that number always be divisible? Justify your answer with a suitable example.
Answer:

If a number is divisible by both 5 and 12, it must be divisible by the Least Common Multiple (LCM) of 5 and 12. This is because the LCM represents the smallest number that is a multiple of both.


Step 1: Find LCM of 5 and 12.
Prime factors of 5: 5
Prime factors of 12: 2 × 2 × 3
LCM = 2 × 2 × 3 × 5 = 60.


Step 2: Identify other divisors.
Since 60 is the LCM, the number will also be divisible by all factors of 60. These include:

  • 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60


Example: Let the number be 120 (divisible by both 5 and 12).
120 ÷ 5 = 24, 120 ÷ 12 = 10 → Valid.
120 is also divisible by 6, 10, 15, etc., as they are factors of 60 (LCM).


Conclusion: Such a number will always be divisible by the common factors of 5 and 12, particularly their LCM and its divisors.

Question 17:
Explain the divisibility rule for 11 with an example. Also, verify the rule using the number 9185.
Answer:

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of its digits in the odd positions (from the right) and the sum of its digits in the even positions is either 0 or a multiple of 11.


Example: Let’s take the number 9185.


Step 1: Identify the digits in odd and even positions (from the right):


  • Odd positions (1st and 3rd from right): 5 (1st) and 1 (3rd)
  • Even positions (2nd and 4th from right): 8 (2nd) and 9 (4th)

Step 2: Calculate the sum of digits in odd positions: 5 + 1 = 6


Step 3: Calculate the sum of digits in even positions: 8 + 9 = 17


Step 4: Find the difference: 17 - 6 = 11


Since 11 is a multiple of 11, the number 9185 is divisible by 11.


Verification: 9185 ÷ 11 = 835, which is a whole number. Hence, the rule is verified.

Question 18:
A number is divisible by both 5 and 12. By which other numbers will that number always be divisible? Justify your answer with an example.
Answer:

If a number is divisible by both 5 and 12, it must also be divisible by the Least Common Multiple (LCM) of 5 and 12. This is because the LCM represents the smallest number that is a multiple of both.


Step 1: Find the LCM of 5 and 12.


  • Prime factors of 5: 5
  • Prime factors of 12: 2 × 2 × 3
  • LCM = 2 × 2 × 3 × 5 = 60

Step 2: Since the number is divisible by 60, it will also be divisible by all the factors of 60.


Example: Let’s take the number 120 (which is divisible by both 5 and 12).


Verification: 120 ÷ 5 = 24, 120 ÷ 12 = 10. Now, 120 is also divisible by 1, 2, 3, 4, 6, 10, 15, 20, 30, 60, etc., as these are factors of 60.


Thus, any number divisible by both 5 and 12 will always be divisible by the LCM (60) and its factors.

Question 19:
Explain the divisibility rule for 11 with examples. Also, verify if the number 90827 is divisible by 11 using this rule.
Answer:

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of its digits in the odd positions (from the right) and the sum of its digits in the even positions is either 0 or a multiple of 11.


Example 1: 121
Sum of odd-positioned digits (1st and 3rd): 1 + 1 = 2
Sum of even-positioned digits (2nd): 2
Difference: 2 - 2 = 0 → Divisible by 11.


Example 2: 1331
Sum of odd-positioned digits: 1 + 3 = 4
Sum of even-positioned digits: 3 + 1 = 4
Difference: 4 - 4 = 0 → Divisible by 11.


Verification for 90827:
Digits: 9 (5th), 0 (4th), 8 (3rd), 2 (2nd), 7 (1st)
Sum of odd-positioned digits (5th, 3rd, 1st): 9 + 8 + 7 = 24
Sum of even-positioned digits (4th, 2nd): 0 + 2 = 2
Difference: 24 - 2 = 22
Since 22 is a multiple of 11 (11 × 2), 90827 is divisible by 11.

Question 20:
A number is divisible by both 5 and 12. By which other numbers will it always be divisible? Justify your answer with reasoning and examples.
Answer:

If a number is divisible by both 5 and 12, it must be divisible by the Least Common Multiple (LCM) of 5 and 12. This is because the LCM is the smallest number that is a multiple of both.


Step 1: Find LCM of 5 and 12
Prime factors of 5: 5
Prime factors of 12: 22 × 3
LCM = 22 × 3 × 5 = 60.


Step 2: Identify divisors of 60
The number will always be divisible by all factors of 60, such as:

  • 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60


Example: 120 (divisible by 5 and 12)
120 ÷ 5 = 24, 120 ÷ 12 = 10
120 is also divisible by 60, 30, 20, etc., as they are factors of 60.


Conclusion: Such a number will always be divisible by the common factors of 5 and 12, particularly their LCM (60) and its factors.

Question 21:
Explain the divisibility rule for 11 with examples. Also, verify the rule using the number 90827.
Answer:

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of its digits in the odd positions (from the right) and the sum of its digits in the even positions is either 0 or a multiple of 11.


Example: Let’s verify the rule for the number 90827.


Step 1: Identify the digits in odd and even positions (from right to left):


  • Odd positions (1st, 3rd, 5th): 7 (1st), 0 (3rd), 9 (5th)
  • Even positions (2nd, 4th): 2 (2nd), 8 (4th)

Step 2: Calculate the sum of digits in odd positions: 7 + 0 + 9 = 16.


Step 3: Calculate the sum of digits in even positions: 2 + 8 = 10.


Step 4: Find the difference: 16 - 10 = 6.


Since 6 is not divisible by 11, the number 90827 is not divisible by 11.


Additional Tip: This rule is helpful for quickly checking divisibility without performing full division.

Question 22:
A number is divisible by both 5 and 12. By which other numbers will it always be divisible? Justify your answer with an example.
Answer:

If a number is divisible by both 5 and 12, it must also be divisible by the Least Common Multiple (LCM) of 5 and 12.


Step 1: Find the LCM of 5 and 12.


  • Prime factors of 5: 5
  • Prime factors of 12: 2 × 2 × 3
  • LCM = 2 × 2 × 3 × 5 = 60

Step 2: Any number divisible by 60 will also be divisible by its factors.


Example: Take the number 120 (which is divisible by both 5 and 12).


  • 120 ÷ 5 = 24 (divisible)
  • 120 ÷ 12 = 10 (divisible)
  • 120 ÷ 60 = 2 (divisible)

Thus, the number will always be divisible by 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.


Key Concept: The LCM of two numbers ensures divisibility by all common and individual factors.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A shopkeeper sells notebooks in packs of 12 and pens in packs of 8. Riya wants to buy the same number of notebooks and pens. What is the minimum number of each item she should buy? Explain using LCM.
Answer:
Problem Interpretation

We need to find the smallest number where notebooks (12 per pack) and pens (8 per pack) match in quantity.

Mathematical Modeling

Our textbook shows that LCM gives the smallest common multiple. Here, we find LCM of 12 and 8.

Solution
  • Prime factors: 12 = 22 × 3, 8 = 23
  • LCM = 23 × 3 = 24

Riya should buy 24 notebooks (2 packs) and 24 pens (3 packs).

Question 2:
A number is divisible by 9 if the sum of its digits is divisible by 9. Check if 4,32,618 is divisible by 9 using this rule.
Answer:
Problem Interpretation

We studied that divisibility by 9 depends on the sum of digits. Let’s verify for 4,32,618.

Mathematical Modeling

Add all digits: 4 + 3 + 2 + 6 + 1 + 8.

Solution
  • Sum = 4 + 3 + 2 + 6 + 1 + 8 = 24
  • 24 ÷ 9 = 2.66 (not a whole number)

Since 24 isn’t divisible by 9, 4,32,618 is not divisible by 9.

Question 3:
A shopkeeper sells notebooks in packs of 12 and pens in packs of 8. Riya wants to buy the same number of notebooks and pens. What is the smallest number of each item she can buy?
Answer:
Problem Interpretation

We need to find the smallest number where notebooks (12 per pack) and pens (8 per pack) match in quantity.

Mathematical Modeling

This is a LCM problem. We studied that LCM gives the smallest common multiple.

Solution
  • Prime factors of 12: 22 × 3
  • Prime factors of 8: 23
  • LCM = 23 × 3 = 24

Riya must buy 24 notebooks (2 packs) and 24 pens (3 packs).

Question 4:
A number is divisible by both 5 and 12. By which other numbers will it always be divisible? Justify your answer using the concept we learned.
Answer:
Problem Interpretation

We must identify numbers that divide any common multiple of 5 and 12.

Mathematical Modeling

Our textbook shows that if a number is divisible by two others, it’s also divisible by their LCM and its factors.

Solution
  • LCM of 5 and 12 = 60
  • Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The number will always be divisible by these factors, like 6 or 15.

Question 5:
A shopkeeper sells notebooks in packs of 12 and pencils in packs of 8. Riya wants to buy the same number of notebooks and pencils. What is the minimum number of each item she should buy? Use the concept of LCM to solve.
Answer:
Problem Interpretation

We need to find the smallest number divisible by both 12 and 8, which is their LCM.

Mathematical Modeling
  • Prime factors of 12: 22 × 3
  • Prime factors of 8: 23
Solution

LCM = highest powers of all primes: 23 × 3 = 24. Riya should buy 24 notebooks (2 packs) and 24 pencils (3 packs).

Question 6:
The product of two numbers is 2160. If their HCF is 12, find the numbers using the relationship: HCF × LCM = Product of numbers.
Answer:
Problem Interpretation

We know HCF and product, so we can find LCM first.

Mathematical Modeling
  • Given: HCF = 12, Product = 2160
  • LCM = Product ÷ HCF = 2160 ÷ 12 = 180
Solution

Possible pairs with HCF 12 and LCM 180 are (12, 180) and (36, 60). Our textbook shows verification: 12×180 = 36×60 = 2160.

Question 7:
A shopkeeper sells notebooks in packs of 12 and pens in packs of 8. Riya wants to buy the same number of notebooks and pens. What is the smallest number of each item she can buy?
Answer:
Problem Interpretation

We need to find the smallest number where the counts of notebooks and pens match. This is a LCM problem.


Mathematical Modeling

Notebooks come in 12s, pens in 8s. We find LCM of 12 and 8.


Solution
  • Prime factors of 12: 22 × 3
  • Prime factors of 8: 23
  • LCM = 23 × 3 = 24

Riya must buy 24 notebooks (2 packs) and 24 pens (3 packs).

Question 8:
A number is divisible by both 5 and 12. By which other numbers will it always be divisible? (Hint: Use common factors)
Answer:
Problem Interpretation

We studied that if a number is divisible by two others, it must also be divisible by their LCM and its factors.


Mathematical Modeling

Find LCM of 5 and 12, then list its factors.


Solution
  • 5 and 12 are co-prime (no common factors)
  • LCM = 5 × 12 = 60
  • Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The number will always be divisible by these factors.

Question 9:
A shopkeeper sells notebooks in packs of 12 and pens in packs of 8. Riya wants to buy the same number of notebooks and pens. What is the minimum number of each item she should buy? Use the concept of LCM to solve.
Answer:
Problem Interpretation

We need to find the smallest number divisible by both 12 and 8, which is their LCM.


Mathematical Modeling

Prime factors of 12 = 22 × 3, and 8 = 23.


Solution

LCM = highest power of each prime: 23 × 3 = 24. So, Riya should buy 24 notebooks (2 packs) and 24 pens (3 packs).

Question 10:
The sum of digits of a two-digit number is 9. When we reverse the digits, the number increases by 27. Find the original number using linear equations.
Answer:
Problem Interpretation

Let the number be 10x + y, where x is the tens digit and y is the units digit.


Mathematical Modeling
  • x + y = 9 (sum of digits)
  • 10y + x = 10x + y + 27 (reversed number condition)

Solution

Solving: y - x = 3. Adding to x + y = 9 gives y = 6, x = 3. Original number = 36.

Question 11:

A school is organizing a math puzzle competition. Students are given a 3-digit number ABC where A, B, and C are digits. The number satisfies the condition: ABC = 100A + 10B + C = 3 × (A + B + C).

(i) Find all possible 3-digit numbers that satisfy this condition.
(ii) Verify your answer by checking divisibility rules.

Answer:

Solution:

(i) To find all 3-digit numbers ABC such that 100A + 10B + C = 3(A + B + C):
Simplify the equation: 100A + 10B + C = 3A + 3B + 3C
Bring like terms together: 97A + 7B - 2C = 0
Rearrange: 97A + 7B = 2C
Since A is from 1 to 9 and B, C are from 0 to 9, test values of A:
For A = 1: 97 + 7B = 2C
Maximum value of 7B is 63 (when B = 9), so 2C ≤ 160 ⇒ C ≤ 80 (invalid, as C must be ≤9).
Thus, no solution exists for A ≥ 1. Wait, this suggests an error in approach.
Alternative method: The original equation implies the number is divisible by 3 (since ABC = 3 × (sum of digits)).
Possible numbers where the number is exactly 3 × sum of digits:
Let’s test numbers from 100 to 999:
108: Sum = 1 + 0 + 8 = 9; 3 × 9 = 27 ≠ 108 ✗
117: Sum = 1 + 1 + 7 = 9; 3 × 9 = 27 ≠ 117 ✗
...
135: Sum = 1 + 3 + 5 = 9; 3 × 9 = 27 ≠ 135 ✗
After testing, we find no 3-digit number satisfies this condition. (Note: The initial approach was flawed; testing reveals no valid numbers.)

(ii) Verification using divisibility:
For a number to be divisible by 3, the sum of its digits must be divisible by 3. However, the given condition requires the number to be exactly 3 × the sum of digits, which is not satisfied by any 3-digit number as shown above.

Question 12:

Riya and Priya are playing a number game. Riya writes a 2-digit number AB (where A is the tens digit and B is the units digit). Priya reverses the digits to form BA. The sum of AB and BA is 132.

(i) Find all possible pairs of AB and BA.
(ii) If the difference between AB and BA is 18, what is the original number?

Answer:

Solution:

(i) Let the original number be AB = 10A + B and reversed number be BA = 10B + A.
Given: AB + BA = 132
Substitute: (10A + B) + (10B + A) = 132
Simplify: 11A + 11B = 132
Divide by 11: A + B = 12
Possible pairs (A, B) where A and B are digits (1 to 9 for A, 0 to 9 for B):

  • (3, 9): AB = 39, BA = 93
  • (4, 8): AB = 48, BA = 84
  • (5, 7): AB = 57, BA = 75
  • (6, 6): AB = 66, BA = 66
  • (7, 5): AB = 75, BA = 57
  • (8, 4): AB = 84, BA = 48
  • (9, 3): AB = 93, BA = 39
All these pairs satisfy A + B = 12.

(ii) Given the difference AB - BA = 18:
Substitute: (10A + B) - (10B + A) = 18
Simplify: 9A - 9B = 18
Divide by 9: A - B = 2
From part (i), A + B = 12 and A - B = 2.
Add the two equations: 2A = 14 ⇒ A = 7
Subtract: B = 12 - 7 = 5
Thus, the original number is AB = 75.

Question 13:

Rahul has a two-digit number. He reverses the digits of this number and adds the original number to the reversed number. The result is 132. He also notices that the original number is 18 more than the sum of its digits. Help Rahul find the original number.

Answer:

Let the original number be 10a + b, where a is the tens digit and b is the units digit.
Reversed number = 10b + a.
According to the problem:
1. (10a + b) + (10b + a) = 132
Simplifying: 11a + 11b = 132 ⇒ a + b = 12.
2. The original number is 18 more than the sum of its digits: 10a + b = a + b + 18 ⇒ 9a = 18 ⇒ a = 2.
Substituting a = 2 in a + b = 12 ⇒ b = 10.
But b must be a single digit (0-9), so this seems invalid. Let’s recheck:
The second condition implies 10a + b = (a + b) + 18 ⇒ 9a = 18 ⇒ a = 2.
Then, b = 12 - 2 = 10, which is not possible. Thus, there might be an error in interpreting the problem.
Alternatively, if the original number is 18 more than the sum of its digits, it implies the number is a two-digit number where the tens digit is 2 (from 9a = 18).
Possible numbers: 20 to 29.
Checking which number satisfies the first condition (number + reversed = 132):
For 20: 20 + 02 = 22 ≠ 132
For 21: 21 + 12 = 33 ≠ 132
...
For 57: 57 + 75 = 132 (but a = 5, b = 7, a + b = 12, and 57 = 5 + 7 + 45 ≠ 18 + sum).
Thus, the correct original number is 57, as it satisfies the first condition, though the second condition seems misstated.

Question 14:

Priya is playing with numbers. She takes a three-digit number, reverses its digits to form another number, and subtracts the smaller number from the larger one. The result is 396. The difference between the hundreds digit and the units digit of the original number is 4. Find all possible original numbers Priya could have taken.

Answer:

Let the original number be 100a + 10b + c, where a, b, and c are its hundreds, tens, and units digits, respectively.
Reversed number = 100c + 10b + a.
Given |(100a + 10b + c) - (100c + 10b + a)| = 396 ⇒ |99a - 99c| = 396 ⇒ |a - c| = 4.
Also, the difference between the hundreds and units digit is 4: a - c = 4 or c - a = 4.
Case 1: a - c = 4
Possible pairs (a, c): (4,0), (5,1), (6,2), (7,3), (8,4), (9,5).
Original number: 100a + 10b + c. Reversed: 100c + 10b + a.
Subtracting smaller from larger: (100a + c) - (100c + a) = 99(a - c) = 396 ⇒ a - c = 4 (consistent).
Thus, possible numbers: 400, 410, ..., 499; 501, ..., 595; ..., 900, ..., 995 (with a - c = 4).
But since the reversed number must be smaller, a > c, so valid numbers are all three-digit numbers where a = c + 4 and b can be 0-9.
Examples: 400, 411, 422, ..., 499; 501, 512, ..., 595; etc.
Case 2: c - a = 4
Possible pairs (a, c): (0,4), (1,5), (2,6), (3,7), (4,8), (5,9).
But a cannot be 0 (as it’s a three-digit number), so valid pairs: (1,5), (2,6), (3,7), (4,8), (5,9).
Original number: 100a + 10b + c. Reversed: 100c + 10b + a.
Subtracting smaller from larger: (100c + a) - (100a + c) = 99(c - a) = 396 ⇒ c - a = 4 (consistent).
Thus, possible numbers: 105, 115, ..., 195; 206, ..., 296; ..., 500, ..., 599 (with c = a + 4).
Final possible original numbers: All three-digit numbers where |a - c| = 4 and b is any digit (0-9).

Question 15:
A school is organizing a math quiz where students are given a number 123456. They are asked to find the smallest number that must be subtracted from it to make it divisible by 9. Explain the steps and reasoning.
Answer:

To solve this, we use the divisibility rule of 9: a number is divisible by 9 if the sum of its digits is divisible by 9.


Step 1: Calculate the sum of digits of 123456.
1 + 2 + 3 + 4 + 5 + 6 = 21.

Step 2: Find the remainder when 21 is divided by 9.
21 ÷ 9 = 2 with remainder 3.

Step 3: The smallest number to subtract is the remainder itself, which is 3.

Verification:
123456 - 3 = 123453.
Sum of digits of 123453: 1 + 2 + 3 + 4 + 5 + 3 = 18, which is divisible by 9.

Thus, the answer is 3.

Question 16:
Riya has a two-digit number. She reverses its digits and adds the original number to the reversed number. The result is 121. Identify all possible original numbers Riya could have, showing your work.
Answer:

Let the original number be 10a + b, where a is the tens digit and b is the units digit. The reversed number is 10b + a.


Step 1: Set up the equation based on the problem.
(10a + b) + (10b + a) = 121.

Step 2: Simplify the equation.
11a + 11b = 121.
Divide both sides by 11: a + b = 11.

Step 3: Find all possible pairs (a, b) where a and b are digits (1 to 9 for a, 0 to 9 for b) satisfying a + b = 11.
  • 2 + 9 = 11 → 29
  • 3 + 8 = 11 → 38
  • 4 + 7 = 11 → 47
  • 5 + 6 = 11 → 56
  • 6 + 5 = 11 → 65
  • 7 + 4 = 11 → 74
  • 8 + 3 = 11 → 83
  • 9 + 2 = 11 → 92

Step 4: Verify one example (56 + 65 = 121).

Thus, the possible numbers are 29, 38, 47, 56, 65, 74, 83, and 92.

Question 17:
A school is organizing a math puzzle competition. Students are given a 3-digit number ABC where A, B, and C are digits. The number satisfies the condition: ABC = 100A + 10B + C = 3 × (A + B + C). Find all possible numbers that meet this condition. Explain your reasoning step-by-step.
Answer:

To solve this, we need to find all 3-digit numbers ABC such that the number itself is equal to three times the sum of its digits.


Step 1: Express the number in terms of its digits.
100A + 10B + C = 3(A + B + C)

Step 2: Simplify the equation.
100A + 10B + C = 3A + 3B + 3C
100A - 3A + 10B - 3B + C - 3C = 0
97A + 7B - 2C = 0

Step 3: Rearrange to express C in terms of A and B.
2C = 97A + 7B
C = (97A + 7B)/2

Step 4: Since C must be a digit (0-9), test possible values of A (1-9) and B (0-9) to find valid combinations.
  • For A = 1: C = (97 + 7B)/2
  • Only B = 1 gives C = 52 (invalid, as C must be ≤ 9). No valid solution for A = 1.
  • For A = 2: C = (194 + 7B)/2
  • Testing B = 0: C = 97 (invalid)
  • Testing B = 2: C = 104 (invalid)
  • No valid solution for A = 2.
  • For A = 3: C = (291 + 7B)/2
  • Testing B = 9: C = (291 + 63)/2 = 177 (invalid)
  • No valid solution for A = 3.

After testing all possible values, we find that there are no 3-digit numbers satisfying the given condition. The equation leads to values of C exceeding 9, which is invalid.

Question 18:
Rahul writes a 2-digit number. He reverses the digits to form another number. The sum of the original number and the reversed number is 110, and the difference is 36. Find the original number. Show your work systematically.
Answer:

Let the original number be 10A + B, where A is the tens digit and B is the units digit. The reversed number is 10B + A.


Step 1: Write the equations based on the given conditions.
Sum: (10A + B) + (10B + A) = 110
Difference: (10A + B) - (10B + A) = 36

Step 2: Simplify the sum equation.
11A + 11B = 110
A + B = 10 (Dividing both sides by 11)

Step 3: Simplify the difference equation.
9A - 9B = 36
A - B = 4 (Dividing both sides by 9)

Step 4: Solve the system of equations.
A + B = 10
A - B = 4
Add the two equations: 2A = 14 → A = 7
Substitute A = 7 into A + B = 10: 7 + B = 10 → B = 3

Step 5: Form the original number.
Original number = 10A + B = 10 × 7 + 3 = 73

Verification:
Reversed number = 37
Sum = 73 + 37 = 110
Difference = 73 - 37 = 36
Both conditions are satisfied.

Question 19:
A school is organizing a math quiz where students are given a number 1234. They need to find all possible two-digit numbers that can be formed using its digits such that the number is divisible by 3. Help them solve this step-by-step.
Answer:

To solve this, we first list all possible two-digit numbers formed by the digits of 1234 and then check divisibility by 3.


Step 1: List all two-digit combinations:
12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43

Step 2: Check divisibility by 3 (sum of digits must be divisible by 3):
  • 12: 1 + 2 = 3 (divisible)
  • 13: 1 + 3 = 4 (not divisible)
  • 14: 1 + 4 = 5 (not divisible)
  • 21: 2 + 1 = 3 (divisible)
  • 23: 2 + 3 = 5 (not divisible)
  • 24: 2 + 4 = 6 (divisible)
  • 31: 3 + 1 = 4 (not divisible)
  • 32: 3 + 2 = 5 (not divisible)
  • 34: 3 + 4 = 7 (not divisible)
  • 41: 4 + 1 = 5 (not divisible)
  • 42: 4 + 2 = 6 (divisible)
  • 43: 4 + 3 = 7 (not divisible)

Final Answer: The valid numbers are 12, 21, 24, 42.
Question 20:
Rahul has a five-digit number 24680. He wants to rearrange its digits to form the smallest possible five-digit number divisible by 5. Explain the steps to achieve this.
Answer:

To form the smallest five-digit number divisible by 5 from 24680, follow these steps:


Step 1: Identify the condition for divisibility by 5 (last digit must be 0 or 5). Since 5 is not in the number, the last digit must be 0.

Step 2: Arrange the remaining digits (2, 4, 6, 8) in ascending order to form the smallest number: 2468.

Step 3: Place 0 at the end to satisfy divisibility: 24680.

Verification:
  • Number formed: 24680
  • Last digit is 0, so divisible by 5.
  • No smaller arrangement is possible with 0 at the end.

Final Answer: The smallest number is 24680.
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