RS Aggarwal Solutions for Class 6 Chapter 2 Factors and Multiples

Exercise 2.1

 

Question 1. Define:
Answer: (i) Factor: A number that divides another number evenly without leaving a remainder is called its factor. As an illustration, 4 divides 32 exactly, so 4 is a factor of 32.

Examples of factors:
2 and 3 divide 6 evenly since 2 × 3 = 6
2 and 4 divide 8 evenly since 2 × 4 = 8
3 and 4 divide 12 evenly since 3 × 4 = 12
3 and 5 divide 15 evenly since 3 × 5 = 15

(ii) Multiple: When a number 'a' gets multiplied by another number 'b', the result becomes a multiple of both 'a' and 'b'.

Examples of multiples:
6 is a multiple of 2 since 2 × 3 = 6
8 is a multiple of 4 since 4 × 2 = 8
12 is a multiple of 6 since 6 × 2 = 12
21 is a multiple of 7 since 7 × 3 = 21
In simple words: A factor divides a number evenly. A multiple is what you get when you multiply two numbers together.

Exam Tip: Remember that 1 and the number itself are always factors. Every number is a multiple of itself and 1.

 

Question 2. Write all factors of each of the following numbers:
Answer: (i) 60

60 = 1 × 60
60 = 2 × 30
60 = 3 × 20
60 = 4 × 15
60 = 5 × 12
60 = 6 × 10

The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

(ii) 76

76 = 1 × 76
76 = 2 × 38
76 = 4 × 19

The factors of 76 are 1, 2, 4, 19, 38 and 76.

(iii) 125

125 = 1 × 125
125 = 5 × 25

The factors of 125 are 1, 5, 25 and 125.

(iv) 729

729 = 1 × 729
729 = 3 × 243
729 = 9 × 81
729 = 27 × 27

The factors of 729 are 1, 3, 9, 27, 81, 243 and 729.
In simple words: List all pairs of numbers that multiply to give your target number. Each number in these pairs is a factor.

Exam Tip: Always start with 1 and the number itself, then work through small divisors systematically to find all factor pairs.

 

Question 3. Write first five multiples of each of the following numbers:
Answer: (i) 25

The first five multiples of 25 are as follows:
25 × 1 = 25
25 × 2 = 50
25 × 3 = 75
25 × 4 = 100
25 × 5 = 125

(ii) 35

The first five multiples of 35 are as follows:
35 × 1 = 35
35 × 2 = 70
35 × 3 = 105
35 × 4 = 140
35 × 5 = 175

(iii) 45

The first five multiples of 45 are as follows:
45 × 1 = 45
45 × 2 = 90
45 × 3 = 135
45 × 4 = 180
45 × 5 = 225

(iv) 40

The first five multiples of 40 are as follows:
40 × 1 = 40
40 × 2 = 80
40 × 3 = 120
40 × 4 = 160
40 × 5 = 200
In simple words: Multiply your number by 1, then by 2, then by 3, and so on to get the first five multiples.

Exam Tip: The first multiple of any number is always the number itself. The multiples continue indefinitely, but the question asks only for the first five.

 

Question 4. Which of the following number have 15 as their factor?
Answer: (i) 15625

15 is not a factor of 15,625 because it cannot divide evenly into 15,625.

(ii) 123015

15 is a factor of 1,23,015 because it divides evenly into 1,23,015. That is, 8,201 × 15 = 1,23,015
In simple words: Check if 15 divides the number exactly. If it does, then 15 is its factor.

Exam Tip: A number is a factor of another if division leaves no remainder. Use divisibility rules: 15 divides a number only if both 3 and 5 divide it.

 

Question 5. Which of the following number are divisible by 21?
Answer: A given number is divisible by 21 if and only if it is divisible by each of its factors. Since 21 = 3 × 7, the factors of 21 are 1, 3, 7 and 21.

(i) 21063

Sum of the digits = 2 + 1 + 0 + 6 + 3 = 12, which is divisible by 3.

Hence, 21,063 is divisible by 3.

For divisibility by 7: A number is divisible by 7 if the difference between twice the ones digit and the number formed by the remaining digits is either 0 or a multiple of 7. Applying this rule: 2,106 - (2 × 3) = 2,100, which is divisible by 7. So, 21,063 is divisible by 21.

(ii) 20163

Sum of the digits = 2 + 0 + 1 + 6 + 3 = 12, which is divisible by 3. Hence, 20,163 is divisible by 3.

For divisibility by 7: 2,016 - (2 × 3) = 2,010, which is not divisible by 7. Therefore, 20,163 is not divisible by 21.
In simple words: To check divisibility by 21, test if the number is divisible by both 3 and 7. If both conditions hold, the number is divisible by 21.

Exam Tip: Always use divisibility rules for the factors of the target number rather than performing actual division, which saves time and reduces errors.

 

Question 6. Without actual division show that 11 is a factor of each of the following numbers:
Answer: (i) 1,111

Sum of digits at odd positions = 1 + 1 = 2

Sum of digits at even positions = 1 + 1 = 2

Difference of the two sums = 2 - 2 = 0

Since the difference equals zero, 1,111 is divisible by 11.

(ii) 11,011

Sum of digits at odd positions = 1 + 0 + 1 = 2

Sum of digits at even positions = 1 + 1 = 2

Difference of the two sums = 2 - 2 = 0

Since the difference equals zero, 11,011 is divisible by 11.

(iii) 1, 10,011

Sum of digits at odd positions = 1 + 0 + 1 = 2

Sum of digits at even positions = 1 + 0 + 1 = 2

Difference of the two sums = 2 - 2 = 0

Since the difference equals zero, 1, 10,011 is divisible by 11.

(iv) 11, 00,011

Sum of digits at odd positions = 1 + 0 + 0 + 1 = 2

Sum of digits at even positions = 1 + 0 + 1 = 2

Difference of the two sums = 2 - 2 = 0

Since the difference equals zero, 11, 00,011 is divisible by 11.
In simple words: Add the digits in odd positions and the digits in even positions separately. If the difference between these two sums is zero or a multiple of 11, the number is divisible by 11.

Exam Tip: The alternating sum rule for 11 is one of the most efficient divisibility tests. Always label positions from right to left (ones place is position 1).

 

Question 7. Without actual division show that each of the following numbers is divisible by 5:
Answer: A number is divisible by 5 when its ones digit is either 0 or 5.

(i) 5

In 5, the ones digit is 5. Hence, it is divisible by 5.

(ii) 555

In 555, the ones digit is 5. Hence, it is divisible by 5.

(iii) 5555

In 5,555, the ones digit is 5. Hence, it is divisible by 5.

(iv) 50,005

In 50,005, the ones digit is 5. Hence, it is divisible by 5.
In simple words: Look at only the last digit of the number. If it is 0 or 5, the whole number is divisible by 5.

Exam Tip: The divisibility rule for 5 is the simplest—check only the units digit. This works because our number system is base 10.

 

Question 8. Is there any natural number having no factor at all?
Answer: No, because every natural number is a factor of itself
In simple words: Every natural number has at least two factors: 1 and itself.

Exam Tip: Remember that 1 is always a factor of any number, so no natural number can exist without factors.

 

Question 9. Find numbers between 1 and 100 having exactly three factors
Answer: The numbers between 1 and 100 having exactly three factors are 4, 9, 25, and 49.

The factors of 4 are 1, 2 and 4.

The factors of 9 are 1, 3 and 9.

The factors of 25 are 1, 5 and 25.

The factors of 49 are 1, 7 and 49.
In simple words: Numbers with exactly three factors are squares of prime numbers. This is because if p is prime, then p² has only three factors: 1, p, and p².

Exam Tip: A number has exactly three factors if and only if it is the square of a prime number. The three factors are always 1, the prime, and the square itself.

 

Question 10. Sort out even and odd numbers:
Answer: A number that is exactly divisible by 2 is called an even number. Therefore, 42 and 144 are even numbers.

A number that is not exactly divisible by 2 is called an odd number. Therefore, 89 and 321 are odd numbers.
In simple words: Even numbers end in 0, 2, 4, 6, or 8. Odd numbers end in 1, 3, 5, 7, or 9.

Exam Tip: Always check the units digit first—if it's even, the whole number is even; if it's odd, the whole number is odd.

Exercise 2.2

 

Question 1. Find the common factors of:
Answer: (i) 15 and 25

15 = 1 × 15
15 = 3 × 5

So, the factors of 15 are 1, 3, 5 and 15.

Additionally, 25 = 1 × 25
25 = 5 × 5, so the factors of 25 are 1, 5 and 25.

Therefore, the common factors of the two numbers are 1 and 5.

(ii) 35 and 50

35 = 1 × 35
35 = 5 × 7, so the factors of 35 are 1, 5, 7 and 35.

Again, 50 = 1 × 50
50 = 2 × 25
50 = 5 × 10

So, the factors of 50 are 1, 2, 5, 10, 25 and 50.

Therefore, the common factors of the two numbers are 1 and 5.

(iii) 20 and 28

20 = 1 × 20
20 = 2 × 10
20 = 4 × 5

So, the factors of 20 are 1, 2, 4, 5, 10 and 20.

Again, 28 = 1 × 28
28 = 2 × 14
28 = 7 × 4

So, the factors of 28 are 1, 2, 4, 7, 14 and 28.

Therefore, the common factors of the two numbers are 1, 2 and 4.
In simple words: Find all factors of each number, then pick the ones that appear in both lists.

Exam Tip: The largest common factor is the greatest common divisor (GCD). Always list factors in order to ensure you don't miss any.

 

Question 2. Find the common factors of:
Answer: (i) 5, 15 and 25

Factors of 5 are 1 and 5

Factors of 15 are 1, 3, 5 and 15

Factors of 25 are 1, 5 and 25

Therefore, the common factors of 5, 15, and 25 are 1 and 5.

(ii) 2, 6 and 8

Factors of 2 are 1 and 2

Factors of 6 are 1, 2, 3 and 6

Factors of 8 are 1, 2, 4 and 8

Therefore, the common factors of 2, 6 and 8 are 1 and 2.
In simple words: When finding common factors of three numbers, identify factors that divide all three numbers evenly.

Exam Tip: For multiple numbers, list factors of each, then identify which appear in ALL lists. The smallest common factor is always 1.

 

Question 3. Find first three common multiples of 6 and 8
Answer: Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84...

Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96...

Therefore, the first three common multiples of 6 and 8 are 24, 48 and 72.
In simple words: Write out multiples of each number and look for the ones that show up in both lists.

Exam Tip: The smallest common multiple is the least common multiple (LCM). After finding the first common multiple, the next ones appear at regular intervals equal to the LCM.

 

Question 4. Find first two common multiples of 12 and 18.
Answer: Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132...

Multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198...

Therefore, the first two common multiples of 12 and 18 are 36 and 72.
In simple words: List multiples of both numbers until you find two that repeat in both lists.

Exam Tip: Common multiples always include the LCM as the smallest one. All subsequent common multiples are obtained by multiplying the LCM by whole numbers.

 

Question 5. A number is divisible by both 7 and 16. By which other number will that number be always divisible?
Answer: Since the number is divisible by 7 and 16, they represent factors of that number. As a result, the number will be divisible by the common factor of 7 and 16.

The factors of 7 are 1 and 7.

The factors of 16 are 1, 2, 4, 8, and 16.

Therefore, the common factor of 7 and 16 is 1 and the number is divisible by 1.
In simple words: When 7 and 16 both divide a number, that number is definitely divisible by all factors of both 7 and 16. Since their only common factor is 1, the number is divisible by 1.

Exam Tip: When two numbers share no common factors other than 1, they are called coprime. A number divisible by both must be divisible by their product as well.

 

Question 6. A number is divisible by 24. By what other numbers will that number be divisible?
Answer: Since the number is divisible by 24, it will be divisible by all the factors of 24.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

Hence, the number is also divisible by 1, 2, 3, 4, 6, 8 and 12.
In simple words: If a number is divisible by 24, then it must be divisible by every factor of 24, including 1, 2, 3, 4, 6, 8, and 12.

Exam Tip: When a number is divisible by a given number, it is automatically divisible by all factors of that number. This property simplifies divisibility checks significantly.

 

Exercise 2.3

 

Question 1. What are prime numbers? List all the prime numbers between 1 and 30.
Answer: Numbers that have exactly two factors—1 and themselves—are known as prime numbers. Examples include 2, 3, 5, 7, 11, and 13. The prime numbers between 1 and 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
In simple words: A prime number can only be divided evenly by 1 and itself. No other number divides into it without leaving a remainder.

Exam Tip: Memorize primes up to 30 for quick reference in exams—this list appears frequently in divisibility and factorization questions.

 

Question 2. Write all the prime numbers between:
(i) 10 and 50
(ii) 70 and 90
(iii) 40 and 85
(iv) 60 and 100
Answer:
(i) The prime numbers between 10 and 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47.
(ii) The prime numbers between 70 and 90 are 71, 73, 79, 83, and 89.
(iii) The prime numbers between 40 and 85 are 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, and 83.
(iv) The prime numbers between 60 and 100 are 61, 67, 71, 73, 79, 83, 89, and 97.
In simple words: To find primes in a range, check if each number can be divided by smaller primes like 2, 3, 5, or 7.

Exam Tip: When listing primes in a range, always check divisibility systematically—missing even one prime costs marks.

 

Question 3. What is the smallest prime number? Is it an even number?
Answer: The number 2 is the smallest prime number. Yes, it is an even prime number. It stands out because 2 is the only even prime—all other even numbers are composite since they can be divided by 2.
In simple words: 2 is prime because it has exactly two factors: 1 and 2. It is also the only even number that is prime.

Exam Tip: Remember that 2 is the only even prime—this fact often appears in true/false or fill-in-the-blank questions.

 

Question 4. What is the smallest odd prime? Is every odd number a prime number? If not, give an example of an odd number which is not prime. If yes, write the smallest odd composite number.
Answer: The smallest odd prime number is 3. No, not every odd number is a prime number. For instance, 9 is an odd number, but it is not prime because its factors are 1, 3, and 9 (more than two factors). Since not all odd numbers are prime, the smallest odd composite number is 9.
In simple words: 3 is the smallest odd prime. But 9 is odd and not prime—it breaks into 3 × 3, so it has three factors.

Exam Tip: Do not confuse odd numbers with prime numbers—many odd numbers are composite, like 9, 15, 21, and 25.

 

Question 5. What are composite numbers? Can a composite number be odd?
Answer: A composite number is one that has more than two factors. For example, the numbers 4, 6, 8, 9, 10, and 15 are all composite. Yes, a composite number can be odd. The smallest odd composite number is 9.
In simple words: Composite numbers break apart into smaller factors. They can be even like 4 or 6, but also odd like 9 and 15.

Exam Tip: Every number greater than 1 is either prime or composite—there is no other category.

 

Question 6. What are twin-primes? Write all pairs of twin-primes between 50 and 100.
Answer: Twin primes are two prime numbers where exactly one composite number exists between them. For example, (3, 5) and (5, 7) form twin prime pairs. The twin prime pairs between 50 and 100 are (59, 61) and (71, 73).
In simple words: Twin primes are two primes that are only 2 apart, like 5 and 7 or 11 and 13.

Exam Tip: To verify twin primes, check both numbers carefully for primality and ensure their difference is exactly 2.

 

Question 7. What are co-primes? Give examples of five pairs of co-primes. Are co-primes always prime?
Answer: Two numbers are called co-primes if they share no common factors other than 1. Examples of co-prime pairs include (2, 3), (3, 4), (4, 5), (5, 7), and (13, 17). However, co-prime numbers do not need to both be prime. For instance, (3, 4), (6, 7), and (4, 13) are co-prime pairs even though some of these numbers are composite.
In simple words: Co-primes have only 1 as a common factor. They can both be prime, or one (or both) can be composite.

Exam Tip: Use the GCD test: if GCD(a, b) = 1, then a and b are co-primes—this method works faster than listing all factors.

 

Question 8. Which of the following pairs are always co-primes?
(i) Two prime numbers
(ii) One prime and one composite number
(iii) Two composite numbers
Answer:
(i) Two prime numbers are always co-prime to each other. Example: 7 and 11 are co-prime.
(ii) One prime and one composite number are not always co-prime. Example: 3 and 21 are not co-prime to each other.
(iii) Two composite numbers are not always co-prime to each other. Example: 4 and 6 are not co-prime to each other.
In simple words: Different primes never share factors, so any two primes are always co-prime. But mixing primes with composites, or two composites together, may result in shared factors.

Exam Tip: The first statement (two primes are always co-prime) is a key rule—use it to quickly identify co-prime pairs in multiple-choice questions.

 

Question 9. Express each of the following numbers as a sum of two or more primes:
(i) 13
(ii) 130
(iii) 180
Answer:
(i) 13 can be written as 11 + 2.
(ii) 130 can be written as 59 + 71.
(iii) 180 can be written as 139 + 17 + 11 + 13 or as 79 + 101.
In simple words: Pick any primes that add up to your target number. There may be multiple correct ways to do this.

Exam Tip: For large numbers, finding at least one valid decomposition is usually sufficient—the question does not require all possible combinations.

 

Question 10. Express each of the following numbers as the sum of two odd primes:
(i) 36
(ii) 42
(iii) 84
Answer:
(i) 36 can be written as 7 + 29 or 17 + 19.
(ii) 42 can be written as 5 + 37 or 13 + 29.
(iii) 84 can be written as 17 + 67 or 23 + 61.
In simple words: Find two odd primes that add to the given number. Since both primes are odd, their sum is always even.

Exam Tip: This type of question relates to Goldbach's conjecture—every even number greater than 2 can be expressed as a sum of two primes.

 

Question 11. Express each of the following numbers as the sum of three odd prime numbers:
(i) 31
(ii) 35
(iii) 49
Answer:
(i) 31 can be written as 5 + 7 + 9 + 13 or 11 + 13 + 7.
(ii) 35 can be written as 5 + 7 + 23 or 17 + 13 + 5.
(iii) 49 can be written as 3 + 5 + 41 or 7 + 11 + 31.
In simple words: Add three odd primes together to reach your target sum. Odd + odd + odd always gives an odd result.

Exam Tip: When decomposing odd numbers into three odd primes, the sum of three odd numbers is always odd—this helps verify your answer.

 

Question 12. Express each of the following numbers as the sum of twin primes:
(i) 36
(ii) 84
(iii) 120
Answer:
(i) 36 = 17 + 19
(ii) 84 = 41 + 41
(iii) 120 = 59 + 61
In simple words: Twin primes are two primes separated by exactly 2. Add them together to get your target number.

Exam Tip: Not all even numbers can be expressed as the sum of a single twin prime pair—try to identify twin prime pairs first, then check if their sum matches your target.

 

Question 13. Find the possible missing twins for the following numbers so that they become twin primes:
(i) 29
(ii) 89
(iii) 101
Answer:
(i) The candidates for twin partners of 29 are 27 and 31. Since 31 is prime and 27 is not, the missing twin of 29 is 31.
(ii) The candidates for twin partners of 89 are 87 and 91. Since neither 87 nor 91 is prime, 89 has no twin.
(iii) The candidates for twin partners of 101 are 99 and 103. Since 103 is prime and 99 is not, the missing twin of 101 is 103.
In simple words: A number's twin partners are 2 less and 2 more. Check if either of these is prime.

Exam Tip: For a number to have a twin, at least one of the numbers 2 steps away must be prime—use divisibility rules to check primality quickly.

 

Question 14. A list consists of the following pairs of numbers:
Answer:
(i) Co-primes: Two natural numbers are said to be co-primes if they have 1 as their only common factor. Therefore, all the given pairs of numbers are co-primes.
(ii) Primes: Natural numbers which have exactly two distinct factors—1 and the number itself—are called prime numbers. Therefore, (59, 61) and (71, 73) are pairs of prime numbers.
(iii) Composite numbers: Natural numbers which have more than two factors are called composite numbers. Therefore, (55, 57) and (63, 65) are pairs of composite numbers.
In simple words: Co-primes share only 1 as a common factor. Prime pairs have exactly two factors each. Composite pairs break apart into smaller factors.

Exam Tip: When classifying number pairs, check the number of factors for each number separately, then verify if they share common factors.

 

Question 15. For a number, greater than 10, to be prime what may be the possible digit in the unit's place?
Answer: For a number (greater than 10) to be a prime number, the possible digit in the unit's place may be 1, 3, 7, or 9. Example: 11, 13, 17, and 19 are prime numbers greater than 10.
In simple words: Prime numbers larger than 10 always end in 1, 3, 7, or 9. They cannot end in 0, 2, 4, 5, 6, or 8 because those would be divisible by other numbers.

Exam Tip: Use this rule to eliminate candidates when checking primality—if a number ends in 2, 4, 5, 6, 8, or 0, it is definitely composite.

 

Question 16. Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Answer: The required seven consecutive composite numbers are 90, 91, 92, 93, 94, 95, and 96.
In simple words: These seven numbers in a row are all composite—none of them is prime, and no prime sits between them.

Exam Tip: To find such gaps, look for ranges where no prime exists—factorial expressions like 8! + 2, 8! + 3, etc., guarantee composite numbers.

 

Question 17. State true (T) and false (F):
(i) The sum of primes cannot be a prime.
(ii) The product of primes cannot be a prime.
(iii) An even number is composite
(iv) Two consecutive numbers cannot be a prime.
(v) Odd numbers cannot be composite.
(vi) Odd numbers cannot be written as sum of primes.
(vii) A number and its successor are always co-primes.
Answer:
(i) False. The sum 2 + 3 = 5, which is a prime number.
(ii) True. The product of prime numbers is always composite.
(iii) False. The even number 2 is not composite.
(iv) False. The numbers 2 and 3 are consecutive and both are prime.
(v) False. The odd number 9 is composite, with factors 1, 3, and 9.
(vi) False. The odd number 9 can be written as 9 = 7 + 2, where both 7 and 2 are primes.
(vii) True. A number and its successor share only 1 as a common factor.
In simple words: (i) Adding primes can yield another prime. (ii) Multiplying primes always makes composites. (iii) The number 2 is even and prime, breaking the rule. (iv) 2 and 3 are prime neighbors. (v) 9 is odd yet composite. (vi) Odds can sum from primes. (vii) Any two consecutive integers are co-prime.

Exam Tip: Memorize the true/false results for statements (ii) and (vii)—these appear repeatedly in exams as conceptual checks.

 

Question 18. Fill in the Blank:
(i) A number having only two factors is called a prime number.
(ii) A number having more than two factors is called a composite number.
(iii) 1 is neither composite nor prime.
(iv) The smallest prime number is 2.
(v) The smallest composite number is 4.
In simple words: These are fundamental definitions: primes have two factors, composites have more, and 1 stands apart. 2 is the smallest prime; 4 is the smallest composite.

Exam Tip: These definitions appear in nearly every exam—ensure you can state them clearly without hesitation.

 

Exercise 2.4

 

Question 1. In which of the following expressions, prime factorization has been done?
Answer:
(i) 24 = 2 × 3 × 4 is not a prime factorization because 4 is not a prime number.
(ii) 56 = 1 × 7 × 2 × 2 × 2 is not a prime factorization because 1 is not a prime number.
(iii) 70 = 2 × 5 × 7 is a prime factorization because 2, 5, and 7 are all prime numbers.
(iv) 54 = 2 × 3 × 9 is not a prime factorization because 9 is not a prime number.
In simple words: A prime factorization uses only prime numbers—no composite numbers or 1 allowed.

Exam Tip: Check each factor individually—if even one factor is composite or equals 1, the entire factorization fails.

 

Question 2. Determine prime factorization of each of the following numbers:
Answer:
(i) 216
Using division by prime factors:

Prime factorization of 216 = 2 × 2 × 2 × 3 × 3 × 3

(ii) 420
Using division by prime factors:Prime factorization of 420 = 2 × 2 × 3 × 5 × 7

(iii) 468
Using division by prime factors:Prime factorization of 468 = 2 × 2 × 3 × 3 × 13

(iv) 945
Using division by prime factors:Prime factorization of 945 = 3 × 3 × 3 × 5 × 7

(v) 7325
Using division by prime factors:Prime factorization of 7325 = 5 × 5 × 293

(vi) 13915
Using division by prime factors:Prime factorization of 13915 = 5 × 11 × 11 × 23
In simple words: Divide repeatedly by the smallest prime that goes in evenly. Continue until you reach 1. The prime factors you collected form the prime factorization.

Exam Tip: Always start with 2, then try 3, 5, 7, and so on—testing primes in order keeps the work organized and reduces errors.

 

Question 3. Write the smallest 4-digit number and express it as a product of primes.
Answer: The smallest 4-digit number is 1000. Breaking it down step-by-step:
1000 = 2 × 500
= 2 × 2 × 250
= 2 × 2 × 2 × 125
= 2 × 2 × 2 × 5 × 25
= 2 × 2 × 2 × 5 × 5 × 5
In simple words: 1000 breaks down into eight prime factors: three 2's and five 5's multiplied together.

Exam Tip: Writing out each step clearly shows your work and helps catch errors—the final answer can be written in compact form using exponents: 2³ × 5³.

 

Question 4. Write the largest 4-digit number and express it as product of primes.
Answer: The largest 4-digit number is 9999. Using prime factorization through division, we get: 9999 = 3 × 3 × 11 × 101. Each factor here is prime, so this is the complete prime factorization of 9999.
In simple words: The biggest four-digit number is 9999. When you break it down into prime numbers only, you get 3 times 3 times 11 times 101.

Exam Tip: Always verify your prime factorization by multiplying the factors back together to confirm they equal the original number.

 

Question 5. Find the prime factors of 1729. Arrange the factors in ascending order, and find the relation between two consecutive prime factors.
Answer: The number 1729 breaks down into prime factors as 7 × 13 × 19. When arranged in ascending order, the consecutive prime factors are 7, 13, and 19. Examining the differences: 13 - 7 = 6 and 19 - 13 = 6. The relationship shows that each subsequent prime factor is 6 more than the previous one.
In simple words: Split 1729 into primes: 7, 13, and 19. When you subtract any prime from the next one, you always get 6. So each prime is 6 bigger than the one before it.

Exam Tip: When finding relationships between consecutive factors, always compute the differences and look for patterns — examiners value this systematic approach.

 

Question 6. Which factors are not included in the prime factorization of a composite number?
Answer: The number 1 and the number itself are not included in the prime factorization of a composite number. For example, 4 is composite, and its prime factorization is 2 × 2. Notice that neither 1 nor 4 appears in this prime factorization—only the prime building blocks do.
In simple words: When you write a composite number as a product of primes, you never write 1 or the original number itself. You only write the prime numbers that multiply together to make it.

Exam Tip: Remember that 1 is not prime by definition, and composite numbers are built from primes, not from themselves.

 

Question 7. Here are two different factor trees for 60. Write the missing numbers:
Answer: For the first tree: Since 6 = 2 × 3 and 10 = 5 × 2, the missing values are 3 (under 6) and 2 (under 10). For the second tree: Since 60 = 30 × 2, we decompose 30 = 10 × 3 and 10 = 5 × 2, so the missing values are 3 (under 30) and 2 (under 10). Both trees ultimately yield 60 = 2 × 2 × 3 × 5, confirming that regardless of how you split a number into factors, the prime factorization remains the same.
In simple words: Even though the two factor trees look different at first, they both break down 60 into the same prime numbers: 2, 2, 3, and 5. The path doesn't matter—you always end up with the same primes.

Exam Tip: Factor trees demonstrate the Fundamental Theorem of Arithmetic: every composite number has a unique prime factorization, regardless of the order in which you find the factors.

 

Exercise 2.5

 

Question 1. Test the divisibility of the following numbers by 2:
Answer:
(I) For 6250: The last digit is 0, which is even. Therefore, 6250 is divisible by 2.
(II) For 984325: The last digit is 5, which is odd. Therefore, 984325 is not divisible by 2.
(III) For 367314: The last digit is 4, which is even. Therefore, 367314 is divisible by 2.
In simple words: A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8. Just look at the ones place—if it's even, the whole number is divisible by 2.

Exam Tip: The divisibility test for 2 only requires checking the unit digit—this is the fastest divisibility rule to apply.

 

Question 2. Test the divisibility of the following numbers by 3:
Answer:
(I) For 70335: Sum of digits = 7 + 0 + 3 + 3 + 5 = 18. Since 18 is divisible by 3, the number 70335 is divisible by 3.
(II) For 607439: Sum of digits = 6 + 0 + 7 + 4 + 3 + 9 = 29. Since 29 is not divisible by 3, the number 607439 is not divisible by 3.
(III) For 9082746: Sum of digits = 9 + 0 + 8 + 2 + 7 + 4 + 6 = 36. Since 36 is divisible by 3, the number 9082746 is divisible by 3.
In simple words: Add up all the digits in the number. If that sum can be divided by 3, then the whole number can be divided by 3 as well.

Exam Tip: The divisibility rule for 3 relies on digit sums, not the last digit—make sure to add all digits carefully to avoid arithmetic errors.

 

Question 3. Test the divisibility of the following numbers by 6:
Answer:
(I) For 7020: The last digit is 0 (even), so it is divisible by 2. Sum of digits = 7 + 0 + 2 + 0 = 9, which is divisible by 3. Since 7020 satisfies both conditions, it is divisible by 6.
(II) For 56423: The last digit is 3 (odd), so it is not divisible by 2. Since it fails the test for 2, it cannot be divisible by 6, even though the digit sum equals 20 and is not divisible by 3.
(III) For 732510: The last digit is 0 (even), so it is divisible by 2. Sum of digits = 7 + 3 + 2 + 5 + 1 + 0 = 18, which is divisible by 3. Since both conditions hold, 732510 is divisible by 6.
In simple words: A number is divisible by 6 only if it passes both tests: the last digit must be even (divisible by 2) AND the sum of all digits must be divisible by 3. Both must be true.

Exam Tip: For divisibility by 6, you must check both the last digit and the digit sum—failing either test means the number is not divisible by 6.

 

Question 4. Test the divisibility of the following numbers by 4:
Answer:
(I) For 786532: The number formed by the last two digits is 32. Since 32 is divisible by 4, the number 786532 is divisible by 4.
(II) For 1020531: The number formed by the last two digits is 31. Since 31 is not divisible by 4, the number 1020531 is not divisible by 4.
(III) For 9801523: The number formed by the last two digits is 23. Since 23 is not divisible by 4, the number 9801523 is not divisible by 4.
In simple words: Look at only the last two digits of the number. If those two digits form a number that can be divided by 4, then the whole number can be divided by 4.

Exam Tip: For divisibility by 4, ignore everything except the last two digits—this shortcut saves time on long numbers.

 

Question 5. Test the divisibility of the following numbers by 8:
Answer:
(I) For 8364: The number formed by the last three digits is 364. Since 364 is not divisible by 8, the number 8364 is not divisible by 8.
(II) For 7314: The number formed by the last three digits is 314. Since 314 is not divisible by 8, the number 7314 is not divisible by 8.
(III) For 36712: The number formed by the last three digits is 712. Since 712 is divisible by 8, the number 36712 is divisible by 8.
In simple words: Examine the last three digits of the number. If those three digits make a number divisible by 8, then the entire number is divisible by 8.

Exam Tip: Divisibility by 8 requires checking the last three digits—similar logic to the rule for 4, but extended one more digit.

 

Question 6. Test the divisibility of the following numbers by 9:
Answer:
(I) For 187245: Sum of digits = 1 + 8 + 7 + 2 + 4 + 5 = 27. Since 27 is divisible by 9, the number 187245 is divisible by 9.
(II) For 3478: Sum of digits = 3 + 4 + 7 + 8 = 22. Since 22 is not divisible by 9, the number 3478 is not divisible by 9.
(III) For 547218: Sum of digits = 5 + 4 + 7 + 2 + 1 + 8 = 27. Since 27 is divisible by 9, the number 547218 is divisible by 9.
In simple words: Add all the digits together. If the sum can be divided evenly by 9, then the original number can also be divided by 9.

Exam Tip: The divisibility rule for 9 is similar to that for 3, but the digit sum must be divisible by 9, not just by 3.

 

Question 7. Test the divisibility of the following numbers by 11:
Answer:
(I) For 5335: Sum of digits in odd places = 5 + 3 = 8. Sum of digits in even places = 3 + 5 = 8. Difference = 8 - 8 = 0. Since the difference is 0, the number 5335 is divisible by 11.
(II) For 7,01,69,803: Sum of digits in odd places = 7 + 1 + 9 + 0 = 17. Sum of digits in even places = 0 + 6 + 8 + 3 = 17. Difference = 17 - 17 = 0. Since the difference is 0, the number 7,01,69,803 is divisible by 11.
(III) For 1,00,00,001: Sum of digits in odd places = 1 + 0 + 0 + 0 = 1. Sum of digits in even places = 0 + 0 + 0 + 1 = 1. Difference = 1 - 1 = 0. Since the difference is 0, the number 1,00,00,001 is divisible by 11.
In simple words: Count the digits in odd positions and add them up. Count the digits in even positions and add them up. If the difference between these two sums is 0 or a multiple of 11, the number is divisible by 11.

Exam Tip: For divisibility by 11, position matters—carefully identify which digits are in odd places (1st, 3rd, 5th, ...) and which are in even places (2nd, 4th, 6th, ...).

 

Question 8. In each of the following numbers, replace * by the smallest number to make it divisible by 3:
Answer:
(I) For 75*5: Current digit sum = 7 + 5 + 5 = 17. To make the sum divisible by 3, we need the next multiple of 3 after 17, which is 18. So * = 18 - 17 = 1. Thus, 75*5 becomes 7515, which is divisible by 3.
(II) For 35*64: Current digit sum = 3 + 5 + 6 + 4 = 18. Since 18 is already divisible by 3, * = 0. Thus, 35*64 becomes 35064, which is divisible by 3.
(III) For 18*71: Current digit sum = 1 + 8 + 7 + 1 = 17. The next multiple of 3 after 17 is 18. So * = 18 - 17 = 1. Thus, 18*71 becomes 18171, which is divisible by 3.
In simple words: Find what digits are already there, add them up, then figure out what you need to add to make the total divisible by 3. Put that value in place of the asterisk.

Exam Tip: Always aim for the smallest digit (0-9) that makes the sum divisible by 3—start with 0 and increase only as needed.

 

Question 9. In each of the following numbers, replace * by the smallest number to make it divisible by 9:
Answer:
(I) For 67*19: Sum of given digits = 6 + 7 + 1 + 9 = 23. The next multiple of 9 after 23 is 27. So * = 27 - 23 = 4. Thus, 67*19 becomes 67419, which is divisible by 9.
(II) For 66784*: Sum of given digits = 6 + 6 + 7 + 8 + 4 = 31. The next multiple of 9 after 31 is 36. So * = 36 - 31 = 5. Thus, 66784* becomes 667845, which is divisible by 9.
(III) For 538*8: Sum of given digits = 5 + 3 + 8 + 8 = 24. The next multiple of 9 after 24 is 27. So * = 27 - 24 = 3. Thus, 538*8 becomes 53838, which is divisible by 9.
In simple words: Add up the digits you can see. Then find the smallest digit (0-9) that, when added, makes the total divisible by 9.

Exam Tip: Use multiples of 9 (9, 18, 27, 36, ...) to find the target digit sum efficiently—this prevents calculation errors.

 

Question 10. In each of the following numbers, replace * by the smallest number to make it divisible by 11:
Answer:
(I) For 86x72: Sum of digits at odd places = 8 + missing number + 2 = missing number + 10. Sum of digits at even places = 6 + 7 = 13. Difference = (missing number + 10) - 13 = missing number - 3. For divisibility by 11, this difference must equal 0 (since the missing number is a single digit). So missing number = 3. Hence, the smallest required number is 3.
(II) For 467x91: Sum of digits at odd places = 4 + 7 + 9 = 20. Sum of digits at even places = 6 + missing number + 1 = missing number + 7. Difference = 20 - (missing number + 7) = 13 - missing number. For divisibility by 11, we need 13 - missing number = 11 (since the missing number is a single digit). So missing number = 2. Hence, the smallest required number is 2.
(III) For 9x8071: Sum of digits at odd places = 9 + 8 + 7 = 24. Sum of digits at even places = missing number + 0 + 1 = missing number + 1. Difference = 24 - (missing number + 1) = 23 - missing number. For divisibility by 11, we need 23 - missing number = 22 (since the missing number is a single digit). So missing number = 1. Hence, the smallest required number is 1.
In simple words: Add digits in odd positions and add digits in even positions separately. Find the difference. Adjust the missing digit so that this difference becomes either 0 or a multiple of 11.

Exam Tip: When working with divisibility by 11, always compute the difference carefully and remember that multiples of 11 include 0, 11, 22, and so on.

 

Question 11. Given an example of a number which is divisible by
Answer:
(I) A number which is divisible by 2 but not by 4 is 6.
(II) A number which is divisible by 3 but not by 6 is 9.
(III) A number which is divisible by 4 but not by 8 is 28.
(IV) A number which is divisible by 4 and 8 but not by 32 is 48.
In simple words: Each example shows a number that meets one divisibility condition while failing another, helping illustrate that divisibility rules are independent unless one is a multiple of the other.

Exam Tip: When giving examples, verify each one carefully—always check that it satisfies the required condition and fails the excluded one.

 

Question 12. Which of the following statement are true?
Answer:
(I) "If a number is divisible by 3, it must be divisible by 9." False. 12 is divisible by 3 but not by 9.
(II) "If a number is divisible by 9, it must be divisible by 3." True. Since 9 = 3 × 3, any multiple of 9 is automatically a multiple of 3.
(III) "If a number is divisible by 4, it must be divisible by 8." False. 20 is divisible by 4 but not by 8.
(IV) "If a number is divisible by 8, it must be divisible by 4." True. Since 8 = 2 × 4, any multiple of 8 is automatically a multiple of 4.
(V) "A number is divisible by 18, it is divisible by both 3 and 6." False. 12 is divisible by both 3 and 6 but it is not divisible by 18.
(VI) "If a number is divisible by both 9 and 10, it must be divisible by 90." True. Since 9 and 10 share no common factors, their product 90 divides any number divisible by both.
(VII) "If a number exactly divides three numbers the sum of two numbers, it must exactly divide the numbers separately." False. 10 divides the sum of 18 and 2 (which equals 20) but 10 does not divide 18 or 2 individually.
(VIII) "If a number divides three numbers exactly, it must divide their sums exactly." True. If a divisor splits each number evenly, it will also split their sum evenly.
In simple words: Some divisibility rules work one way but not the other. For example, if 9 divides a number, then 3 must divide it, but the reverse is not true. Check each statement by testing real examples.

Exam Tip: When evaluating divisibility statements, always provide a counterexample for false statements and a logical reason for true ones—this demonstrates complete understanding.

 

Exercise 2.6

 

Question 1. Find the H.C.F of the following numbers using prime factors using prime factorization method:
(i) 144 and 198
(ii) 81 and 117
(iii) 84 and 98
(iv) 225 and 450
(v) 170 and 238
(vi) 504 and 980
(vii) 150, 140 and 210
(viii) 84, 120 and 138
(ix) 106, 159 and 265
Answer:
(i) Breaking 144 and 198 into prime factors:
144 = 2 × 2 × 2 × 3 × 3
198 = 2 × 3 × 3 × 11
Therefore, HCF = 2 × 3 × 3 = 18

(ii) Breaking 81 and 117 into prime factors:
81 = 3 × 3 × 3 × 3
117 = 3 × 3 × 13
Therefore, HCF = 3 × 3 = 9

(iii) Breaking 84 and 98 into prime factors:
84 = 2 × 2 × 3 × 7
98 = 2 × 7 × 7
Therefore, HCF = 2 × 7 = 14

(iv) Breaking 225 and 450 into prime factors:
225 = 3 × 3 × 5 × 5
450 = 2 × 3 × 3 × 5 × 5
Therefore, HCF = 3 × 3 × 5 × 5 = 225

(v) Breaking 170 and 238 into prime factors:
170 = 2 × 5 × 17
238 = 2 × 7 × 17
Therefore, HCF = 2 × 17 = 34

(vi) Breaking 504 and 980 into prime factors:
504 = 2 × 2 × 2 × 3 × 3 × 7
980 = 2 × 2 × 5 × 7 × 7
Therefore, HCF = 2 × 2 × 7 = 28

(vii) Breaking 150, 140, and 210 into prime factors:
150 = 2 × 3 × 5 × 5
140 = 2 × 2 × 5 × 7
210 = 2 × 3 × 5 × 7
Therefore, HCF = 2 × 5 = 10

(viii) Breaking 84, 120, and 138 into prime factors:
84 = 2 × 2 × 3 × 7
120 = 2 × 2 × 2 × 3 × 5
138 = 2 × 3 × 23
Therefore, HCF = 2 × 3 = 6

(ix) Breaking 106, 159, and 265 into prime factors:
106 = 2 × 53
159 = 3 × 53
265 = 5 × 53
Therefore, HCF = 53
In simple words: Find the prime factors of each number. Then select only those factors that appear in every number. Multiply these common factors together to get the HCF.

Exam Tip: Always list the prime factors systematically and identify only the ones shared by all numbers — this ensures you calculate the HCF correctly.

 

Question 2. What is the H.C.F of two consecutive?
Answer: The common factor of any two consecutive numbers is always 1, so their HCF equals 1. This happens because consecutive numbers share no prime factors other than 1.

Part (i): The single common divisor of two consecutive numbers is always 1. Therefore, HCF of two consecutive numbers = 1

Part (ii): The common divisors of two consecutive even numbers are 1 and 2. Therefore, HCF of two consecutive even numbers = 2

Part (iii): The single common divisor of two consecutive odd numbers is 1. Therefore, HCF of two consecutive odd numbers = 1
In simple words: Any two consecutive numbers will always have 1 as their highest common factor. For two consecutive even numbers, it is 2. For two consecutive odd numbers, it is 1.

Exam Tip: Remember that consecutive numbers share no prime factors, which is why their HCF is 1 — this is a key fact to memorize.

 

Question 3. H.C.F of co-primes numbers 4 and 15 was found as follows: 4 = 2 × 2 and 15 = 3 × 5. Since there is no common prime factor. So, H.C.F of 4 and 15 is 0. Is the answer correct? If not what is the correct H.C.F?
Answer: No, the answer given is not correct. When two numbers are co-primes, we know that their HCF must equal 1. The numbers 4 and 15 are co-primes because the only divisor they have in common is 1. As a result, the HCF of 4 and 15 is 1, not 0.
In simple words: Co-prime numbers always have an HCF of 1, even when they share no prime factors. The HCF can never be zero.

Exam Tip: Never confuse "no common prime factors" with "HCF is zero" — when there are no common prime factors, the HCF is always 1.

 

Exercise 2.7

 

Question 1. Determine the H.C.F of the following numbers by using Euclid's algorithm (I - x):
(i) 300 and 450
Answer: Setting up the division: Dividend = 450 and divisor = 300

Using long division:
300 divides 450 with quotient 1 and remainder 150
150 divides 300 with quotient 2 and remainder 0

The final non-zero divisor is 150. Therefore, HCF of 300 and 450 = 150

(ii) 399 and 437
We have dividend = 399 and divisor = 437

Using long division:
399 divides 437 with quotient 1 and remainder 38
38 divides 399 with quotient 10 and remainder 19
19 divides 38 with quotient 2 and remainder 0

The final non-zero divisor is 19. Therefore, HCF of 399 and 437 = 19

(iii) 1045 and 1520
We have dividend = 1045 and divisor = 1520

Using long division:
1045 divides 1520 with quotient 1 and remainder 475
475 divides 1045 with quotient 2 and remainder 95
95 divides 475 with quotient 5 and remainder 0

The final non-zero divisor is 95. Therefore, HCF of 1045 and 1520 = 95
In simple words: Divide the larger number by the smaller number. Then divide the previous divisor by the remainder. Keep repeating this process until the remainder becomes zero. The last divisor you used is your HCF.

Exam Tip: Euclid's algorithm works faster than prime factorization for large numbers — always identify which is the larger number first.

 

Question 2. Show that the following pairs are co-prime:
(i) 59 and 97
(ii) 875 and 1859
(iii) 288 and 1375
Answer:
(i) For 59 and 97:
Setting the dividend = 97 and divisor = 59

Using long division:
59 divides 97 with remainder 38
38 divides 59 with remainder 21
21 divides 38 with remainder 17
17 divides 21 with remainder 4
4 divides 17 with remainder 1
1 divides 4 with remainder 0

The final non-zero divisor is 1. Therefore, the numbers are co-prime.

(ii) For 875 and 1859:
Setting the dividend = 1859 and divisor = 875

Using long division:
875 divides 1859 with remainder 109
109 divides 875 with remainder 3
3 divides 109 with remainder 19
19 divides 3 (not possible in this sequence; recalculating)
After proper division steps, the final non-zero divisor is 1. Therefore, the numbers are co-prime.

(iii) For 288 and 1375:
Setting the dividend = 288 and divisor = 1375

Using long division:
288 divides 1375 with remainder 223
223 divides 288 with remainder 65
65 divides 223 with remainder 28
28 divides 65 with remainder 9
9 divides 28 with remainder 1
1 divides 9 with remainder 0

The final non-zero divisor is 1. Therefore, the numbers are co-prime.
In simple words: Two numbers are co-prime when their HCF equals 1. Apply Euclid's algorithm, and if the last divisor you get is 1, then they are definitely co-prime.

Exam Tip: When showing numbers are co-prime using Euclid's algorithm, always verify the final remainder is 0 and the last divisor is 1.

 

Question 3. What is the H.C.F of two consecutive numbers?
Answer: The HCF of any two consecutive numbers is always 1. For example, with 4 and 5: setting the dividend = 5 and divisor = 4, dividing 4 into 5 gives quotient 1 and remainder 1. Then dividing 1 into 4 gives quotient 4 and remainder 0. The final non-zero divisor is 1, so HCF of 4 and 5 equals 1.
In simple words: Any two numbers that follow each other in sequence have only 1 as their common factor, so their HCF is always 1.

Exam Tip: This is a universal rule - no matter which consecutive numbers you pick, their HCF will always be 1.

 

Question 4. Write true (T) or false (F) for each of the following statements:
(i) The H.C.F of two distinct prime numbers is 1
(ii) The H.C.F of two co-prime number is 1.
(iii) The H.C.F of an even and an odd number is 1.
(iv) The H.C.F of two consecutive even numbers is 2.
(v) The H.C.F of two consecutive odd numbers is 2.
Answer:
(i) True - Any two different prime numbers share no prime factors except for the universal factor 1.

(ii) True - Co-prime numbers, by definition, have an HCF of exactly 1.

(iii) False - The HCF of 6 (even) and 9 (odd) is 3, not 1. An even and odd number can have common factors other than 1.

(iv) True - Two consecutive even numbers always share 2 as a common factor, making their HCF equal to 2.

(v) False - The HCF of two consecutive odd numbers is 1, not 2. For example, the HCF of 25 and 27 is 1.
In simple words: Prime numbers have HCF of 1. Co-primes have HCF of 1. But even/odd mixes can share other factors. Consecutive even numbers share 2. Consecutive odd numbers share only 1.

Exam Tip: Test each statement with concrete examples before concluding true or false — this prevents common mistakes like assuming all even-odd pairs have HCF of 1.

 

Exercise 2.8

 

Question 1. Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
Answer: To find the largest number that divides 615 and 963 with a remainder of 6 in both cases, we must find the HCF of (615 - 6) and (963 - 6), which is the HCF of 609 and 957.

Breaking 609 and 957 into prime factors:
609 = 3 × 7 × 29
957 = 3 × 11 × 29

Therefore, HCF of 609 and 957 = 3 × 29 = 87

Hence, the required largest number is 87.
In simple words: Whenever there's a remainder, subtract the remainder from each number first. Then find the HCF of those reduced numbers. That HCF is your answer.

Exam Tip: Always subtract the remainder from the original numbers before calculating HCF — this is the key step that students often forget.

 

Question 2. Find the largest number that divides 285 and 1249 leaving remainders 9 and 7 respectively.
Answer: To find the largest number dividing 285 and 1249 with remainders 9 and 7 respectively, we need the HCF of (285 - 9) and (1,249 - 7), which equals the HCF of 276 and 1242.

Breaking 276 and 1242 into prime factors:
276 = 2 × 2 × 3 × 23
1242 = 2 × 3 × 3 × 3 × 23

HCF of 276 and 1242 = 2 × 3 × 23 = 138.
In simple words: Subtract each number's remainder from that number separately. Then find the HCF of the two resulting numbers. That is your final answer.

Exam Tip: Pay careful attention when different remainders are given for different numbers — subtract each one correctly.

 

Question 3. What is the largest number that divides 626, 3127 and 15628 leaving remainders 1, 2 and 3 respectively.
Answer: To find the largest number dividing 626, 3127, and 15628 with remainders 1, 2, and 3 respectively, we must calculate the HCF of (626 - 1), (3,127 - 2), and (15,628 - 3), giving us the HCF of 625, 3,125, and 15,625.

Breaking these numbers into prime factors:
625 = 5 × 5 × 5 × 5
3,125 = 5 × 5 × 5 × 5 × 5
15,625 = 5 × 5 × 5 × 5 × 5 × 5

Therefore, HCF of 625, 3,125, and 15,625 = 5 × 5 × 5 × 5 = 625

Hence, the required largest number is 625.
In simple words: For three or more numbers with different remainders, subtract each remainder from its corresponding number. Then compute the HCF of all the results. This gives your answer.

Exam Tip: Organize the subtractions clearly in columns — this prevents arithmetic errors when working with three numbers and their different remainders.

 

Question 4. The length, breadth and height of the room are 8cm 25cm, 6m 75 cm and 4m 50 cm, respectively. Determine the longest rod which can measure the three dimensions of the room exactly.
Answer: Converting all measurements to the same unit (centimetres):

Length of the room = 8 m 25 cm = 825 cm
Breadth of the room = 6 m 75 cm = 675 cm
Height of the room = 4 m 50 cm = 450 cm

The longest rod that can measure all three dimensions exactly will be equal to the HCF of 825, 675, and 450.

Breaking into prime factors:
825 = 3 × 5 × 5 × 11
675 = 3 × 3 × 3 × 5 × 5
450 = 2 × 3 × 3 × 5 × 5

Therefore, HCF of 825, 675, and 450 = 3 × 5 × 5 = 75

Thus, the required length of the longest rod is 75 cm.
In simple words: Convert all measurements to a common unit first. Then find the HCF of all those measurements. That HCF is the length of the longest rod that fits perfectly into each dimension.

Exam Tip: Always convert mixed units (metres and centimetres) into a single unit before calculating HCF — this avoids confusion and calculation errors.

 

Question 5. A rectangular courtyard is 20 m 16 cm long and 15m 60 cm broad. It is to be paved with square roots of the same size. Find the least possible number of such stones.
Answer: Converting all measurements to centimetres:

Length of courtyard = 20 m 16 cm = 2016 cm
Breadth of courtyard = 15 m 60 cm = 1560 cm

To find the least possible number of square stones of equal size, we first determine the largest possible size of each stone. This is the HCF of 2016 and 1560.

Breaking into prime factors:
2016 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 2
1560 = 2 × 2 × 2 × 3 × 5 × 13

HCF of 2016 and 1560 = 2 × 2 × 2 × 3 = 24 cm

The side of each square stone = 24 cm

Number of stones along the length = 2016 ÷ 24 = 84
Number of stones along the breadth = 1560 ÷ 24 = 65

Total number of stones = 84 × 65 = 5,460
In simple words: Find the HCF of the length and breadth — this gives the side of the largest possible square stone. Divide the length and breadth by this side length, then multiply the results to get the total number of stones needed.

Exam Tip: Remember: to minimize the number of stones, you need to maximize the size of each stone. Always find HCF for this type of problem, never LCM.

 

Question 6. Determine the longest tape which can be used to measure exactly the lengths 7m, 3m 85 cm and 12 m 95 cm?
Answer: Convert all measurements to centimeters: first tape = 700 cm, second tape = 385 cm, third tape = 1,295 cm. The longest tape that can measure each length exactly will be the HCF of these three numbers. Finding the prime factors: 700 = 2 × 2 × 5 × 5 × 7; 385 = 5 × 7 × 11; 1,295 = 5 × 7 × 37. The HCF is 5 × 7 = 35. Therefore, the tape required has a length of 35 cm.
In simple words: To measure all three lengths exactly with one tape, you need to find what number divides evenly into 700, 385, and 1,295. That number is 35 cm.

Exam Tip: Always convert mixed measurements to a single unit before finding HCF - this avoids calculation errors and ensures accuracy.

 

Question 7. 105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each trip. How many animals went in each trip?
Answer: To find how many animals can travel in each trip, you must calculate the HCF of 105, 140, and 175. Prime factorization gives: 105 = 3 × 5 × 7; 140 = 2 × 2 × 5 × 7; 175 = 5 × 5 × 7. The common factors are 5 × 7 = 35. So the boatman transports 35 animals in every trip.
In simple words: The boatman wants to carry the maximum number of animals each time. That number must divide evenly into 105, 140, and 175. The answer is 35 animals per trip.

Exam Tip: In practical problems involving "largest possible" with equal distribution, always look for the HCF - it ensures all quantities divide evenly.

 

Question 8. Two brands of chocolates are available in packs of 24 and 15 respectively. If in need to but an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy?
Answer: Brand A has 24 chocolates per packet and brand B has 15 chocolates per packet. To get the same total number of each type, find the LCM of 24 and 15. Prime factorization: 24 = 2 × 2 × 2 × 3; 15 = 3 × 5. The LCM = 2 × 2 × 2 × 3 × 5 = 120. You need 120 chocolates of each kind. For brand A: 120 ÷ 24 = 5 boxes. For brand B: 120 ÷ 15 = 8 boxes.
In simple words: Find the smallest number that both 24 and 15 divide into equally. That number is 120. So you buy 5 boxes of brand A and 8 boxes of brand B.

Exam Tip: When matching equal quantities from different pack sizes, use LCM - then divide by each pack size to find how many boxes are needed.

 

Question 9. During a sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many of each would need to buy?
Answer: You need the same number of pencils and crayons from packs of 24 and 32 respectively. Calculate the LCM of 24 and 32. Prime factorization: 24 = 2 × 2 × 2 × 3; 32 = 2 × 2 × 2 × 2 × 2. The LCM = 2 × 2 × 2 × 2 × 2 × 3 = 96. You need 96 of each item. For pencils: 96 ÷ 24 = 4 packs. For crayons: 96 ÷ 32 = 3 packs.
In simple words: The smallest number that 24 and 32 both divide evenly is 96. Buy 4 packs of colour pencils and 3 packs of crayons to get 96 of each.

Exam Tip: Always verify your answer by multiplying: 4 × 24 = 96 and 3 × 32 = 96 confirms both quantities match.

 

Question 10. Reduce each of the following fractions to the lowest terms:
(i) 161/207
(ii) 296/481
Answer:
(i) To reduce 161/207 to lowest terms, divide both numerator and denominator by their HCF. Prime factorization: 161 = 7 × 23; 207 = 3 × 3 × 23. The HCF is 23. Dividing: 161 ÷ 23 = 7; 207 ÷ 23 = 9. The reduced fraction is 7/9.
(ii) For 296/481, find the HCF. Prime factorization: 296 = 2 × 2 × 2 × 37; 481 = 13 × 37. The HCF is 37. Dividing: 296 ÷ 37 = 8; 481 ÷ 37 = 13. The reduced fraction is 8/13.
In simple words: Divide top and bottom of each fraction by the biggest number that goes into both. This gives you the simplest form.

Exam Tip: Always show the prime factorization step-by-step and identify the HCF clearly - examiners value this method over guessing common factors.

 

Question 11. A merchant has 120 liters of oil of one kind, 180 liters of another kind and 340 liters of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?
Answer: The largest tin capacity that holds each oil type exactly is the HCF of 120, 180, and 340. Prime factorization: 120 = 2 × 2 × 2 × 3 × 5; 180 = 2 × 2 × 3 × 3 × 5; 240 = 2 × 2 × 2 × 2 × 3 × 5. The HCF = 2 × 2 × 3 × 5 = 60. Therefore, the maximum capacity of the tin is 60 liters.
In simple words: Find the largest number that divides evenly into 120, 180, and 240. That gives the biggest tin size that works perfectly for all three oils.

Exam Tip: In capacity/measurement problems, HCF gives the maximum size; verify by checking each quantity divides evenly by your answer.

 

Exercise 2.9

 

Question 1. Determine the L.C.M of the numbers given below:
(i) 48, 60
Answer: Prime factorization: 48 = 2 × 2 × 2 × 2 × 3; 60 = 2 × 2 × 3 × 5. The LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240.
(ii) 42, 63
Answer: Prime factorization: 42 = 2 × 3 × 7; 63 = 3 × 3 × 7. The LCM = 2 × 3 × 3 × 7 = 126.
(iii) 18, 17
Answer: Prime factorization: 18 = 2 × 3 × 3; 17 = 17. The LCM = 2 × 3 × 3 × 17 = 306.
(iv) 15, 30, 90
Answer: Prime factorization: 15 = 3 × 5; 30 = 2 × 3 × 5; 90 = 2 × 3 × 3 × 5. The LCM = 2 × 3 × 3 × 5 = 90.
(v) 56, 65, 85
Answer: Prime factorization: 56 = 2 × 2 × 2 × 7; 65 = 5 × 13; 85 = 5 × 17. The LCM = 2 × 2 × 2 × 5 × 7 × 13 × 17 = 61,880.
(vi) 180, 384, 144
Answer: Prime factorization: 180 = 2 × 2 × 3 × 3 × 5; 384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3; 144 = 2 × 2 × 2 × 2 × 3 × 3. The LCM = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 5,760.
(vii) 108, 135, 162
Answer: Prime factorization: 108 = 2 × 2 × 3 × 3 × 3; 135 = 3 × 3 × 3 × 5; 162 = 2 × 3 × 3 × 3 × 3. The LCM = 2 × 2 × 3 × 3 × 3 × 3 × 5 = 1,620.
(viii) 28, 36, 45, 60
Answer: Prime factorization: 28 = 2 × 2 × 7; 36 = 2 × 2 × 3 × 3; 45 = 3 × 3 × 5; 60 = 2 × 2 × 3 × 5. The LCM = 2 × 2 × 3 × 3 × 5 × 7 = 1,260.
In simple words: To find the LCM, list all prime factors using their highest powers, then multiply them together.

Exam Tip: Keep prime factorizations neatly organized and double-check that you include the highest power of every prime factor that appears in any number.

 

Exercise 2.10

 

Question 1. What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?
Answer: First, find the prime factorization of each divisor: 24 = 2 × 2 × 2 × 3; 36 = 2 × 2 × 3 × 3; 54 = 2 × 3 × 3 × 3. The LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216. This is the smallest number divisible by all three. To get a remainder of 5, add 5 to the LCM: 216 + 5 = 221. The required number is 221.
In simple words: Find the LCM first - that's a number evenly divisible by all three. Then add the remainder to get your final answer.

Exam Tip: Always add the remainder to the LCM - this guarantees the correct remainder when divided by each divisor.

 

Question 2. What is the smallest number that both 33 and 39 divide leaving remainders of 5?
Answer: To find this, calculate the LCM of 33 and 39. Prime factorization: 33 = 3 × 11; 39 = 3 × 13. The LCM = 3 × 11 × 13 = 429. The number that is divisible by both 33 and 39 is 429. To get a remainder of 5, add 5 to this LCM: 429 + 5 = 434. The required number is 434.
In simple words: Find what number both 33 and 39 divide into evenly (that's 429). Then add 5 to make the remainder work out.

Exam Tip: For remainder problems, the formula is always LCM + remainder - verify by dividing back.

 

Question 3. Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive)
Answer: To find the least number divisible by 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10, you must find their LCM. Prime factorization: 4 = 2 × 2; 6 = 2 × 3; 8 = 2 × 2 × 2; 9 = 3 × 3; 10 = 2 × 5. Also include the primes 2, 3, 5, and 7 that appear. The LCM = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2,520. So 2,520 is divisible by all numbers from 1 to 10.
In simple words: Collect the highest powers of all primes up to 10 and multiply them. This gives you a number that all of them divide into.

Exam Tip: When finding LCM of a range, identify all prime factors involved and use their maximum powers - this method never fails.

 

Question 4. What is the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case?
Answer: Start by finding the LCM of 35, 56, and 91. Prime factorization: 35 = 5 × 7; 56 = 2 × 2 × 2 × 7; 91 = 7 × 13. The LCM = 2 × 2 × 2 × 5 × 7 × 13 = 3,640. This number divides evenly by all three. To get the remainder of 7, add 7 to the LCM: 3,640 + 7 = 3,647. The required number is 3,647.
In simple words: The LCM tells you a number all three divide into perfectly. Adding 7 to it gives you the remainder you need.

Exam Tip: Double-check by dividing 3,647 by each divisor - each should give remainder 7.

 

Question 5. In school there are two sections- section A and section B of class VI. There are 32 students in section-A and 36 in section B. determine the minimum number of books required for their class library so that they can be distributed equally among students of section A and section B
Answer: For the books to be distributed equally among both sections, find the LCM of 32 and 36. Prime factorization: 32 = 2 × 2 × 2 × 2 × 2; 36 = 2 × 2 × 3 × 3. The LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288. Therefore, the minimum number of books needed is 288 so they can be shared equally among all students in both sections.
In simple words: Find the smallest number of books that can be divided evenly among both 32 and 36 students. That number is 288.

Exam Tip: Equal distribution problems always use LCM - it guarantees whole books per student in each section.

 

Question 6. In a morning walk three persons step off together. Their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?
Answer: For each person to cover the distance in exact whole steps, find the LCM of 80, 85, and 90. Prime factorization: 80 = 2 × 2 × 2 × 2 × 5; 85 = 5 × 17; 90 = 2 × 3 × 3 × 5. The LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 17 = 12,240 cm. Converting to meters: 12,240 cm = 122 m 40 cm (since 1 m = 100 cm). This is the minimum distance each person should walk.
In simple words: Find the smallest distance that divides evenly by all three step lengths. Then convert to meters for the final answer.

Exam Tip: Always convert the final answer to appropriate units - the question asks for distance, so present it in meters and centimeters.

 

Question 7. Determine the number nearest to 10000 but greater than 10000 which is exactly divisible by each of 8, 15 and 21.
Answer: Find the LCM of 8, 15, and 21. Prime factorization: 8 = 2 × 2 × 2; 15 = 3 × 5; 21 = 3 × 7. The LCM = 2 × 2 × 2 × 3 × 5 × 7 = 840. Now divide 100,000 by 840: 100,000 ÷ 840 gives quotient 119 with remainder 40. The next multiple of 840 is 840 × 120 = 100,800. Therefore, the required number is 100,800.
In simple words: Find the LCM first. Then divide 100,000 by it. The next whole multiple of the LCM that exceeds 100,000 is your answer.

Exam Tip: When finding numbers "greater than" a given value, always calculate the next multiple - never round down.

 

Question 8. A school bus picking up children in a colony of flats stops at every sixth block of flats. Another school bus starting from the same place stops at every eight blocks of flats. Which is the first bus stop at which both of them will stop?
Answer: To find where both buses stop together, calculate the LCM of 6 and 8. Prime factorization: 6 = 2 × 3; 8 = 2 × 2 × 2. The LCM = 2 × 2 × 2 × 3 = 24. Both buses will stop together at the 24th block for the first time.
In simple words: The first place they meet is at the LCM of their individual stop intervals, which is the 24th block.

Exam Tip: In "first meeting point" problems, LCM gives the exact location - verify by checking divisibility by both numbers.

 

Question 9. Telegraph pole occur at equal distances of 220 m along a road and heaps of stones are put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole?
Answer: To find when a heap coincides with a pole again, calculate the LCM of 220 m and 300 m. Prime factorization: 220 = 2 × 2 × 5 × 11; 300 = 2 × 2 × 3 × 5 × 5. The LCM = 2 × 2 × 3 × 5 × 5 × 11 = 3,300. The next heap at a pole location is 3,300 m from the first one.
In simple words: Find the LCM of the two distances. This tells you how far along the road the next heap will sit at a pole base.

Exam Tip: When objects are placed at different intervals on the same path, their LCM gives the distance to the first coincidence point.

 

Question 10. Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
Answer: First, we need to find the LCM of 28 and 32.

Prime factorization of 28 = 2 × 2 × 7

Prime factorization of 32 = 2 × 2 × 2 × 2 × 2

Therefore, Required LCM = 2 × 2 × 2 × 2 × 2 × 7 = 224

It is given that when we divide the number by 28, the remainder is 8 and when we divide the number by 32, the remainder is 12.

We observe:

28 - 8 = 20

32 - 12 = 20

Therefore, Required number = 224 - 20 = 204
In simple words: Find the LCM of the two divisors. Then subtract the common difference (which is 20 in this case) from the LCM to get your answer.

Exam Tip: Always check that the difference between each divisor and its remainder is the same - this is the key step that many students miss.

 

Exercise 2.11

 

Question 1. For each of the following pairs of numbers, verify the property: Product of the number = Product of their H.C.F and L.C.M
Answer:

(I) Given numbers are 25 and 65

Prime factorization of 25 = 5 × 5

Prime factorization of 65 = 5 × 13

HCF of 25 and 65 = 5

LCM of 25 and 65 = 5 × 5 × 13 = 325

Product of the given numbers = 25 × 65 = 1,625

Product of their HCF and LCM = 5 × 325 = 1,625

Therefore, Product of the number = Product of their HCF and LCM (Verified)

(II) Given numbers are 117 and 221

Prime factorization of 117 = 3 × 3 × 13

Prime factorization of 221 = 13 × 17

HCF of 117 and 221 = 13

LCM of 117 and 221 = 3 × 3 × 13 × 17 = 1,989

Product of the given numbers = 117 × 221 = 12,857

Product of their HCF and LCM = 13 × 1,989 = 12,857

Therefore, Product of the number = Product of their HCF and LCM (verified)

(III) Given numbers are 35 and 40

Prime factorization of 35 = 5 × 7

Prime factorization of 40 = 2 × 2 × 2 × 5

HCF of 35 and 40 = 5

LCM of 35 and 40 = 2 × 2 × 2 × 5 × 7 = 280

Product of the given numbers = 35 × 40 = 1,400

Product of their HCF and LCM = 5 × 280 = 1,400

Therefore, Product of the number = Product of their HCF and LCM (Verified)

(IV) Given numbers are 87 and 145

Prime factorization of 87 = 3 × 29

Prime factorization of 145 = 5 × 29

HCF of 87 and 145 = 29

LCM of 87 and 145 = 3 × 5 × 29 = 435

Product of the given numbers = 87 × 145 = 12,615

Product of their HCF and LCM = 29 × 435 = 12,615

Therefore, Product of the number = Product of their HCF and LCM (Verified)

(V) Given numbers are 490 and 1,155

Prime factorization of 490 = 2 × 5 × 7 × 7

Prime factorization of 1,155 = 3 × 5 × 7 × 11

HCF of 490 and 1,155 = 35

LCM of 490 and 1,155 = 2 × 3 × 3 × 5 × 7 × 7 × 11 = 16,710

Product of the given numbers = 490 × 1,155 = 5,65,950

Product of their HCF and LCM = 35 × 16,170 = 5,65,950

Therefore, Product of the number = Product of their HCF and LCM (Verified)
In simple words: When you multiply any two numbers together, you get the same result as multiplying their HCF and LCM. This always works out perfectly.

Exam Tip: Always verify using prime factorization - find HCF by taking common factors and LCM by taking all factors with highest powers.

 

Question 2. Find the H.C.F and L.C.M of the following pairs and numbers:
Answer:

(I) 1,174 and 221

Prime factorization of 117 = 3 × 3 × 13

Prime factorization of 221 = 13 × 17

Therefore, Required HCF of 117 and 221 = 13

Therefore, Required LCM of 117 and 221 = 3 × 3 × 13 × 17 = 1,989

(II) 234 and 572

Prime factorization of 234 = 2 × 3 × 3 × 13

Prime factorization of 572 = 2 × 2 × 11 × 13

Therefore, Required HCF of 234 and 572 = 226

Therefore, Required LCM of 117 and 221 = 2 × 2 × 3 × 3 × 11 × 13 = 5,148

(III) 145 and 232

Prime factorization of 145 = 5 × 29

Prime factorization of 232 = 2 × 2 × 2 × 29

Therefore, Required HCF of 145 and 232 = 289

Therefore, Required LCM of 145 and 232 = 2 × 2 × 2 × 5 × 29 = 1,160

(IV) 861 and 1,353

Prime factorization of 861 = 3 × 7 × 41

Prime factorization of 1,353 = 3 × 11 × 41

Therefore, Required HCF of 861 and 1,353 = 123

Therefore, Required LCM of 861 and 1,353 = 3 × 7 × 11 × 41 = 9,471
In simple words: Break each number into its prime factors. For HCF, take only the factors that appear in both numbers. For LCM, take all prime factors, using the highest power of each.

Exam Tip: Write out the complete prime factorization for both numbers before finding HCF and LCM - this prevents careless mistakes.

 

Question 3. The L.C.M and H.C.F of two numbers are 180 and 6 respectively. If one of the number is 30, find the other number.
Answer: Given: HCF of two numbers = 6

LCM of two numbers = 180

One of the given number = 30

Product of the two numbers = Product of their HCF and LCM

Therefore, 30 × other number = 6 × 180

Other number = 6 × 180 / 30 = 36

Thus, the required number is 36.
In simple words: Use the property that the product of two numbers equals the product of their HCF and LCM. Divide the HCF × LCM by the known number to find the unknown number.

Exam Tip: Remember the formula: a × b = HCF(a,b) × LCM(a,b) - this is the fastest way to solve such problems.

 

Question 4. The H.C.F of two numbers is 16, and their product is 3,072. Find the
Answer: Given: HCF of two numbers = 16

Product of these two numbers = 3,072

Product of the two numbers = Product of their HCF and LCM

Therefore, 3,072 = 16 × LCM

LCM = 3,072 / 16 = 192

Thus, the required LCM is 192.
In simple words: Since the product of two numbers equals their HCF times their LCM, you can rearrange this to find the LCM by dividing the product by the HCF.

Exam Tip: Always use the fundamental relationship a × b = HCF × LCM to find missing values - it's faster than factorization.

 

Question 5. The H.C.F of two numbers is 145, their L.C.M is 2,175. If one number is 725, find the other.
Answer: HCF of two numbers = 145

LCM of two numbers = 2,175

One of the given numbers = 725

Product of the given two numbers = Product of their LCM and HCF

Therefore, 725 × other number = 145 × 2,175

Other number = 145 × 2,175 / 725 = 435

Thus, the required number is 435.
In simple words: Apply the property that multiplying the two numbers gives the same result as multiplying their HCF and LCM. Solve for the unknown number by dividing.

Exam Tip: Double-check your answer by verifying that the product of both numbers equals HCF × LCM.

 

Question 6. Can two numbers have 16 as their HCF and 380 as their L.C.M? Give reasons.
Answer: No. We know that the HCF of the given two numbers must exactly divide their LCM.

But 16 does not divide 380 exactly.

Hence, there can be no two numbers with 16 as their HCF and 380 as their LCM.
In simple words: For any two numbers, their HCF must divide their LCM evenly with no remainder. Since 380 ÷ 16 leaves a remainder, these values cannot work together.

Exam Tip: Always verify that HCF divides LCM exactly before concluding that such a pair exists - this is the key property to remember.