Prime Numbers, Factors and Multiples

(i) A number having exactly two factors (1 and itself) is called a prime number.

(ii) A number having more than two factors is called a composite number.

(iii) 1 is neither prime nor composite.

(iv) The smallest prime number is 2.

(v) The smallest odd prime number is 3.

(vi) The smallest composite number is 4.

(vii) The smallest odd composite number is 9.

(viii) All prime numbers (except 2) are odd numbers.

(i) False.

Reason: The sum of three odd numbers is always odd. For example, 1 + 3 + 5 = 9 (odd).

(ii) True.

Reason: Sum of two odd numbers is even, and even + even = even. For example, 3 + 5 + 4 = 12 (even).

(iii) True.

Reason: The product of any two even numbers always contains 2 as a factor. For example, 4 × 6 = 24 (even).

(iv) True.

Reason: The product of three odd numbers is always odd. For example, 3 × 5 × 7 = 105 (odd).

(v) False.

Reason: If an even number is divided by 2, the quotient may be even or odd. For example, 8 ÷ 2 = 4 (even).

(vi) False.

Reason: 2 is a prime number which is even.

(vii) False.

Reason: 2 is even but it is a prime number, not composite.

(viii) False.

Reason: Prime numbers have exactly two factors, namely 1 and the number itself.

(ix) False.

Reason: 2 and 3 are two consecutive natural numbers and both are prime.

(x) True.

Reason: Two prime numbers have no common factor other than 1, so they are always co-prime.

(i) 68

68 = 1 × 68

68 = 2 × 34

68 = 4 × 17

Hence, factors of 68 are 1, 2, 4, 17, 34 and 68.

(ii) 27

27 = 1 × 27

27 = 3 × 9

Hence, factors of 27 are 1, 3, 9 and 27.

(iii) 210

210 = 1 × 210

210 = 2 × 105

210 = 3 × 70

210 = 5 × 42

210 = 6 × 35

210 = 7 × 30

210 = 10 × 21

210 = 14 × 15

Hence, factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.

(i) The first six multiples of 3 are:

3 × 1 = 3, 3 × 2 = 6, 3 × 3 = 9, 3 × 4 = 12, 3 × 5 = 15, 3 × 6 = 18

Hence, the first six multiples of 3 are 3, 6, 9, 12, 15 and 18.

(ii) The first six multiples of 5 are:

5 × 1 = 5, 5 × 2 = 10, 5 × 3 = 15, 5 × 4 = 20, 5 × 5 = 25, 5 × 6 = 30

Hence, the first six multiples of 5 are 5, 10, 15, 20, 25 and 30.

(iii) The first six multiples of 12 are:

12 × 1 = 12, 12 × 2 = 24, 12 × 3 = 36, 12 × 4 = 48, 12 × 5 = 60, 12 × 6 = 72

Hence, the first six multiples of 12 are 12, 24, 36, 48, 60 and 72.

(i) The seventh multiple of 36 = 36 × 7 = 252.

Hence, the seventh multiple of 36 is 252.

(ii) Even multiples of 9 are obtained by multiplying 9 by even numbers.

9 × 2 = 18, 9 × 4 = 36, 9 × 6 = 54, 9 × 8 = 72, 9 × 10 = 90

(9 × 12 = 108, which is greater than 100.)

Hence, the even multiples of 9 less than 100 are 18, 36, 54, 72 and 90.

(iii) Odd multiples of 17 are obtained by multiplying 17 by odd numbers.

17 × 1 = 17, 17 × 3 = 51, 17 × 5 = 85, 17 × 7 = 119

(17 × 9 = 153, which is greater than 150.)

Hence, the odd multiples of 17 less than 150 are 17, 51, 85 and 119.

(iv) The smallest 3-digit number is 100. We find the smallest multiple of 7 which is a 3-digit even number.

7 × 15 = 105 (odd)

7 × 16 = 112 (even)

Hence, the first 3-digit even multiple of 7 is 112.

(v) Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, ...

Hence, the multiples of 6 between 40 and 80 are 42, 48, 54, 60, 66, 72 and 78.

(vi) The greatest 2-digit number is 99. An even multiple of 5 must end in 0.

The greatest 2-digit number ending in 0 is 90, and 90 = 5 × 18.

Hence, the greatest 2-digit even multiple of 5 is 90.

(i) The factors of 20 are 1, 2, 4, 5, 10 and 20.

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

Hence, the common factors of 20 and 28 are 1, 2 and 4.

(ii) The factors of 35 are 1, 5, 7 and 35.

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

Hence, the common factors of 35 and 50 are 1 and 5.

(iii) The factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.

The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.

Hence, the common factors of 56 and 120 are 1, 2, 4 and 8.

(i) The factors of 4 are 1, 2 and 4.

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

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

Hence, the common factors of 4, 8 and 12 are 1, 2 and 4.

(ii) The factors of 10 are 1, 2, 5 and 10.

The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

The factors of 45 are 1, 3, 5, 9, 15 and 45.

Hence, the common factors of 10, 30 and 45 are 1 and 5.

The common multiples of 3 and 4 are the multiples of LCM(3, 4) = 12.

Multiples of 12 less than 100 are: 12, 24, 36, 48, 60, 72, 84, 96.

Hence, the natural numbers less than 100 which are common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84 and 96.

(i) The odd numbers between 36 and 53 are 37, 39, 41, 43, 45, 47, 49 and 51.

(ii) The even numbers between 232 and 251 are 234, 236, 238, 240, 242, 244, 246, 248 and 250.

(i) The odd numbers immediately after 79 are obtained by adding 2 successively.

79 + 2 = 81, 81 + 2 = 83, 83 + 2 = 85, 85 + 2 = 87.

Hence, the four consecutive odd numbers succeeding 79 are 81, 83, 85 and 87.

(ii) The even numbers immediately before 124 are obtained by subtracting 2 successively.

124 − 2 = 122, 122 − 2 = 120, 120 − 2 = 118.

Hence, the three consecutive even numbers preceding 124 are 118, 120 and 122.

The prime numbers between 1 and 15 are 2, 3, 5, 7, 11 and 13.

Hence, the greatest prime number between 1 and 15 is 13.

(i) 29

As 5 × 5 = 25 < 29 and 6 × 6 = 36 > 29, so 5 is the largest number such that 5 × 5 ≤ 29.

The prime numbers less than or equal to 5 are 2, 3 and 5.

29 is not divisible by 2, 3 or 5.

Hence, 29 is a prime number.

(ii) 57

Sum of digits of 57 = 5 + 7 = 12, which is divisible by 3.

So, 57 is divisible by 3 (57 = 3 × 19).

Hence, 57 is not a prime number.

(iii) 43

As 6 × 6 = 36 < 43 and 7 × 7 = 49 > 43, so 6 is the largest number such that 6 × 6 ≤ 43.

The prime numbers less than or equal to 6 are 2, 3 and 5.

43 is not divisible by 2, 3 or 5.

Hence, 43 is a prime number.

(iv) 61

As 7 × 7 = 49 < 61 and 8 × 8 = 64 > 61, so 7 is the largest number such that 7 × 7 ≤ 61.

The prime numbers less than or equal to 7 are 2, 3, 5 and 7.

61 is not divisible by 2, 3, 5 or 7.

Hence, 61 is a prime number.

(i) The factors of 12 are 1, 2, 3, 4, 6 and 12.

The factors of 35 are 1, 5, 7 and 35.

The only common factor is 1.

Hence, 12 and 35 are co-prime.

(ii) The factors of 15 are 1, 3, 5 and 15.

The factors of 37 are 1 and 37 (since 37 is prime).

The only common factor is 1.

Hence, 15 and 37 are co-prime.

(iii) The factors of 27 are 1, 3, 9 and 27.

The factors of 32 are 1, 2, 4, 8, 16 and 32.

The only common factor is 1.

Hence, 27 and 32 are co-prime.

(iv) The factors of 17 are 1 and 17.

The factors of 85 are 1, 5, 17 and 85.

The common factors are 1 and 17.

Hence, 17 and 85 are not co-prime.

(v) 515 and 516 are two consecutive natural numbers.

Any two consecutive natural numbers are always co-prime.

Hence, 515 and 516 are co-prime.

(vi) The factors of 215 are 1, 5, 43 and 215.

The factors of 415 are 1, 5, 83 and 415.

The common factors are 1 and 5.

Hence, 215 and 415 are not co-prime.

(i) 24 = 5 + 19

(Other ways: 24 = 7 + 17, 24 = 11 + 13)

(ii) 36 = 7 + 29

(Other ways: 36 = 13 + 23, 36 = 17 + 19, 36 = 5 + 31)

(iii) 84 = 17 + 67

(Other ways: 84 = 11 + 73, 84 = 13 + 71, 84 = 5 + 79, 84 = 23 + 61, 84 = 31 + 53, 84 = 37 + 47, 84 = 41 + 43)

(iv) 98 = 19 + 79

(Other ways: 98 = 31 + 67)

Twin primes are pairs of prime numbers whose difference is 2. The twin-primes less than 100 are: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73).

(i) 24 = 11 + 13

(ii) 36 = 17 + 19

(iii) 84 = 41 + 43

(iv) 120 = 59 + 61

(i) 21 = 3 + 7 + 11

(Other ways: 21 = 5 + 5 + 11, 21 = 7 + 7 + 7, 21 = 3 + 5 + 13)

(ii) 35 = 5 + 11 + 19

(Other ways: 35 = 5 + 7 + 23, 35 = 3 + 13 + 19, 35 = 5 + 13 + 17, 35 = 7 + 11 + 17, 35 = 11 + 11 + 13)

(iii) 49 = 7 + 11 + 31

(Other ways: 49 = 5 + 7 + 37, 49 = 7 + 13 + 29, 49 = 13 + 13 + 23)

(iv) 63 = 7 + 13 + 43

(Other ways: 63 = 5 + 11 + 47, 63 = 3 + 7 + 53, 63 = 11 + 23 + 29, 63 = 17 + 23 + 23)

A number is divisible by 5 if its last digit is 0 or 5.

A number is divisible by 10 if its last digit is 0.

(i) 3725 — last digit is 5.

Hence, 3725 is divisible by 5 only.

(ii) 48970 — last digit is 0.

Hence, 48970 is divisible by both 5 and 10.

(iii) 56823 — last digit is 3.

Hence, 56823 is divisible neither by 5 nor by 10.

(iv) 760035 — last digit is 5.

Hence, 760035 is divisible by 5 only.

(v) 7893217 — last digit is 7.

Hence, 7893217 is divisible neither by 5 nor by 10.

(vi) 4500010 — last digit is 0.

Hence, 4500010 is divisible by both 5 and 10.

A number is divisible by 2 if its last digit is 0, 2, 4, 6 or 8.

A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

(i) 54014 — last digit 4, divisible by 2.

Last two digits = 14, not divisible by 4.

Hence, 54014 is divisible by 2 only.

(ii) 723840 — last digit 0, divisible by 2.

Last two digits = 40, divisible by 4.

Last three digits = 840, and 840 ÷ 8 = 105, divisible by 8.

Hence, 723840 is divisible by 2, 4 and 8.

(iii) 6531088 — last digit 8, divisible by 2.

Last two digits = 88, divisible by 4.

Last three digits = 088 = 88, and 88 ÷ 8 = 11, divisible by 8.

Hence, 6531088 is divisible by 2, 4 and 8.

(iv) 75689604 — last digit 4, divisible by 2.

Last two digits = 04 = 4, divisible by 4.

Last three digits = 604, and 604 is not divisible by 8 because 8 × 75 = 600 and remainder is 4.

Hence, 75689604 is divisible by 2 and 4 only.

(v) 786235 — last digit 5, not divisible by 2.

Hence, 786235 is not divisible by 2, 4 or 8.

(vi) 5321048 — last digit 8, divisible by 2.

Last two digits = 48, divisible by 4.

Last three digits = 048 = 48, and 48 ÷ 8 = 6, divisible by 8.

Hence, 5321048 is divisible by 2, 4 and 8.

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

A number is divisible by 9 if the sum of its digits is divisible by 9.

(i) 7341 — sum of digits = 7 + 3 + 4 + 1 = 15.

15 is divisible by 3 but not by 9.

Hence, 7341 is divisible by 3 only.

(ii) 59031 — sum of digits = 5 + 9 + 0 + 3 + 1 = 18.

18 is divisible by both 3 and 9.

Hence, 59031 is divisible by both 3 and 9.

(iii) 12345678 — sum of digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36.

36 is divisible by both 3 and 9.

Hence, 12345678 is divisible by both 3 and 9.

(iv) 560319 — sum of digits = 5 + 6 + 0 + 3 + 1 + 9 = 24.

24 is divisible by 3 but not by 9.

Hence, 560319 is divisible by 3 only.

(v) 720634 — sum of digits = 7 + 2 + 0 + 6 + 3 + 4 = 22.

22 is not divisible by 3.

Hence, 720634 is not divisible by 3 or 9.

(vi) 3721509 — sum of digits = 3 + 7 + 2 + 1 + 5 + 0 + 9 = 27.

27 is divisible by both 3 and 9.

Hence, 3721509 is divisible by both 3 and 9.

A number is divisible by 11 if the difference between the sum of digits at odd places and the sum of digits at even places is either 0 or divisible by 11.

(i) 10428

Sum of digits at odd places (from right) = 8 + 4 + 1 = 13

Sum of digits at even places (from right) = 2 + 0 = 2

Difference = 13 − 2 = 11, which is divisible by 11.

Hence, 10428 is divisible by 11.

(ii) 70169803

Sum of digits at odd places (from right) = 3 + 8 + 6 + 0 = 17

Sum of digits at even places (from right) = 0 + 9 + 1 + 7 = 17

Difference = 17 − 17 = 0.

Hence, 70169803 is divisible by 11.

(iii) 7136985

Sum of digits at odd places (from right) = 5 + 9 + 3 + 7 = 24

Sum of digits at even places (from right) = 8 + 6 + 1 = 15

Difference = 24 − 15 = 9, which is not divisible by 11.

Hence, 7136985 is not divisible by 11.

A number is divisible by 6 if it is divisible by both 2 and 3.

(i) 93573 — last digit 3, not divisible by 2.

Hence, 93573 is not divisible by 6.

(ii) 217944 — last digit 4, divisible by 2.

Sum of digits = 2 + 1 + 7 + 9 + 4 + 4 = 27, divisible by 3.

Hence, 217944 is divisible by 6.

(iii) 5034126 — last digit 6, divisible by 2.

Sum of digits = 5 + 0 + 3 + 4 + 1 + 2 + 6 = 21, divisible by 3.

Hence, 5034126 is divisible by 6.

(iv) 901352 — last digit 2, divisible by 2.

Sum of digits = 9 + 0 + 1 + 3 + 5 + 2 = 20, not divisible by 3.

Hence, 901352 is not divisible by 6.

(v) 639210 — last digit 0, divisible by 2.

Sum of digits = 6 + 3 + 9 + 2 + 1 + 0 = 21, divisible by 3.

Hence, 639210 is divisible by 6.

(vi) 1790184 — last digit 4, divisible by 2.

Sum of digits = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30, divisible by 3.

Hence, 1790184 is divisible by 6.

For a number to be divisible by 9, the sum of its digits must be divisible by 9.

(i) 4710*82

Sum of given digits (except *) = 4 + 7 + 1 + 0 + 8 + 2 = 22.

We need 22 + * to be divisible by 9.

22 + 5 = 27, which is divisible by 9.

Hence, * should be replaced by 5. The number formed is 4710582.

(ii) 70*356722

Sum of given digits (except *) = 7 + 0 + 3 + 5 + 6 + 7 + 2 + 2 = 32.

We need 32 + * to be divisible by 9.

32 + 4 = 36, which is divisible by 9.

Hence, * should be replaced by 4. The number formed is 704356722.

For a number to be divisible by 3, the sum of its digits must be divisible by 3.

(a) 4*672

Sum of given digits (except *) = 4 + 6 + 7 + 2 = 19.

We need 19 + * to be divisible by 3.

The values of * which make 19 + * divisible by 3 are 2 (gives 21), 5 (gives 24) and 8 (gives 27).

(i) The smallest digit is 2. The number formed is 42672.

(ii) The greatest digit is 8. The number formed is 48672.

(b) 4756*2

Sum of given digits (except *) = 4 + 7 + 5 + 6 + 2 = 24.

We need 24 + * to be divisible by 3.

The values of * which make 24 + * divisible by 3 are 0 (gives 24), 3 (gives 27), 6 (gives 30) and 9 (gives 33).

(i) The smallest digit is 0. The number formed is 475602.

(ii) The greatest digit is 9. The number formed is 475692.

A number is divisible by 11 if the difference between the sum of digits at odd places (from ones) and the sum of digits at even places (from tens) is either 0 or divisible by 11.

(i) 8*9484

Digits from right: 4, 8, 4, 9, *, 8.

Sum of digits at odd places (from right) = 4 + 4 + * = 8 + *.

Sum of digits at even places (from right) = 8 + 9 + 8 = 25.

Difference = 25 − (8 + *) = 17 − *.

For divisibility by 11, 17 − * = 0 (impossible) or 17 − * = 11 ⇒ * = 6.

Hence, * should be replaced by 6. The number formed is 869484.

(ii) 9*53762

Digits from right: 2, 6, 7, 3, 5, *, 9.

Sum of digits at odd places (from right) = 2 + 7 + 5 + 9 = 23.

Sum of digits at even places (from right) = 6 + 3 + * = 9 + *.

Difference = 23 − (9 + *) = 14 − *.

For divisibility by 11, 14 − * = 0 (impossible) or 14 − * = 11 ⇒ * = 3.

Hence, * should be replaced by 3. The number formed is 9353762.

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

(a) 2*4706

Last digit is 6, so the number is divisible by 2.

Sum of given digits (except *) = 2 + 4 + 7 + 0 + 6 = 19.

We need 19 + * to be divisible by 3.

The values of * which make 19 + * divisible by 3 are 2, 5 and 8.

(i) The smallest digit is 2. The number formed is 224706.

(ii) The greatest digit is 8. The number formed is 284706.

(b) 5825*34

Last digit is 4, so the number is divisible by 2.

Sum of given digits (except *) = 5 + 8 + 2 + 5 + 3 + 4 = 27.

We need 27 + * to be divisible by 3.

The values of * which make 27 + * divisible by 3 are 0, 3, 6 and 9.

(i) The smallest digit is 0. The number formed is 5825034.

(ii) The greatest digit is 9. The number formed is 5825934.

(i) 101

As 10 × 10 = 100 < 101 and 11 × 11 = 121 > 101, so 10 is the largest number such that 10 × 10 ≤ 101.

The prime numbers less than or equal to 10 are 2, 3, 5 and 7.

101 is not divisible by 2, 3, 5 or 7.

Hence, 101 is a prime number.

(ii) 251

As 15 × 15 = 225 < 251 and 16 × 16 = 256 > 251, so 15 is the largest number such that 15 × 15 ≤ 251.

The prime numbers less than or equal to 15 are 2, 3, 5, 7, 11 and 13.

251 is not divisible by 2, 3, 5, 7, 11 or 13.

Hence, 251 is a prime number.

(iii) 323

Note that 17 × 19 = 323, so 17 and 19 are factors of 323.

Hence, 323 is not a prime number.

(iv) 397

As 19 × 19 = 361 < 397 and 20 × 20 = 400 > 397, so 19 is the largest number such that 19 × 19 ≤ 397.

The prime numbers less than or equal to 19 are 2, 3, 5, 7, 11, 13, 17 and 19.

397 is not divisible by 2, 3, 5, 7, 11, 13, 17 or 19.

Hence, 397 is a prime number.

To determine if 372645 is divisible by 45, we test it for divisibility by 5 and 9 both (since 45 = 5 × 9 and 5, 9 are co-prime).

The number 372645 has digit 5 at the unit's place, so it is divisible by 5.

Sum of digits = 3 + 7 + 2 + 6 + 4 + 5 = 27, which is divisible by 9, so 372645 is divisible by 9.

Since 5 and 9 are co-prime, the number 372645 is divisible by their product, i.e., by 45.

Hence, 372645 is divisible by 45.

If a number is divisible by 12, then it is divisible by each of the factors of 12.

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

Hence, the number is also divisible by 1, 2, 3, 4 and 6.

Let a natural number, say n, be divisible by both 3 and 8.

Since 3 and 8 are co-prime numbers and if a number is divisible by two co-prime numbers, then it is divisible by their product.

So, n is divisible by 3 × 8 = 24.

We know that,

If a number is divisible by another number, then it is divisible by each of the factors of that number

Thus, n is 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 always divisible by 1, 2, 4, 6, 12 and 24.

(i) False.

Reason: For example, 12 is divisible by 4 but not by 8.

(ii) False.

Reason: For example, 6 is divisible by 3 but not by 9.

(iii) True.

Reason: Since 3 is a factor of 9, every number divisible by 9 is also divisible by 3.

(iv) True.

Reason: 9 and 10 are co-prime, so any number divisible by both is divisible by 9 × 10 = 90.

(v) False.

Reason: For example, 5 divides the sum 3 + 7 = 10, but 5 does not divide 3 or 7 separately.

(vi) True.

Reason: Since 3 and 8 are co-prime, the number is divisible by 3 × 8 = 24, and 12 is a factor of 24, so the number is divisible by 12 as well.

(vii) False.

Reason: 6 and 15 are not co-prime (HCF = 3). The LCM of 6 and 15 is 30. For example, 30 is divisible by both 6 and 15 but not by 90.

(i) In the first factor tree, 90 = 10 × 9.

10 = 2 × ?, so the missing number is 5 (since 10 = 2 × 5).

9 = ? × 3, so the missing number is 3 (since 9 = 3 × 3).

(ii) In the second factor tree, 90 = 45 × ?, so the missing number is 2 (since 90 = 45 × 2).

45 = 15 × ?, so the missing number is 3 (since 45 = 15 × 3).

15 = ? × ?, so the missing numbers are 3 and 5 (since 15 = 3 × 5).

(i) 72

2	72
2	36
2	18
3	9
3	3
	1

Hence, 72 = 2 × 2 × 2 × 3 × 3.

(ii) 172

2	172
2	86
43	43
	1

Hence, 172 = 2 × 2 × 43.

(iii) 450

2	450
3	225
3	75
5	25
5	5
	1

Hence, 450 = 2 × 3 × 3 × 5 × 5.

(iv) 980

2	980
2	490
5	245
7	49
7	7
	1

Hence, 980 = 2 × 2 × 5 × 7 × 7.

(v) 8712

2	8712
2	4356
2	2178
3	1089
3	363
11	121
11	11
	1

Hence, 8712 = 2 × 2 × 2 × 3 × 3 × 11 × 11.

(vi) 13500

2	13500
2	6750
3	3375
3	1125
3	375
5	125
5	25
5	5
	1

Hence, 13500 = 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5.

The smallest 3-digit number is 100.

2	100
2	50
5	25
5	5
	1

So, 100 = 2 × 2 × 5 × 5.

The greatest 3-digit number is 999.

3	999
3	333
3	111
37	37
	1

So, 999 = 3 × 3 × 3 × 37.

The smallest 5-digit number is 10000.

2	10000
2	5000
2	2500
2	1250
5	625
5	125
5	25
5	5
	1

Hence, 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5.

The smallest four prime numbers are 2, 3, 5 and 7.

The smallest number with four distinct prime factors is the product of the smallest four prime numbers.

Required number = 2 × 3 × 5 × 7 = 210.

Hence, the smallest number having four different prime factors is 210.

(i) 28, 36

2	28
2	14
7	7
	1

So, 28 = 2 × 2 × 7.

2	36
2	18
3	9
3	3
	1

So, 36 = 2 × 2 × 3 × 3.

The common prime factor is 2, occurring twice in both numbers.

Hence, HCF of 28 and 36 = 2 × 2 = 4.

(ii) 54, 72, 90

2	54
3	27
3	9
3	3
	1

So, 54 = 2 × 3 × 3 × 3.

2	72
2	36
2	18
3	9
3	3
	1

So, 72 = 2 × 2 × 2 × 3 × 3.

2	90
3	45
3	15
5	5
	1

So, 90 = 2 × 3 × 3 × 5.

The common prime factors are 2 (one time) and 3 (two times) in all three numbers.

Hence, HCF of 54, 72 and 90 = 2 × 3 × 3 = 18.

(iii) 105, 140, 175

3	105
5	35
7	7
	1

So, 105 = 3 × 5 × 7.

2	140
2	70
5	35
7	7
	1

So, 140 = 2 × 2 × 5 × 7.

5	175
5	35
7	7
	1

So, 175 = 5 × 5 × 7.

The common prime factors are 5 (one time) and 7 (one time) in all three numbers.

Hence, HCF of 105, 140 and 175 = 5 × 7 = 35.

(i) 198, 429

$$\begin{array}{l} 198\overline{\smash{\big)}429\smash{\big(}}2\phantom{)} \\ \phantom{x}\phantom{))}\underline{-396} \\ \phantom{{x^2 } (()))}33\overline{\smash{\big)}198\smash{\big(}}\phantom{)}6 \\ \phantom{{x} +1xa)}\underline{-198} \\ \phantom{{x^2 a} + 2x+()} 0 \\ \end{array}$$

Hence, HCF of 198 and 429 = 33.

(ii) 20, 64, 104

First, find HCF of 20 and 64:

$$\begin{array}{l} 20\overline{\smash{\big)}64\smash{\big(}}\phantom{)}3 \\ \phantom{x}\phantom{}\underline{-60} \\ \phantom{{x^2 } 1)}4\overline{\smash{\big)}20\smash{\big(}}\phantom{}5 \\ \phantom{{x} +(}\underline{-20} \\ \phantom{{x^2 a} + 2} 0 \\ \end{array}$$

So, HCF of 20 and 64 = 4.

Now, find HCF of 4 and 104:

$$\begin{array}{l} 4\overline{\smash{\big)}104\smash{\big(}}\phantom{)}26 \\ \phantom{(}\phantom{}\underline{-104} \\ \phantom{{x^2 } +)}0 \\ \end{array}$$

So, HCF of 4 and 104 = 4.

Hence, HCF of 20, 64 and 104 = 4.

(iii) 120, 144, 204

First, find HCF of 120 and 144:

$$\begin{array}{l} 120\overline{\smash{\big)}144\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{))}\underline{-120} \\ \phantom{{x^2 } ()1)}24\overline{\smash{\big)}120\smash{\big(}}\phantom{}5 \\ \phantom{{x} +1xa}\underline{-120} \\ \phantom{{x^2 a} + 2x+} 0 \\ \end{array}$$

So, HCF of 120 and 144 = 24.

Now, find HCF of 24 and 204:

$$\begin{array}{l} 24\overline{\smash{\big)}204\smash{\big(}}\phantom{}8 \\ \phantom{x}\phantom{}\underline{-192} \\ \phantom{{x^2 } 1)}12\overline{\smash{\big)}24\smash{\big(}}\phantom{}2 \\ \phantom{{x} +1(}\underline{-24} \\ \phantom{{x^2 a} + 2x} 0 \\ \end{array}$$

So, HCF of 24 and 204 = 12.

Hence, HCF of 120, 144 and 204 = 12.

(i) Two consecutive natural numbers do not share any common factor other than 1.

Hence, HCF of two consecutive natural numbers is 1.

(ii) Two consecutive odd numbers (e.g., 5 and 7, or 9 and 11) have no common factor other than 1.

Hence, HCF of two consecutive odd numbers is 1.

(iii) Two consecutive even numbers (e.g., 6 and 8, or 10 and 12) are both divisible by 2, but not by any other common factor.

Hence, HCF of two consecutive even numbers is 2.

When 257 is divided by the required number, 5 is left as a remainder. So 257 − 5 = 252 is exactly divisible by that number.

Similarly, 329 − 5 = 324 is exactly divisible by that number.

Therefore, the required number is the HCF of 252 and 324.

$$\begin{array}{l} 252\overline{\smash{\big)}324\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{))}\underline{-252} \\ \phantom{{x^2 }() 1)}72\overline{\smash{\big)}252\smash{\big(}}\phantom{}3 \\ \phantom{{x} +1xa}\underline{-216} \\ \phantom{{x^2 a} + 2x()} 36\overline{\smash{\big)}72\smash{\big(}}\phantom{)}2 \\ \phantom{{x} +1xa+()}\underline{-72} \\ \phantom{{x^2 a} + 2x++()} 0 \\ \end{array}$$

Hence, the required greatest number is 36.

When 623 is divided by the required number, 3 is left as a remainder. So 623 − 3 = 620 is exactly divisible by that number.

Similarly, 729 − 9 = 720 and 841 − 1 = 840 are exactly divisible by that number.

Therefore, the required number is the HCF of 620, 720 and 840.

First, find HCF of 620 and 720:

$$\begin{array}{l} 620\overline{\smash{\big)}720\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{))}\underline{-620} \\ \phantom{{x^2 } 1()}100\overline{\smash{\big)}620\smash{\big(}}\phantom{}6 \\ \phantom{{x} +1xa}\underline{-600} \\ \phantom{{x^2 a} + 2x()} 20\overline{\smash{\big)}100\smash{\big(}}\phantom{}5 \\ \phantom{{x} +1xa++}\underline{-100} \\ \phantom{{x^2 a} + 2x+++} 0 \\ \end{array}$$

So, HCF of 620 and 720 = 20.

Now, find HCF of 20 and 840:

$$\begin{array}{l} 20\overline{\smash{\big)}840\smash{\big(}}\phantom{}42 \\ \phantom{x}\phantom{)}\underline{-840} \\ \phantom{{x^2 } 1+)}0 \\ \end{array}$$

So, HCF of 20 and 840 = 20.

Hence, the largest required number is 20.

The required maximum weight that can measure the weight of rice in both bags an exact number of times is the HCF of 75 and 69.

$$\begin{array}{l} 69\overline{\smash{\big)}75\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{}\underline{-69} \\ \phantom{{x^2 } 1)}6\overline{\smash{\big)}69\smash{\big(}}\phantom{}11 \\ \phantom{{x} +(}\underline{-66} \\ \phantom{{x^2 a} + 2)} 3\overline{\smash{\big)}6\smash{\big(}}\phantom{}2 \\ \phantom{{x} +1xa(}\underline{-6} \\ \phantom{{x^2 a} + 2x()} 0 \\ \end{array}$$

So, HCF of 75 and 69 = 3.

Hence, the maximum weight which can measure the rice exactly is 3 kg.

The required maximum capacity of a container that can measure the diesel in all three tankers exactly is the HCF of 403, 434 and 465.

First, find HCF of 403 and 434:

$$\begin{array}{l} 403\overline{\smash{\big)}434\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{()}\underline{-403} \\ \phantom{{x^2 } 1())}31\overline{\smash{\big)}403\smash{\big(}}\phantom{}13 \\ \phantom{{x} +1xa}\underline{-403} \\ \phantom{{x^2 a} + 2x+} 0 \\ \end{array}$$

So, HCF of 403 and 434 = 31.

Now, find HCF of 31 and 465:

$$\begin{array}{l} 31\overline{\smash{\big)}465\smash{\big(}}\phantom{}15 \\ \phantom{x}\phantom{}\underline{-465} \\ \phantom{{x^2 } 1()}0 \\ \end{array}$$

So, HCF of 31 and 465 = 31.

Hence, the maximum capacity of the container is 31 litres.

(i) 28, 98

Prime factorisation of the given numbers are:

28 = 2 × 2 × 7

98 = 2 × 7 × 7

2 occurs as a prime factor maximum 2 times and 7 also 2 times.

Hence, LCM of 28 and 98 = 2 × 2 × 7 × 7 = 196.

(ii) 36, 40, 126

Prime factorisation of the given numbers are:

36 = 2 × 2 × 3 × 3

40 = 2 × 2 × 2 × 5

126 = 2 × 3 × 3 × 7

2 occurs as a prime factor maximum 3 times, 3 occurs 2 times, 5 occurs 1 time and 7 also 1 time.

Hence, LCM of 36, 40 and 126 = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520.

(iii) 108, 135, 162

Prime factorisation of the given numbers are:

108 = 2 × 2 × 3 × 3 × 3

135 = 3 × 3 × 3 × 5

162 = 2 × 3 × 3 × 3 × 3

3 occurs as a prime factor maximum 4 times, 2 occurs 2 times and 5 occurs 1 time.

Hence, LCM of 108, 135 and 162 = 2 × 2 × 3 × 3 × 3 × 3 × 5 = 1620.

(iv) 24, 28, 196

Prime factorisation of the given numbers are:

24 = 2 × 2 × 2 × 3

28 = 2 × 2 × 7

196 = 2 × 2 × 7 × 7

2 occurs as a prime factor maximum 3 times, 3 occurs 1 time and 7 also 2 times.

Hence, LCM of 24, 28 and 196 = 2 × 2 × 2 × 3 × 7 × 7 = 1176.

(i) 480, 672

2	480, 672
2	240, 336
2	120, 168
2	60, 84
2	30, 42
3	15, 21
5	5, 7
7	1, 7
	1, 1

Hence, LCM of 480 and 672 = 2 × 2 × 2 × 2 × 2 × 3 × 5 × 7 = 3360.

(ii) 6, 8, 45

2	6, 8, 45
2	3, 4, 45
2	3, 2, 45
3	3, 1, 45
3	1, 1, 15
5	1, 1, 5
	1, 1, 1

Hence, LCM of 6, 8 and 45 = 2 × 2 × 2 × 3 × 3 × 5 = 360.

(iii) 24, 40, 84

2	24, 40, 84
2	12, 20, 42
2	6, 10, 21
3	3, 5, 21
5	1, 5, 7
7	1, 1, 7
	1, 1, 1

Hence, LCM of 24, 40 and 84 = 2 × 2 × 2 × 3 × 5 × 7 = 840.

(iv) 20, 36, 63, 77

2	20, 36, 63, 77
2	10, 18, 63, 77
3	5, 9, 63, 77
3	5, 3, 21, 77
5	5, 1, 7, 77
7	1, 1, 7, 77
11	1, 1, 1, 11
	1, 1, 1, 1

Hence, LCM of 20, 36, 63 and 77 = 2 × 2 × 3 × 3 × 5 × 7 × 11 = 13860.

First, we find the LCM of 15, 35 and 48.

2	15, 35, 48
2	15, 35, 24
2	15, 35, 12
2	15, 35, 6
3	15, 35, 3
5	5, 35, 1
7	1, 7, 1
	1, 1, 1

LCM of 15, 35 and 48 = 2 × 2 × 2 × 2 × 3 × 5 × 7 = 1680.

Thus, 1680 is the least number exactly divisible by 15, 35 and 48.

According to given condition, the required number when increased by 15 gives 1680.

The required number = 1680 − 15 = 1665.

Hence, the required least number is 1665.

First, we find the LCM of 6, 15 and 18.

2	6, 15, 18
3	3, 15, 9
3	1, 5, 3
5	1, 5, 1
	1, 1, 1

LCM of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90.

Thus, 90 is the least number exactly divisible by 6, 15 and 18.

The required least number which leaves remainder 5 when divided by these numbers = 90 + 5 = 95.

Hence, the required least number is 95.

First, we find the LCM of 24, 36, 45 and 54.

2	24, 36, 45, 54
2	12, 18, 45, 27
2	6, 9, 45, 27
3	3, 9, 45, 27
3	1, 3, 15, 9
3	1, 1, 5, 3
5	1, 1, 5, 1
	1, 1, 1, 1

LCM of 24, 36, 45 and 54 = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 1080.

The required least number which leaves remainder 3 = 1080 + 3 = 1083.

Hence, the required least number is 1083.

First, we find the LCM of 8, 20 and 24.

2	8, 20, 24
2	4, 10, 12
2	2, 5, 6
3	1, 5, 3
5	1, 5, 1
	1, 1, 1

LCM of 8, 20 and 24 = 2 × 2 × 2 × 3 × 5 = 120.

The greatest 3-digit number is 999.

We divide 999 by 120 and find the remainder.

$$\begin{array}{l} \phantom{x^2 1}{\quad 8} \\ 120\overline{\smash{\big)}999} \\ \phantom{x^ + }\phantom{}\underline{-960} \\ \phantom{{x^2 } + (} 39 \\ \end{array}$$

The remainder is 39.

The required greatest 3-digit number = 999 − 39 = 960.

Hence, the greatest 3-digit number divisible by 8, 20 and 24 is 960.

First, we find the LCM of 32, 36 and 48.

2	32, 36, 48
2	16, 18, 24
2	8, 9, 12
2	4, 9, 6
2	2, 9, 3
3	1, 9, 3
3	1, 3, 1
	1, 1, 1

LCM of 32, 36 and 48 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288.

The smallest 4-digit number is 1000.

We divide 1000 by 288 and find the remainder.

$$\begin{array}{l} \phantom{x^2 1}{\quad 3} \\ 288\overline{\smash{\big)}1000} \\ \phantom{x^ + )}\phantom{)}\underline{-864} \\ \phantom{{x^2 } + 2} 136 \\ \end{array}$$

The remainder is 136.

The required smallest 4-digit number = 1000 + (288 − 136) = 1000 + 152 = 1152.

Hence, the smallest 4-digit number divisible by 32, 36 and 48 is 1152.

First, we find the LCM of 8, 12 and 20.

2	8, 12, 20
2	4, 6, 10
2	2, 3, 5
3	1, 3, 5
5	1, 1, 5
	1, 1, 1

LCM of 8, 12 and 20 = 2 × 2 × 2 × 3 × 5 = 120.

The greatest 4-digit number is 9999.

We divide 9999 by 120 and find the remainder.

$$\begin{array}{l} \phantom{x^2 }{\quad 83} \\ 120\overline{\smash{\big)}9999} \\ \phantom{x^(}\phantom{)}\underline{-960} \\ \phantom{{x^2 } + (} 399 \\ \phantom{{x} +}\underline{-360} \\ \phantom{{x^2 } + 2(} 39 \\ \end{array}$$

The remainder is 39.

The required greatest 4-digit number = 9999 − 39 = 9960.

Hence, the greatest 4-digit number divisible by 8, 12 and 20 is 9960.

First, we find the LCM of 32, 36 and 45.

2	32, 36, 45
2	16, 18, 45
2	8, 9, 45
2	4, 9, 45
2	2, 9, 45
3	1, 9, 45
3	1, 3, 15
5	1, 1, 5
	1, 1, 1

LCM of 32, 36 and 45 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440.

The smallest 5-digit number is 10000.

We divide 10000 by 1440 and find the remainder.

$$\begin{array}{l} \phantom{x^2 1+}{\quad 6} \\ 1440\overline{\smash{\big)}10000} \\ \phantom{x^ + )}\phantom{))}\underline{-8640} \\ \phantom{{x^2 } + 2a} 1360 \\ \end{array}$$

The remainder is 1360.

The required least 5-digit number = 10000 + (1440 − 1360) = 10000 + 80 = 10080.

Hence, the least 5-digit number divisible by 32, 36 and 45 is 10080.

The required minimum distance is the LCM of 63, 70 and 77.

2	63, 70, 77
3	63, 35, 77
3	21, 35, 77
5	7, 35, 77
7	7, 7, 77
11	1, 1, 11
	1, 1, 1

LCM of 63, 70 and 77 = 2 x 3 x 3 x 5 x 7 x 11 = 6930 cm.

Converting to metres and centimetres: 6930 cm = 69 m 30 cm.

Hence, the minimum distance each should cover is 6930 cm or 69 m 30 cm.

The required time interval after which the traffic lights will change together again is the LCM of 48, 72 and 108.

2	48, 72, 108
2	24, 36, 54
2	12, 18, 27
2	6, 9, 27
3	3, 9, 27
3	1, 3, 9
3	1, 1, 3
	1, 1, 1

LCM of 48, 72 and 108 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds.

Converting to minutes: 432 seconds = 7 minutes 12 seconds.

The traffic lights change simultaneously at 7 A.M. So they will change together again at:

7 A.M. + 7 min 12 sec = 7:07:12 A.M.

Hence, they will change together again at 7 minutes 12 seconds past 7 A.M.

We know that:

HCF × LCM = Product of two numbers

⇒ 12 × LCM = 4032

⇒ LCM = $\frac{4032}{12}$

⇒ LCM = 336.

Hence, their LCM is 336.

We know that:

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

⇒ 9 × 270 = 45 × (other number)

⇒ Other number = $\frac{9 \times 270}{45}$

⇒ Other number = $\frac{2430}{45}$

⇒ Other number = 54.

Hence, the other number is 54.

To find HCF of 180 and 336 by division method:

$$\begin{array}{l} 180\overline{\smash{\big)}336\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{()}\underline{-180} \\ \phantom{{x^2 } 1()}156\overline{\smash{\big)}180\smash{\big(}}\phantom{}1 \\ \phantom{{x} +1xa}\underline{-156} \\ \phantom{{x^2 a} + 2x()} 24\overline{\smash{\big)}156\smash{\big(}}\phantom{}6 \\ \phantom{{x} +1xa+()}\underline{-144} \\ \phantom{{x^2 a} + 2x++()} 12\overline{\smash{\big)}24\smash{\big(}}\phantom{}2 \\ \phantom{{x} +1xa++++}\underline{-24} \\ \phantom{{x^2 a} + 2x++++()} 0 \\ \end{array}$$

So, HCF of 180 and 336 = 12.

Now, LCM = $\frac{\text{Product of numbers}}{\text{HCF}}$

⇒ LCM = $\frac{180 \times 336}{12}$

⇒ LCM = 180 × 28

⇒ LCM = 5040.

Hence, HCF = 12 and LCM = 5040.

We know that the LCM of two numbers is always exactly divisible by their HCF.

On dividing 110 by 15:

$$\begin{array}{l} \phantom{x^2 1}{\quad 7} \\ 15\overline{\smash{\big)}110} \\ \phantom{x^(}\phantom{}\underline{-105} \\ \phantom{{x^2 } + 2 } 5 \\ \end{array}$$

We get quotient = 7 and remainder = 5.

Since the remainder ≠ 0, 110 is not exactly divisible by 15.

Hence, two numbers cannot have 15 as their H.C.F. and 110 as their L.C.M. and L.C.M. of two numbers is always exactly divisible by H.C.F.

(i) The only natural number which has exactly one factor is 1.

(ii) The only prime number which is even is 2.

(iii) The HCF of two co-prime numbers is 1.

(iv) Two perfect numbers are 6 and 28.

(v) The only prime-triplet is 3, 5, 7.

(vi) The LCM of two or more given numbers is the lowest of their common multiples.

(vii) The HCF of two or more given numbers is the highest of their common factors.

(i) True.

Reason: Every natural number has a limited number of factors, with the largest factor being the number itself.

(ii) True.

Reason: Every natural number can be multiplied by infinitely many natural numbers giving infinite multiples.

(iii) True.

Reason: It has been proven (since the time of Euclid) that there are infinitely many prime numbers.

(iv) False.

Reason: The HCF of two numbers is less than or equal to the smaller of the two numbers. For example, HCF(6, 12) = 6, which is equal to the smaller number.

(v) False.

Reason: The LCM of two numbers is greater than or equal to the larger of the two numbers. For example, LCM(6, 12) = 12, which is equal to the larger number.

(i) False.

Reason: For example, 2 and 7 both are prime numbers but their sum = 2 + 7 = 9, which is an odd number.

(ii) False.

Reason: For example, 3 and 5 both are prime numbers but their sum = 3 + 5 = 8, which is a composite number.

(iii) False.

Reason: For example, 2 and 5 both are prime numbers but their sum = 2 + 5 = 7, which is a prime number.

(iv) False.

Reason: For example,13 is an odd number and 13 = 2 + 11, which is the sum of two prime numbers.

(v) False.

Reason: For example, 8 and 15 are co-prime but neither 8 is prime nor 15 is prime.

(vi) True.

Reason: Since 3 and 6 are factors of 18, any number divisible by 18 is divisible by both 3 and 6.

(vii) False.

Reason: For example, 20 is divisible by 2 and 4 both but not divisible by 8.

(viii) False.

Reason: 12 is divisible by 3 and 6 both but 12 is not divisible by 18.

(ix) False.

Reason: For example, 6 is even and 9 is odd but HCF(6, 9) = 3.

The factors of 6 are obtained from:

6 = 1 × 6 and 6 = 2 × 3.

So, the factors of 6 are 1, 2, 3 and 6.

Hence, option 4 is the correct option.

7 is a prime number.

9 = 3 × 3, so it has factors 1, 3 and 9 (more than two factors). Hence, 9 is an odd composite number.

11 is a prime number.

12 is an even composite number.

Hence, option 2 is the correct option.

The even numbers between 68 and 90 (exclusive) are: 70, 72, 74, 76, 78, 80, 82, 84, 86 and 88.

Counting these, we get 10 even numbers.

Hence, option 1 is the correct option.

69 = 3 × 23, so 69 is not prime.

87 = 3 × 29, so 87 is not prime.

91 = 7 × 13, so 91 is not prime.

97: As 9 × 9 = 81 < 97 and 10 × 10 = 100 > 97, primes ≤ 9 are 2, 3, 5, 7. None of these divide 97. So 97 is prime.

Hence, option 4 is the correct option.

Twin primes are pairs of primes whose difference is 2.

19, 21: 21 = 3 × 7 is not prime.

43, 47: difference is 4, not twin primes.

59, 61: both are primes and difference is 2. ✓

73, 79: difference is 6, not twin primes.

Hence, option 3 is the correct option.

The largest 4-digit number is 9999.

By prime factorisation:

9999 = 3 × 3 × 11 × 101.

The distinct prime factors are 3, 11 and 101 — that is 3 distinct prime factors.

Hence, option 2 is the correct option.

The smallest 5-digit number is 10000.

Prime factorisation: 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5.

The distinct prime factors are 2 and 5 — that is 2 distinct prime factors.

Hence, option 1 is the correct option.

The prime factorisation of 1729 is 7 × 13 × 19.

Sum of prime factors = 7 + 13 + 19 = 39.

Hence, option 4 is the correct option.

8, 45: 8 = 2³, 45 = 3² × 5. No common factor other than 1. Co-prime ✓

3, 18: 18 = 2 × 3². Common factor is 3. Not co-prime.

5, 35: 35 = 5 × 7. Common factor is 5. Not co-prime.

6, 39: 6 = 2 × 3, 39 = 3 × 13. Common factor is 3. Not co-prime.

Hence, option 1 is the correct option.

Every natural number has an infinite number of multiples.

Hence, option 3 is the correct option.

A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

1. 308594: last two digits = 94, not divisible by 4.

2. 506784: last two digits = 84, and 84 ÷ 4 = 21, divisible by 4. ✓

3. 732106: last two digits = 06 = 6, not divisible by 4.

4. 9301538: last two digits = 38, not divisible by 4.

Hence, option 2 is the correct option.

A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

1. 503786: last three digits = 786, not divisible by 8.

2. 505268: last three digits = 268, not divisible by 8.

3. 305678: last three digits = 678, not divisible by 8.

4. 703568: last three digits = 568, and 568 ÷ 8 = 71, divisible by 8. ✓

Hence, option 4 is the correct option.

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

1. 50762: sum = 5 + 0 + 7 + 6 + 2 = 20, not divisible by 3.

2. 42063: sum = 4 + 2 + 0 + 6 + 3 = 15, divisible by 3. ✓

3. 52871: sum = 5 + 2 + 8 + 7 + 1 = 23, not divisible by 3.

4. 37036: sum = 3 + 7 + 0 + 3 + 6 = 19, not divisible by 3.

Hence, option 2 is the correct option.

A number is divisible by 9 if the sum of its digits is divisible by 9.

1. 972063: sum = 9 + 7 + 2 + 0 + 6 + 3 = 27, divisible by 9. ✓

2. 730542: sum = 7 + 3 + 0 + 5 + 4 + 2 = 21, not divisible by 9.

3. 785423: sum = 7 + 8 + 5 + 4 + 2 + 3 = 29, not divisible by 9.

4. 5612844: sum = 5 + 6 + 1 + 2 + 8 + 4 + 4 = 30, not divisible by 9.

Hence, option 1 is the correct option.

A number is divisible by 6 if it is divisible by both 2 and 3.

1. 560324: even. Sum = 5 + 6 + 0 + 3 + 2 + 4 = 20, not divisible by 3.

2. 650374: even. Sum = 6 + 5 + 0 + 3 + 7 + 4 = 25, not divisible by 3.

3. 798653: ends in 3, odd. Not divisible by 2.

4. 750972: even. Sum = 7 + 5 + 0 + 9 + 7 + 2 = 30, divisible by 3. ✓

Hence, option 4 is the correct option.

For divisibility by 9, the sum of digits must be divisible by 9.

Sum of given digits (except *) = 5 + 4 + 2 + 8 + 1 = 20.

We need 20 + * to be divisible by 9.

20 + 7 = 27, which is divisible by 9.

So, * = 7.

Hence, option 2 is the correct option.

22 = 2 × 11, and 2, 11 are co-prime. So the number must be divisible by both 2 and 11.

Last digit is 8, so divisible by 2 for any value of *.

For divisibility by 11:

Digits from right: 8, 9, *, 4, 5, 2, 7.

Sum of digits at odd places (from right) = 8 + * + 5 + 7 = 20 + *.

Sum of digits at even places (from right) = 9 + 4 + 2 = 15.

Difference = (20 + *) − 15 = 5 + *.

For divisibility by 11: 5 + * = 11 ⇒ * = 6.

Hence, option 4 is the correct option.

5 and 6 are co-prime, so any number divisible by both is divisible by 5 × 6 = 30.

So the number is divisible by every factor of 30, namely 1, 2, 3, 5, 6, 10, 15 and 30.

But 60 is not a factor of 30, so the number may not be divisible by 60. (For example, 30 itself is divisible by 5 and 6 but not by 60.)

Hence, option 4 is the correct option.

Prime factorisation of the given numbers are:

60 = 2 × 2 × 3 × 5

75 = 3 × 5 × 5

105 = 3 × 5 × 7

The common prime factors of 60, 75 and 105 are 3 and 5 — that is 2 common prime factors.

Hence, option 1 is the correct option.

By division method:

$$\begin{array}{l} 144\overline{\smash{\big)}198\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{))}\underline{-144} \\ \phantom{{x^2 } 1(()}54\overline{\smash{\big)}144\smash{\big(}}\phantom{}2 \\ \phantom{{x} +1xa}\underline{-108} \\ \phantom{{x^2 a} + 2x()} 36\overline{\smash{\big)}54\smash{\big(}}\phantom{}1 \\ \phantom{{x} +1xa+()}\underline{-36} \\ \phantom{{x^2 a} + 2x++(} 18\overline{\smash{\big)}36\smash{\big(}}\phantom{}2 \\ \phantom{{x} +1xa+++(}\underline{-36} \\ \phantom{{x^2 a} + 2x++++(} 0 \\ \end{array}$$

So, HCF of 144 and 198 = 18.

Hence, option 4 is the correct option.

Prime factorisation of the given numbers are:

30 = 2 × 3 × 5

45 = 3 × 3 × 5

LCM = 2 × 3 × 3 × 5 = 90.

Hence, option 4 is the correct option.

We know that:

HCF × LCM = Product of two numbers

⇒ 15 × LCM = 1575

⇒ LCM = $\frac{1575}{15}$

⇒ LCM = 105.

Hence, option 2 is the correct option.

LCM of two numbers is always exactly divisible by their HCF. So the HCF must be a factor of the LCM.

180 ÷ 45 = 4 ✓ (45 is a factor)

180 ÷ 60 = 3 ✓ (60 is a factor)

180 ÷ 75 = 2.4 ✗ (75 is not a factor)

180 ÷ 90 = 2 ✓ (90 is a factor)

So 75 cannot be the HCF.

Hence, option 3 is the correct option.

Statement I: Since 4 × 5 = 20, both 4 and 5 are factors of 20, and 20 is a multiple of both 4 and 5.

∴ Statement I is true.

Statement II: This is the correct definition. If a is a factor of c (i.e., a divides c), then c is called a multiple of a.

∴ Statement II is true.

Both statements are true and Statement II correctly explains Statement I.

Hence, option 3 is the correct option.

Statement I: Each of 5, 7, 11, 13 and 17 has exactly two factors (1 and itself). So they are all prime numbers.

∴ Statement I is true.

Statement II: The set of natural numbers is {1, 2, 3, ...}, so the smallest natural number is 1.

∴ Statement II is true.

Both statements are true.

Hence, option 3 is the correct option.

Statement I: 2 is a prime number, not a composite number (its only factors are 1 and 2). So 2, 4, 6 and 9 are not all composite.

∴ Statement I is false.

Statement II: A composite number is one which has more than two different factors (i.e., more than just 1 and itself). The given definition is incomplete/incorrect — every prime number also has prime factors (itself), but is not composite.

∴ Statement II is false.

Hence, option 4 is the correct option.

Statement I: Both 41 and 43 are prime numbers, and their difference is 43 − 41 = 2. So they form a pair of twin primes.

∴ Statement I is true.

Statement II: This is the correct definition of twin prime numbers.

∴ Statement II is true.

Hence, option 3 is the correct option.

A prime triplet is a set of three consecutive prime numbers of the form p, p + 2, p + 4.

The only such set is 3, 5, 7.

Sum = 3 + 5 + 7 = 15.

∴ Both Statement I and statement II are true.

Hence, option 3 is the correct option.

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

Sum = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60.

Twice the number = 2 × 24 = 48.

Since 60 ≠ 48, 24 is not a perfect number.

∴ Statement I is false.

Statement II: This is the correct definition of a perfect number.

∴ Statement II is true.

Hence, option 2 is the correct option.

(i) 88

88 = 1 × 88

88 = 2 × 44

88 = 4 × 22

88 = 8 × 11

Hence, factors of 88 are 1, 2, 4, 8, 11, 22, 44 and 88.

(ii) 105

105 = 1 × 105

105 = 3 × 35

105 = 5 × 21

105 = 7 × 15

Hence, factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105.

(iii) 96

96 = 1 × 96

96 = 2 × 48

96 = 3 × 32

96 = 4 × 24

96 = 6 × 16

96 = 8 × 12

Hence, factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96.

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

The multiples of 12 are: 12, 24, 36, 48, 60, 72, 84, 96, ...

Hence, the common multiples of 8 and 12 are 24, 48, 72, 96, ... (i.e., multiples of 24).

(i) 25 and 105

The factors of 25 are 1, 5 and 25.

The factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105.

The common factors are 1 and 5.

Hence, 25 and 105 are not co-prime.

(ii) 59 and 97

Both 59 and 97 are prime numbers.

The only common factor is 1.

Hence, 59 and 97 are co-prime.

(iii) 161 and 192

161 = 7 × 23, so factors of 161 are 1, 7, 23 and 161.

192 = 2⁶ × 3, so factors of 192 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96 and 192.

The only common factor is 1.

Hence, 161 and 192 are co-prime.

(i) 197244

Divisibility by 4: Last two digits = 44, and 44 ÷ 4 = 11. Divisible by 4.

Divisibility by 6: Last digit 4 (even, divisible by 2). Sum of digits = 1 + 9 + 7 + 2 + 4 + 4 = 27 (divisible by 3). Divisible by 6.

Divisibility by 8: Last three digits = 244, and 244 ÷ 8 = 30.5. Not divisible by 8.

Divisibility by 9: Sum of digits = 27, divisible by 9. Divisible by 9.

Divisibility by 11: Sum at odd places (from right) = 4 + 2 + 9 = 15. Sum at even places = 4 + 7 + 1 = 12. Difference = 15 − 12 = 3, not divisible by 11. Not divisible by 11.

Hence, 197244 is divisible by 4, 6 and 9.

(ii) 613440

Divisibility by 4: Last two digits = 40, divisible by 4. Divisible by 4.

Divisibility by 6: Last digit 0 (divisible by 2). Sum of digits = 6 + 1 + 3 + 4 + 4 + 0 = 18 (divisible by 3). Divisible by 6.

Divisibility by 8: Last three digits = 440, and 440 ÷ 8 = 55. Divisible by 8.

Divisibility by 9: Sum of digits = 18, divisible by 9. Divisible by 9.

Divisibility by 11: Sum at odd places (from right) = 0 + 4 + 1 = 5. Sum at even places = 4 + 3 + 6 = 13. Difference = 13 − 5 = 8, not divisible by 11. Not divisible by 11.

Hence, 613440 is divisible by 4, 6, 8 and 9.

(iii) 4100448

Divisibility by 4: Last two digits = 48, divisible by 4. Divisible by 4.

Divisibility by 6: Last digit 8 (divisible by 2). Sum of digits = 4 + 1 + 0 + 0 + 4 + 4 + 8 = 21 (divisible by 3). Divisible by 6.

Divisibility by 8: Last three digits = 448, and 448 ÷ 8 = 56. Divisible by 8.

Divisibility by 9: Sum of digits = 21, not divisible by 9. Not divisible by 9.

Divisibility by 11: Sum at odd places (from right) = 8 + 4 + 0 + 4 = 16. Sum at even places = 4 + 0 + 1 = 5. Difference = 16 − 5 = 11, divisible by 11. Divisible by 11.

Hence, 4100448 is divisible by 4, 6, 8 and 11.

In 92*389:

Digits from right: 9, 8, 3, *, 2, 9.

Sum of digits at odd places (from right) = 9 + 3 + 2 = 14.

Sum of digits at even places (from right) = 8 + * + 9 = 17 + *.

Difference = (17 + *) − 14 = 3 + *.

For divisibility by 11, 3 + * = 0 (impossible) or 3 + * = 11 ⇒ * = 8.

Hence, * should be replaced by 8. The number formed is 928389.

(i) 168

2	168
2	84
2	42
3	21
7	7
	1

Hence, 168 = 2 × 2 × 2 × 3 × 7.

(ii) 2304

2	2304
2	1152
2	576
2	288
2	144
2	72
2	36
2	18
3	9
3	3
	1

Hence, 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3.

(i) 24, 45

Prime factorisation of the given numbers are:

24 = 2 × 2 × 2 × 3

45 = 3 × 3 × 5

The only common prime factor is 3 (appearing once in each).

Hence, GCD of 24 and 45 = 3.

(ii) 180, 252, 324

Prime factorisation of the given numbers are:

180 = 2 × 2 × 3 × 3 × 5

252 = 2 × 2 × 3 × 3 × 7

324 = 2 × 2 × 3 × 3 × 3 × 3

The common prime factors are 2 (twice) and 3 (twice) in all three numbers.

Hence, GCD of 180, 252 and 324 = 2 × 2 × 3 × 3 = 36.

(i) 54, 82

$$\begin{array}{l} 54\overline{\smash{\big)}82\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{}\underline{-54} \\ \phantom{{x^2 } 1}28\overline{\smash{\big)}54\smash{\big(}}\phantom{}1 \\ \phantom{{x} +1}\underline{-28} \\ \phantom{{x^2 a} + 2} 26\overline{\smash{\big)}28\smash{\big(}}\phantom{}1 \\ \phantom{{x} +1xa()}\underline{-26} \\ \phantom{{x^2 a} + 2x+(} 2\overline{\smash{\big)}26\smash{\big(}}\phantom{}13 \\ \phantom{{x} +1xa++(}\underline{-26} \\ \phantom{{x^2 a} + 2x+++} 0 \\ \end{array}$$

Hence, HCF of 54 and 82 = 2.

(ii) 84, 120, 156

First, find HCF of 84 and 120:

$$\begin{array}{l} 84\overline{\smash{\big)}120\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{)}\underline{-84} \\ \phantom{{x^2 } 1)}36\overline{\smash{\big)}84\smash{\big(}}\phantom{}2 \\ \phantom{{x} +1(}\underline{-72} \\ \phantom{{x^2 a} + 2(} 12\overline{\smash{\big)}36\smash{\big(}}\phantom{}3 \\ \phantom{{x} +1xa()}\underline{-36} \\ \phantom{{x^2 a} + 2x+(} 0 \\ \end{array}$$

So, HCF of 84 and 120 = 12.

Now, find HCF of 12 and 156:

$$\begin{array}{l} 12\overline{\smash{\big)}156\smash{\big(}}\phantom{}13 \\ \phantom{x}\phantom{)}\underline{-156} \\ \phantom{{x^2 } 1+)}0 \\ \end{array}$$

So, HCF of 12 and 156 = 12.

Hence, HCF of 84, 120 and 156 = 12.

(i) 27, 90

Prime factorisation of the given numbers are:

27 = 3 × 3 × 3

90 = 2 × 3 × 3 × 5

Maximum power of 2 is 1 (in 90), 3 is 3 (in 27), and 5 is 1 (in 90).

Hence, LCM of 27 and 90 = 2 × 3 × 3 × 3 × 5 = 270.

(ii) 36, 48, 210

Prime factorisation of the given numbers are:

36 = 2 × 2 × 3 × 3

48 = 2 × 2 × 2 × 2 × 3

210 = 2 × 3 × 5 × 7

Maximum power of 2 is 4 (in 48), 3 is 2 (in 36), 5 is 1 (in 210), and 7 is 1 (in 210).

Hence, LCM of 36, 48 and 210 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 5040.

(i) 48, 60

2	48, 60
2	24, 30
2	12, 15
2	6, 15
3	3, 15
5	1, 5
	1, 1

Hence, LCM of 48 and 60 = 2 × 2 × 2 × 2 × 3 × 5 = 240.

(ii) 112, 168, 266

2	112, 168, 266
2	56, 84, 133
2	28, 42, 133
2	14, 21, 133
3	7, 21, 133
7	7, 7, 133
19	1, 1, 19
	1, 1, 1

Hence, LCM of 112, 168 and 266 = 2 × 2 × 2 × 2 × 3 × 7 × 19 = 6384.

When 2706 is divided by the required number, 6 is left as remainder. So 2706 − 6 = 2700 is exactly divisible by that number.

Similarly, 7041 − 21 = 7020 and 8250 − 42 = 8208 are exactly divisible by that number.

Therefore, the required number is the HCF of 2700, 7020 and 8208.

First, find HCF of 2700 and 7020:

$$\begin{array}{l} 2700\overline{\smash{\big)}7020\smash{\big(}}\phantom{}2 \\ \phantom{x}\phantom{())}\underline{-5400} \\ \phantom{{x^2 } 1+)}1620\overline{\smash{\big)}2700\smash{\big(}}\phantom{}1 \\ \phantom{{x} +1xa()}\underline{-1620} \\ \phantom{{x^2 a} + 2x+} 1080\overline{\smash{\big)}1620\smash{\big(}}\phantom{}1 \\ \phantom{{x} +1xa+++}\underline{-1080} \\ \phantom{{x^2 a} + 2x++++} 540\overline{\smash{\big)}1080\smash{\big(}}\phantom{}2 \\ \phantom{{x} +1xa++++(())}\underline{-1080} \\ \phantom{{x^2 a} + 2x+++++++} 0 \\ \end{array}$$

So, HCF of 2700 and 7020 = 540.

Now, find HCF of 540 and 8208:

$$\begin{array}{l} 540\overline{\smash{\big)}8208\smash{\big(}}\phantom{}15 \\ \phantom{x}\phantom{))}\underline{-8100} \\ \phantom{{x^2 } 1+)}108\overline{\smash{\big)}540\smash{\big(}}\phantom{}5 \\ \phantom{{x} +1xa(}\underline{-540} \\ \phantom{{x^2 a} + 2x+)} 0 \\ \end{array}$$

So, HCF of 540 and 8208 = 108.

Hence, the required greatest number is 108.

First, we find the LCM of 18, 21, 28 and 30.

2	18, 21, 28, 30
2	9, 21, 14, 15
3	9, 21, 7, 15
3	3, 7, 7, 5
5	1, 7, 7, 5
7	1, 7, 7, 1
	1, 1, 1, 1

LCM of 18, 21, 28 and 30 = 2 × 2 × 3 × 3 × 5 × 7 = 1260.

According to given condition, the required number, when decreased by 20, gives 1260.

The required number = 1260 + 20 = 1280.

Hence, the required least number is 1280.

The required maximum capacity of a bag is the HCF of 120, 144 and 204.

First, find HCF of 120 and 144:

$$\begin{array}{l} 120\overline{\smash{\big)}144\smash{\big(}}\phantom{}1 \\ \phantom{x}\phantom{))}\underline{-120} \\ \phantom{{x^2 } 1())}24\overline{\smash{\big)}120\smash{\big(}}\phantom{}5 \\ \phantom{{x} +1xa}\underline{-120} \\ \phantom{{x^2 a} + 2x+} 0 \\ \end{array}$$

So, HCF of 120 and 144 = 24.

Now, find HCF of 24 and 204:

$$\begin{array}{l} 24\overline{\smash{\big)}204\smash{\big(}}\phantom{}8 \\ \phantom{x}\phantom{}\underline{-192} \\ \phantom{{x^2 } 1)}12\overline{\smash{\big)}24\smash{\big(}}\phantom{)}2 \\ \phantom{{x} +1)}\underline{-24} \\ \phantom{{x^2 a} + 2x} 0 \\ \end{array}$$

So, HCF of 24 and 204 = 12.

Hence, the maximum capacity of a bag is 12 kg.

The required time interval is the LCM of 30, 36 and 45.

2	30, 36, 45
2	15, 18, 45
3	15, 9, 45
3	5, 3, 15
5	5, 1, 5
	1, 1, 1

LCM of 30, 36 and 45 = 2 × 2 × 3 × 3 × 5 = 180 minutes.

Converting to hours: 180 minutes = 3 hours.

The bells ring simultaneously at 8 a.m. So they will ring together again at:

8 a.m. + 3 hours = 11 a.m.

Hence, the bells will ring together again at 11 a.m.

The minimum equal number of biscuits of each brand the shopkeeper should buy is the LCM of 12, 15 and 21.

2	12, 15, 21
2	6, 15, 21
3	3, 15, 21
5	1, 5, 7
7	1, 1, 7
	1, 1, 1

LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 = 420.

So, the shopkeeper needs 420 biscuits of each brand.

Number of packets of brand A = 420 ÷ 12 = 35.

Number of packets of brand B = 420 ÷ 15 = 28.

Number of packets of brand C = 420 ÷ 21 = 20.

Hence, the shopkeeper should buy 35 packets of brand A, 28 packets of brand B and 20 packets of brand C.

If two numbers are co-prime, their LCM equals the product of the numbers.

⇒ Product of the numbers = LCM = 4940

⇒ 65 × (other number) = 4940

⇒ Other number = $\frac{4940}{65}$

⇒ Other number = 76.

Hence, the other number is 76.

For a 2-digit number to be odd, the unit's digit must be odd (1, 3, 5, 7 or 9).

We need digits with sum 8:

• Unit's digit 1: ten's digit = 7. Number = 71.
• Unit's digit 3: ten's digit = 5. Number = 53.
• Unit's digit 5: ten's digit = 3. Number = 35.
• Unit's digit 7: ten's digit = 1. Number = 17.

(Unit's digit 9 gives ten's digit −1, not valid.)

Hence, the 2-digit odd numbers whose sum of digits is 8 are 17, 35, 53 and 71.

The 2-digit twin primes are: (11, 13), (17, 19), (29, 31), (41, 43), (59, 61) and (71, 73).

We check each pair by reversing the digits:

• (11, 13): Reversed → (11, 31). Both 11 and 31 are prime. ✓
• (17, 19): Reversed → (71, 91). 91 = 7 × 13, not prime. ✗
• (29, 31): Reversed → (92, 13). 92 is even, not prime. ✗
• (41, 43): Reversed → (14, 34). Both even, not prime. ✗
• (59, 61): Reversed → (95, 16). 95 = 5 × 19, 16 = 2⁴, not prime. ✗
• (71, 73): Reversed → (17, 37). Both 17 and 37 are prime. ✓

Hence, the required pairs of 2-digit twin primes are (11, 13) and (71, 73).

A natural number has exactly three factors only if it is the square of a prime number (the factors are 1, the prime, and the square).

For 25 = 5², the factors are 1, 5 and 25.

We look for squares of primes less than 100:

• 2² = 4 (factors: 1, 2, 4) ✓
• 3² = 9 (factors: 1, 3, 9) ✓
• 5² = 25 (factors: 1, 5, 25) ✓ (given)
• 7² = 49 (factors: 1, 7, 49) ✓
• 11² = 121 > 100 ✗

Hence, the other three numbers are 4, 9 and 49. These numbers are squares of prime numbers.

1,00,000 = 10⁵ = (2 × 5)⁵ = 2⁵ × 5⁵.

To ensure neither number contains 0 as a digit, we group all the 2's into one number and all the 5's into the other:

2⁵ = 32

5⁵ = 3125

Verification: 32 × 3125 = 1,00,000. ✓

Neither 32 nor 3125 contains 0 as a digit.

Hence, the two positive integers are 32 and 3125.

The common multiples shown are 15 and 30. So the two circles represent multiples of two numbers whose common multiples (up to 30) include 15 and 30.

The two numbers must be factors of 15 (excluding 1 and 15 themselves, as the circles need separate non-common multiples in the highest range up to 30).

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

Multiples of 3 (up to 30, excluding common multiples): 3, 6, 9, 12, 18, 21, 24, 27.

Multiples of 5 (up to 30, excluding common multiples): 5, 10, 20, 25.

Common multiples (already given): 15, 30.

Hence, the left circle represents Multiples of 3 (containing 3, 6, 9, 12, 18, 21, 24, 27), the right circle represents Multiples of 5 (containing 5, 10, 20, 25), and the common region contains 15 and 30.

(i) This depends on the student's year of birth. The student should list all years which are multiples of 4 (and not exceptions) from their birth year up to the present year.

For example, if the student was born in 2014 and the present year is 2026, the leap years are: 2016, 2020, 2024.

Hence, list the multiples of 4 between your birth year and the present year, excluding any year divisible by 100 but not by 400.

(ii) From 2024 till 2099 (both inclusive):

Multiples of 4 from 2024 to 2096:

2024, 2028, 2032, ..., 2096.

This forms an arithmetic sequence with first term 2024, common difference 4 and last term 2096.

Number of terms = $\frac{2096 - 2024}{4} + 1 = \frac{72}{4} + 1 = 18 + 1 = 19$.

We must check exceptions: years divisible by 100 but not 400 are not leap years. The only century year in this range is 2100, but 2100 > 2099, so it's not in our range. There are no exceptions in 2024–2099.

Hence, there are 19 leap years from 2024 till 2099.