Playing With Numbers

Solving,

⇒ 24 - 20 ÷ 5 × 2

⇒ 24 - 4 × 2     [Performing division]

⇒ 24 - 8     [Performing multiplication]

⇒ 16

Hence, 24 - 20 ÷ 5 × 2 = 16.

Solving,

⇒ 19 + 12 ÷ 6 - 2 × 3

⇒ 19 + 2 - 2 × 3     [Performing division]

⇒ 19 + 2 - 6     [Performing multiplication]

⇒ 21 - 6

⇒ 15

Hence, 19 + 12 ÷ 6 - 2 × 3 = 15.

Solving,

⇒ 21 - 12 ÷ 4 - 3 × (-7)

⇒ 21 - 3 - 3 × (-7)     [Performing division]

⇒ 21 - 3 + 21     [Performing multiplication]

⇒ 21 + 18

⇒ 39

Hence, 21 - 12 ÷ 4 - 3 × (-7) = 39.

Solving,

⇒ 23 + 18 ÷ 6 + 3 × (-4)

⇒ 23 + 3 + 3 × (-4)     [Performing division]

⇒ 23 + 3 - 12     [Performing multiplication]

⇒ 26 - 12

⇒ 14

Hence, 23 + 18 ÷ 6 + 3 × (-4) = 14.

Solving,

⇒ 30 - {8 ÷ (2 × 3 - 4)}

⇒ 30 - {8 ÷ (6 - 4)}     [Removing × in ()]

⇒ 30 - {8 ÷ 2}     [Removing ()]

⇒ 30 - 4     [Removing {}]

⇒ 26.

Hence, 30 - {8 ÷ (2 × 3 - 4)} = 26.

Solving,

⇒ 9 - {5 × 3 + 12 × 16 ÷ (-8)}

⇒ 9 - {5 × 3 + 12 × (-2)}     [Performing division in {}]

⇒ 9 - {15 - 24}     [Performing multiplication in {}]

⇒ 9 + 9     [Removing {}]

⇒ 18

Hence, 9 - {5 × 3 + 12 × 16 ÷ (-8)} = 18.

Solving,

⇒ 16 - {9 - 24 ÷ (8 + 6 × 2 - 16)}

⇒ 16 - {9 - 24 ÷ (8 + 12 - 16)}     [Removing × in ()]

⇒ 16 - {9 - 24 ÷ (20 - 16)}

⇒ 16 - {9 - 24 ÷ 4}     [Removing ()]

⇒ 16 - {9 - 6}     [Performing ÷ in {}]

⇒ 16 - 3     [Removing {}]

⇒ 13

Hence, 16 - {9 - 24 ÷ (8 + 6 × 2 - 16)} = 13.

Solving,

⇒ 12 - [13 - {9 + 10 ÷ (7 - 2)}]

⇒ 12 - [13 - {9 + 10 ÷ 5}]     [Removing ()]

⇒ 12 - [13 - {9 + 2}]     [Performing ÷ in {}]

⇒ 12 - [13 - 11]     [Removing {}]

⇒ 12 - 2     [Removing []]

⇒ 10

Hence, 12 - [13 - {9 + 10 ÷ (7 - 2)}] = 10.

Solving,

⇒ 4 - [16 - {6 + 12 ÷ (9 - 6)}]

⇒ 4 - [16 - {6 + 12 ÷ 3}]     [Removing ()]

⇒ 4 - [16 - {6 + 4}]     [Performing ÷ in {}]

⇒ 4 - [16 - 10]     [Removing {}]

⇒ 4 - 6     [Removing []]

⇒ -2

Hence, 4 - [16 - {6 + 12 ÷ (9 - 6)}] = -2.

Solving,

⇒ 28 - [15 - {8 + 20 ÷ (7 - $\overline{8 - 6})$}]

⇒ 28 - [15 - {8 + 20 ÷ (7 - 2)}]     [Removing bar]

⇒ 28 - [15 - {8 + 20 ÷ 5}]     [Removing ()]

⇒ 28 - [15 - {8 + 4}]     [Performing ÷ in {}]

⇒ 28 - [15 - 12]     [Removing {}]

⇒ 28 - 3     [Removing []]

⇒ 25

Hence, 28 - [15 - {8 + 20 ÷ (7 - $\overline{8 - 6})$}] = 25.

On dividing 1065 by 15, we get:

$$\begin{array}{l} \phantom{22 -)}{71} \\ 15\overline{\smash{\big)}1065} \\ \phantom{22}\phantom{)}\mathllap{-,}\underline{105\phantom{+1}} \\ \phantom{{22 - 1}}15 \\ \phantom{22 - 1}\mathllap{-,}\underline{15} \\ \phantom{22 - 1)}\underline{0} \\ \end{array}$$

On dividing 1065 by 15, the remainder is 0.

∴ 1065 is exactly divisible by 15.

Hence, 15 is a factor of 1065.

On dividing 1241 by 17, we get:

$$\begin{array}{l} \phantom{22 -)}{73} \\ 17\overline{\smash{\big)}1241} \\ \phantom{22}\phantom{)}\mathllap{-,}\underline{119\phantom{+1}} \\ \phantom{{22 - 1}}51 \\ \phantom{22 - 1}\mathllap{-,}\underline{51} \\ \phantom{22 - 1)}\underline{0} \\ \end{array}$$

On dividing 1241 by 17, the remainder is 0.

∴ 1241 is exactly divisible by 17.

Hence, 17 is a factor of 1241.

On dividing 2186 by 14, we get:

$$\begin{array}{l} \phantom{--}{156} \\ 14\overline{\smash{\big)}2186} \\ \phantom{22}\phantom{)}\mathllap{-,}\underline{14\phantom{+1}} \\ \phantom{1234.}78\ \phantom{22 - }\mathllap{-,}\underline{70\phantom{1}} \\ \phantom{12356.}86\ \phantom{22 - 1}\mathllap{-,}\underline{84\phantom{1}} \\ \phantom{22 -12 }\underline{2\phantom{1}} \\ \end{array}$$

On dividing 2186 by 14, the remainder is 2.

∴ 2186 is not exactly divisible by 14.

Hence, 14 is not a factor of 2186.

On dividing 2789 by 23, we get:

$$\begin{array}{l} \phantom{--}{121} \\ 23\overline{\smash{\big)}2789} \\ \phantom{22}\phantom{)}\mathllap{-,}\underline{23\phantom{+1}} \\ \phantom{1234.}48\ \phantom{22 - }\mathllap{-,}\underline{46\phantom{1}} \\ \phantom{12356.}29\ \phantom{22 - 1}\mathllap{-,}\underline{23\phantom{1}} \\ \phantom{22 -12 }\underline{6\phantom{1}} \\ \end{array}$$

On dividing 2789 by 23, the remainder is 6.

∴ 2789 is not exactly divisible by 23.

Hence, 23 is not a factor of 2789.

(i) We have,

40 = 1 × 40

40 = 2 × 20

40 = 4 × 10

40 = 5 × 8

Hence, possible factors of 40 are 1, 2, 4, 5, 8, 10, 20, 40.

(ii) We have,

56 = 1 × 56

56 = 2 × 28

56 = 4 × 14

56 = 7 × 8

Hence, possible factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56.

(iii) We have,

63 = 1 × 63

63 = 3 × 21

63 = 7 × 9

Hence, possible factors of 63 are 1, 3, 7, 9, 21, 63.

(iv) We have,

84 = 1 × 84

84 = 2 × 42

84 = 3 × 28

84 = 4 × 21

84 = 6 × 14

84 = 7 × 12

Hence, possible factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84.

(i) Multiples of 7 are obtained by multiplying 7 with natural numbers.

7 × 1 = 7

7 × 2 = 14

7 × 3 = 21

7 × 4 = 28

7 × 5 = 35

7 × 6 = 42

The first six multiples of 7 are 7, 14, 21, 28, 35, 42.

(ii) Multiples of 13 are obtained by multiplying 13 with natural numbers.

13 × 1 = 13

13 × 2 = 26

13 × 3 = 39

13 × 4 = 52

13 × 5 = 65

13 × 6 = 78

13 × 7 = 91

The first seven multiples of 13 are 13, 26, 39, 52, 65, 78, 91.

(iii) Multiples of 19 are obtained by multiplying 19 with natural numbers.

19 × 1 = 19

19 × 2 = 38

19 × 3 = 57

19 × 4 = 76

19 × 5 = 95

The first five multiples of 19 are 19, 38, 57, 76, 95.

A prime number has exactly 2 factors, 1 and itself.

Hence, the prime numbers are 19, 29, 37, 47, 73, 97.

The prime numbers between 30 and 60 are 31, 37, 41, 43, 47, 53, 59.

The prime numbers between 70 and 100 are 71, 73, 79, 83, 89, 97.

(i) By successive division of 105 by prime factors, we get:

$$\begin{array}{r|l} 3 & 105 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Hence, 105 = 3 × 5 × 7.

(ii) By successive division of 180 by prime factors, we get:

$$\begin{array}{r|l} 2 & 180 \\ \hline 2 & \phantom{1}90 \\ \hline 3 & \phantom{1}45 \\ \hline 3 & \phantom{1}15 \\ \hline 5 & \phantom{11}5 \\ \hline & \phantom{11} 1 \end{array}$$

Hence, 180 = 2 × 2 × 3 × 3 × 5.

(iii) By successive division of 420 by prime factors, we get:

$$\begin{array}{r|l} 2 & 420 \\ \hline 2 & 210 \\ \hline 3 & 105 \\ \hline 5 & \phantom{1}35 \\ \hline 7 & \phantom{11}7 \\ \hline & \phantom{11} 1 \end{array}$$

Hence, 420 = 2 × 2 × 3 × 5 × 7.

(iv) By successive division of 462 by prime factors, we get:

$$\begin{array}{r|l} 2 & 462 \\ \hline 3 & 231 \\ \hline 7 & \phantom{1}77 \\ \hline 11 & \phantom{1} 11 \\ \hline & \phantom{11} 1 \end{array}$$

Hence, 462 = 2 × 3 × 7 × 11.

(v) By successive division of 1035 by prime factors, we get:

$$\begin{array}{r|l} 3 & 1035 \\ \hline 3 & \phantom{1}345 \\ \hline 5 & \phantom{1}115 \\ \hline 23 & \phantom{11} 23 \\ \hline & \phantom{111} 1 \end{array}$$

Hence, 1035 = 3 × 3 × 5 × 23.

(vi) By successive division of 1197 by prime factors, we get:

$$\begin{array}{r|l} 3 & 1197 \\ \hline 3 & \phantom{1}399 \\ \hline 7 & \phantom{1}133 \\ \hline 19 & \phantom{11} 19 \\ \hline & \phantom{111} 1 \end{array}$$

Hence, 1197 = 3 × 3 × 7 × 19.

(vii) By successive division of 2535 by prime factors, we get:

$$\begin{array}{r|l} 3 & 2535 \\ \hline 5 & \phantom{1}845 \\ \hline 13 & \phantom{1}169 \\ \hline 13 & \phantom{11} 13 \\ \hline & \phantom{111} 1 \end{array}$$

Hence, 2535 = 3 × 5 × 13 × 13.

(viii) By successive division of 4641 by prime factors, we get:

$$\begin{array}{r|l} 3 & 4641 \\ \hline 7 & 1547 \\ \hline 13 & \phantom{1}221 \\ \hline 17 & \phantom{11} 17 \\ \hline & \phantom{111} 1 \end{array}$$

Hence, 4641 = 3 × 7 × 13 × 17.

(i) 1 is a factor of every number.

Reason: Every number is divisible by 1 without any remainder.

(ii) 0 is a multiple of every number.

Reason: When any number is multiplied by 0, the product is always 0.

(iii) 2 is the only even prime number.

Reason: 2 is an even number and is divisible only by 1 and 2. So, it is also a prime number.

(iv) 1 is neither prime nor composite.

Reason: A prime number has exactly two different factors, 1 and itself. A composite number has more than two factors. Since 1 has only one factor i.e. 1, it does not fit either definition.

(v) The largest 2-digit prime number is 97.

Reason: 97 has only two factors, 1 and 97, and there is no larger 2-digit prime number.

(vi) There are 25 prime numbers between 1 and 100.

Reason: By using the Sieve of Eratosthenes method to cross out multiples of 2, 3, 5, and 7, we are left with exactly 25 numbers that have no other factors besides 1 and themselves.

Rule: A number is divisible by 3, only if the sum of its digits is divisible by 3.

(i) The sum of the digits of the number 57432 is,

5 + 7 + 4 + 3 + 2 = 21

Since the sum of the digits, 21, is divisible by 3, the number 57432 is also divisible by 3.

(ii) The sum of the digits of the number 693503 is,

6 + 9 + 3 + 5 + 0 + 3 = 26

Since the sum of the digits, 26, is not divisible by 3, the number 693503 is not divisible by 3.

(iii) The sum of the digits of the number 3002002 is,

3 + 0 + 0 + 2 + 0 + 0 + 2 = 7

Since the sum of the digits, 7, is not divisible by 3, the number 3002002 is not divisible by 3.

(iv) The sum of the digits of the number 777771 is,

7 + 7 + 7 + 7 + 7 + 1 = 36

Since the sum of the digits, 36, is divisible by 3, the number 777771 is also divisible by 3.

Rule: A number is divisible by 9, only if the sum of its digits is divisible by 9.

(i) The sum of the digits of the number 666657 is,

6 + 6 + 6 + 6 + 5 + 7 = 36

Since the sum of the digits, 36, is divisible by 9, the number 666657 is also divisible by 9.

(ii) The sum of the digits of the number 999083 is,

9 + 9 + 9 + 0 + 8 + 3 = 38

Since the sum of the digits, 38, is not divisible by 9, the number 999083 is not divisible by 9.

(iii) The sum of the digits of the number 756432 is,

7 + 5 + 6 + 4 + 3 + 2 = 27

Since the sum of the digits, 27, is divisible by 9, the number 756432 is also divisible by 9.

(iv) The sum of the digits of the number 876869 is,

8 + 7 + 6 + 8 + 6 + 9 = 44

Since the sum of the digits, 44, is not divisible by 9, the number 876869 is not divisible by 9.

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

(i) Consider the number 22222.

Sum of its digits at odd places = 2 + 2 + 2 = 6

Sum of its digits at even places = 2 + 2 = 4

Their difference = 6 - 4 = 2

Since 2 is not 0 nor a multiple of 11, the number 22222 is not divisible by 11.

(ii) Consider the number 3303033.

Sum of its digits at odd places = 3 + 0 + 0 + 3 = 6

Sum of its digits at even places = 3 + 3 + 3 = 9

Their difference = 9 - 6 = 3

Since 3 is not 0 nor a multiple of 11, the number 3303033 is not divisible by 11.

(iii) Consider the number 1589148.

Sum of its digits at odd places = 8 + 1 + 8 + 1 = 18

Sum of its digits at even places = 4 + 9 + 5 = 18

Their difference = 18 - 18 = 0

Since the difference is 0, the number 1589148 is divisible by 11.

(iv) Consider the number 1302444.

Sum of its digits at odd places = 4 + 4 + 0 + 1 = 9

Sum of its digits at even places = 4 + 2 + 3 = 9

Their difference = 9 - 9 = 0

Since the difference is 0, the number 1302444 is divisible by 11.

Rule: A number is divisible by 4 if the number formed by its last two digits (tens and ones digits) is divisible by 4.

(i) Consider the number 36518. The number formed by its last two digits is 18 which is not divisible by 4.

Hence, the number 36518 is not divisible by 4.

(ii) Consider the number 579012. The number formed by its last two digits is 12 which is clearly divisible by 4.

Hence, the number 579012 is divisible by 4.

(iii) Consider the number 236754. The number formed by its last two digits is 54 which is not divisible by 4.

Hence, the number 236754 is not divisible by 4.

(iv) Consider the number 907246. The number formed by its last two digits is 46 which is not divisible by 4.

Hence, the number 907246 is not divisible by 4.

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

(i) Consider the number 13756. The number formed by its last three digits is 756 which is not divisible by 8.

So the number 13756 is not divisible by 8.

(ii) Consider the number 467132. The number formed by its last three digits is 132 which is not divisible by 8.

So the number 467132 is not divisible by 8.

(iii) Consider the number 6572904. The number formed by its last three digits is 904 which is divisible by 8.

So the number 6572904 is divisible by 8.

(iv) Consider the number 972972. The number formed by its last three digits is 972 which is not divisible by 8.

So the number 972972 is not divisible by 8.

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

(i) Consider the number 835704.

Its unit digit is 4. So, it is divisible by 2.

The sum of its digits, 8 + 3 + 5 + 7 + 0 + 4 = 27, which is divisible by 3.

As the number 835704 is divisible by both 2 and 3, it is divisible by 6.

Hence, 835704 is divisible by 6.

(ii) Consider the number 367038.

Its unit digit is 8. So, it is divisible by 2.

The sum of its digits, 3 + 6 + 7 + 0 + 3 + 8 = 27, which is divisible by 3.

As the number 367038 is divisible by both 2 and 3, it is divisible by 6.

Hence, 367038 is divisible by 6.

(iii) Consider the number 630663.

Its unit digit is 3. So, it is not divisible by 2.

As the number 630663 is not divisible by 2, it is not divisible by 6.

Hence, 630663 is not divisible by 6.

(iv) Consider the number 626266.

Its unit digit is 6. So, it is divisible by 2.

The sum of its digits, 6 + 2 + 6 + 2 + 6 + 6 = 28, which is not divisible by 3.

As the number 626266 is not divisible by 3, it is not divisible by 6.

Hence, 626266 is not divisible by 6.

Rule: A number is divisible by 7 if the difference between double the digit at ones place and the number formed by the rest of its digits is divisible by 7.

(i) Consider the number 67102.

6710 - (2 × 2) = 6710 - 4 = 6706

On dividing 6706 by 7, we get 958 with remainder 0.

Thus, 6706 is divisible by 7.

∴ 67102 is also divisible by 7.

(ii) Consider the number 73056.

7305 - (2 × 6) = 7305 - 12 = 7293

On dividing 7293 by 7, we get 1041 with remainder 6.

Thus, 7293 is not divisible by 7.

∴ 73056 is also not divisible by 7.

(iii) Consider the number 102543.

10254 - (2 × 3) = 10254 - 6 = 10248

On dividing 10248 by 7, we get 1464 with remainder 0.

Thus, 10248 is divisible by 7.

∴ 102543 is also divisible by 7.

(iv) Consider the number 203647.

20364 - (2 × 7) = 20364 - 14 = 20350

On dividing 20350 by 7, we get 2907 with remainder 1.

Thus, 20350 is not divisible by 7.

∴ 203647 is also not divisible by 7.

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

(i) The last digit of the number 55513 is 3.

Hence, 55513 is not divisible by 5.

(ii) The last digit of the number 689015 is 5.

Hence, 689015 is divisible by 5.

(iii) The last digit of the number 37490 is 0.

Hence, 37490 is divisible by 5.

(iv) The last digit of the number 1010106 is 6.

Hence, 1010106 is not divisible by 5.

Rule: Even numbers end in 0, 2, 4, 6, 8 whereas odd numbers end in 1, 3, 5, 7, 9.

By prime factorisation method, we get:

$$\begin{array}{r|r} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

So,

32 = 2 × 2 × 2 × 2 × 2;

56 = 2 × 2 × 2 × 7

The common factors are 2, 2 and 2.

∴ H.C.F. = 2 × 2 × 2 = 8

Hence, the H.C.F. of 32 and 56 is 8.

By prime factorisation method, we get:

$$\begin{array}{r|r} 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

So,

42 = 2 × 3 × 7;

63 = 3 × 3 × 7

The common factors are 3 and 7.

∴ H.C.F. = 3 × 7 = 21

Hence, the H.C.F. of 42 and 63 is 21.

By prime factorisation method, we get:

$$\begin{array}{r|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

So,

72 = 2 × 2 × 2 × 3 × 3;

96 = 2 × 2 × 2 × 2 × 2 × 3

The common factors are 2, 2, 2 and 3.

∴ H.C.F. = 2 × 2 × 2 × 3= 24

Hence, the H.C.F. of 72 and 96 is 24.

By prime factorisation method, we get:

$$\begin{array}{r|r} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \qquad \text{and} \quad \begin{array}{r|r} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

So,

81 = 3 × 3 × 3 × 3;

108 = 2 × 2 × 3 × 3 × 3

The common factors are 3, 3 and 3.

∴ H.C.F. = 3 × 3 × 3 = 27

Hence, the H.C.F. of 81 and 108 is 27.

By prime factorisation method, we get:

$$\begin{array}{r|r} 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

So,

135 = 3 × 3 × 3 × 5;

180 = 2 × 2 × 3 × 3 × 5

The common factors are 3, 3 and 5.

∴ H.C.F. = 3 × 3 × 5 = 45

Hence, the H.C.F. of 135 and 180 is 45.

By prime factorisation method, we get:

$$\begin{array}{r|r} 2 & 168 \\ \hline 2 & 84 \\ \hline 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

So,

168 = 2 × 2 × 2 × 3 × 7;

216 = 2 × 2 × 2 × 3 × 3 × 3

The common factors are 2, 2, 2, and 3.

∴ H.C.F. = 2 × 2 × 2 × 3 = 24

Hence, the H.C.F. of 168 and 216 is 24.

By prime factorisation method, we get:

$$\begin{array}{r|r} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 198 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$$

So,

144 = 2 × 2 × 2 × 2 × 3 × 3;

180 = 2 × 2 × 3 × 3 × 5;

198 = 2 × 3 × 3 × 11

The common factors are 2, 3 and 3.

∴ H.C.F. = 2 × 3 × 3 = 18

Hence, the H.C.F. of 144, 180 and 198 is 18.

By prime factorisation method, we get:

$$\begin{array}{r|r} 2 & 66 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 102 \\ \hline 3 & 51 \\ \hline 17 & 17 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 138 \\ \hline 3 & 69 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$$

So,

66 = 2 × 3 × 11;

102 = 2 × 3 × 17;

138 = 2 × 3 × 23

The common factors are 2 and 3.

∴ H.C.F. = 2 × 3 = 6

Hence, the H.C.F. of 66, 102 and 138 is 6.

Rule : To find the H.C.F., we perform long division by dividing the larger number by the smaller one and then repeatedly dividing each divisor by its remainder until the remainder is zero. The last divisor is the H.C.F.

We have:

$$\begin{array}{l} 144\overline{\smash{\big)}312 \smash{\big(}}2 \\ \phantom{144}\phantom{)}{288 } \\ \phantom{14444}\overline{24\smash{\big)}144 \smash{\big(}6} \\ \phantom{space 143 }\underline{144\phantom{1}} \\ \phantom{space 1431}0\ \end{array}$$

Hence, H.C.F. of 144 and 312 is 24.

We have:

$$\begin{array}{l} 252\overline{\smash{\big)}576 \smash{\big(}}2 \\ \phantom{144}\phantom{)}{504 } \\ \phantom{14444}\overline{72\smash{\big)}252 \smash{\big(}3} \\ \phantom{space 143 }{216\phantom{1}} \\ \phantom{space 1444}\overline{36\smash{\big)}72 \smash{\big(}2} \\ \phantom{space 1444 545 }\underline{72\phantom{1}} \\ \phantom{space 1444 545 )}0\ \end{array}$$

Hence, H.C.F. of 252 and 576 is 36.

We have:

$$\begin{array}{l} 245\overline{\smash{\big)}315 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{245 } \\ \phantom{14444}\overline{70\smash{\big)}245 \smash{\big(}3} \\ \phantom{space 143 }{210\phantom{1}} \\ \phantom{space 1444}\overline{35\smash{\big)}70 \smash{\big(}2} \\ \phantom{space 1444 545 }\underline{70\phantom{1}} \\ \phantom{space 1444 545 )}0\ \end{array}$$

Hence, H.C.F. of 245 and 315 is 35.

We have:

$$\begin{array}{l} 575\overline{\smash{\big)}920 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{575} \\ \phantom{1444}\overline{345\smash{\big)}575 \smash{\big(}1} \\ \phantom{space 14) }{345\phantom{1}} \\ \phantom{space 144}\overline{230\smash{\big)}345 \smash{\big(}1} \\ \phantom{space 1444 545 }{230\phantom{1}} \\ \phantom{space 144 5456 }\overline{115\smash{\big)}230 \smash{\big(}2} \\ \phantom{space 1444 5456778 }\underline{230\phantom{1}} \\ \phantom{space 1444 5456778 )}0\ \end{array}$$

Hence, H.C.F. of 575 and 920 is 115.

We have:

$$\begin{array}{l} 605\overline {\smash{\big)}935 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{605} \\ \phantom{1444}\overline{330\smash{\big)}605 \smash{\big(}1} \\ \phantom{space 14) }{330\phantom{1}} \\ \phantom{space 144}\overline{275\smash{\big)}330 \smash{\big(}1} \\ \phantom{space 1444 545 }{275\phantom{1}} \\ \phantom{space 144 54567 }\overline{55\smash{\big)}275 \smash{\big(}5} \\ \phantom{space 1444 5456778 }\underline{275\phantom{1}} \\ \phantom{space 1444 5456778 )}0\ \end{array}$$

Hence, H.C.F. of 605 and 935 is 55.

First, we will find the H.C.F. of 60 and 96.

$$\begin{array}{l} 60\overline{\smash{\big)}96 \smash{\big(}}1 \\ \phantom{14}\phantom{)}{60 } \\ \phantom{144}\overline{36\smash{\big)}60 \smash{\big(}1} \\ \phantom{space 1 }{36\phantom{1}} \\ \phantom{space 4}\overline{24\smash{\big)}36 \smash{\big(}1} \\ \phantom{space 1441 }{24\phantom{1}} \\ \phantom{space 4444}\overline{12\smash{\big)}24 \smash{\big(}2} \\ \phantom{space 1444 545 }\underline{24\phantom{1}} \\ \phantom{space 1444 545 )}0\ \end{array}$$

∴ H.C.F. of 60 and 96 is 12.

Next, we find the H.C.F. of 12 and 150.

$$\begin{array}{l} 12\overline{\smash{\big)}150 \smash{\big(}}12 \\ \phantom{14}\phantom{)}{144} \\ \phantom{1444)}\overline{6\smash{\big)}12 \smash{\big(}2} \\ \phantom{space 14 }\underline{12\phantom{1}} \\ \phantom{space 14)}0\ \end{array}$$

The H.C.F. of 12 and 150 is 6.

Hence, H.C.F. of 60, 96 and 150 is 6.

First, we will find the H.C.F. of 112 and 140.

$$\begin{array}{l} 112\overline{\smash{\big)}140 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{112} \\ \phantom{1444)}\overline{28\smash{\big)}112 \smash{\big(}4} \\ \phantom{space 144 }\underline{112\phantom{1}} \\ \phantom{space 1444)}0\ \end{array}$$

∴ H.C.F. of 112 and 140 is 28.

Next, we find the H.C.F. of 28 and 168.

$$\begin{array}{l} 28\overline{\smash{\big)}168 \smash{\big(}}6 \\ \phantom{14}\phantom{)}\underline{168} \\ \phantom{ 114)}0\ \end{array}$$

The H.C.F. of 28 and 168 is 28.

Hence, H.C.F. of 112, 140 and 168 is 28.

First, we will find the H.C.F. of 147 and 210.

$$\begin{array}{l} 147\overline{\smash{\big)}210 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{147} \\ \phantom{1444)}\overline{63\smash{\big)}147 \smash{\big(}2} \\ \phantom{144 4444}\phantom{)}{126} \\ \phantom{1444 5678)}\overline{21\smash{\big)}63 \smash{\big(}3} \\ \phantom{space 144 5678 }\underline{63\phantom{1}} \\ \phantom{space 1444 567)}0\ \end{array}$$

∴ H.C.F. of 147 and 210 is 21.

Next, we find the H.C.F. of 21 and 294.

$$\begin{array}{l} 21\overline{\smash{\big)}294 \smash{\big(}}14 \\ \phantom{14}\phantom{)}\underline{21} \\ \phantom{14}\phantom{1)}{84} \\ \phantom{14}\phantom{2)}\underline{84} \\ \phantom{ 1141)}0\ \end{array}$$

The H.C.F. of 21 and 294 is 21.

Hence, H.C.F. of 147, 210 and 294 is 21.

The required number is the H.C.F. of 385 and 735.

By long division, we get:

$$\begin{array}{l} 385\overline{\smash{\big)}735 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{385} \\ \phantom{1444}\overline{350\smash{\big)}385 \smash{\big(}1} \\ \phantom{144 4444}\phantom{)}{350} \\ \phantom{1444 5678)}\overline{35\smash{\big)}350 \smash{\big(}10} \\ \phantom{space 144 5678 }\underline{350\phantom{1}} \\ \phantom{space 1444 567)}0\ \end{array}$$

So the H.C.F. of 385 and 735 is 35.

Hence, the required number is 35.

The required number is the H.C.F. of 306, 450 and 540.

First, we will find the H.C.F. of 306 and 450.

$$\begin{array}{l} 306\overline{\smash{\big)}450 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{306} \\ \phantom{1444}\overline{144\smash{\big)}306 \smash{\big(}2} \\ \phantom{144 4444}\phantom{)}{288} \\ \phantom{1444 5678)}\overline{18\smash{\big)}144 \smash{\big(}8} \\ \phantom{space 144 5678 }\underline{144\phantom{1}} \\ \phantom{space 1444 567)}0\ \end{array}$$

Therefore, the H.C.F. of 306 and 450 is 18.

Next, we need to find the H.C.F. of 18 and 540.

$$\begin{array}{l} 18\overline{\smash{\big)}540 \smash{\big(}}30 \\ \phantom{14}\phantom{)}\underline{54} \\ \phantom{14}\phantom{1)}{00} \\ \phantom{14}\phantom{2)}\underline{00} \\ \phantom{ 1141)}0\ \end{array}$$

So the H.C.F. of 306, 450 and 540 is 18.

Hence, the required number is 18.

The required number is the H.C.F. of {(37 - 5) , (53 - 5)}. This implies that we need to find the H.C.F. of 32 and 48.

$$\begin{array}{l} 32\overline{\smash{\big)}48 \smash{\big(}}1 \\ \phantom{144}{32} \\ \phantom{144}\overline{16\smash{\big)}32 \smash{\big(}2} \\ \phantom{144 12) }\underline{32\phantom{1}} \\ \phantom{space 14}0\ \end{array}$$

The H.C.F. of 32 and 48 is 16.

Hence, the required number is 16.

The required number is the H.C.F. of {(138 - 3), (183 - 3), (423 - 3)}. This implies that we need to find the H.C.F. of 135, 180 and 420.

First find H.C.F. of 135 and 180:

$$\begin{array}{l} 135\overline{\smash{\big)}180 \smash{\big(}}1 \\ \phantom{1441}{135} \\ \phantom{144 12}\overline{45\smash{\big)}135 \smash{\big(}3} \\ \phantom{144 1223) }\underline{135\phantom{1}} \\ \phantom{space 1444}0\ \end{array}$$

The H.C.F. of 135 and 180 is 45. Next, find the H.C.F. of 45 and 420.

$$\begin{array}{l} 45\overline{\smash{\big)}420 \smash{\big(}}9 \\ \phantom{144}{405} \\ \phantom{144 1}\overline{15\smash{\big)}45 \smash{\big(}3} \\ \phantom{144 122) }\underline{45\phantom{1}} \\ \phantom{space 144}0\ \end{array}$$

The H.C.F. of 45 and 420 is 15. Therefore, the H.C.F. of 135, 180 and 420 is 15.

Hence, the required number is 15.

The required number = H.C.F. of {(76 - 2), (114 - 3), (152 - 4)} = H.C.F. of { 74, 111, 148}.

First find H.C.F. of 74 and 111:

$$\begin{array}{l} 74\overline{\smash{\big)}111 \smash{\big(}}1 \\ \phantom{1441}{74} \\ \phantom{144 1}\overline{37\smash{\big)}74 \smash{\big(}2} \\ \phantom{144 1221 }\underline{74\phantom{1}} \\ \phantom{space 144}0\ \end{array}$$

The H.C.F. of 74 and 111 is 37.

Now find H.C.F. of 37 and 148:

$$\begin{array}{l} 37\overline{\smash{\big)}148 \smash{\big(}}4 \\ \phantom{144 }\underline{148\phantom{1}} \\ \phantom{space }0\ \end{array}$$

The H.C.F. of 37 and 148 is 37. Therefore the H.C.F. of 74, 111 and 148 is 37.

Hence, the required number is 37.

The required measure of the bucket = H.C.F. of 104 and 91

By long division method, we get:

$$\begin{array}{l} 91\overline{\smash{\big)}104 \smash{\big(}}1 \\ \phantom{1441}{91} \\ \phantom{144 1}\overline{13\smash{\big)}91 \smash{\big(}7} \\ \phantom{144 1221 }\underline{91\phantom{1}} \\ \phantom{space 144}0\ \end{array}$$

The H.C.F. of 104 and 91 is 13.

Hence, the required measure of the bucket is 13 litres.

Two numbers are co-prime if their Highest Common Factor is 1.

By prime factorisation method, we get:

$$\begin{array}{r|r} 3 & 357 \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 5 & 1625 \\ \hline 5 & 325 \\ \hline 5 & 65 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$$

Here,

357 = 3 × 7 × 17;

1625 = 5 × 5 × 5 × 13

By comparing the prime factors, we can see there are no common factors other than 1. So H.C.F. of 357 and 1625 is 1.

Since the H.C.F. of 357 and 1625 is 1, the numbers are co-primes.

Hence, proved that 357 and 1625 are co-primes.

Rule :

To find the L.C.M. using prime factorization, express each number as a product of primes and then find the product of all prime factors taken the maximum number of times they appear in any single factorization.

By prime factorisation, we have:

$$\begin{array}{r|r} 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

18 = 2 × 3 × 3;

24 = 2 × 2 × 2 × 3

Here, 2 appears once in the factorisation of 18 and three times in that of 24. Also, 3 appears twice in the factorisation of 18 and once in that of 24. Thus, 2 appears maximum three times and 3 appears maximum two times.

So, L.C.M. of 18 and 24 = 2 × 2 × 2 × 3 × 3 = 72

Hence, the L.C.M. of 18 and 24 is 72.

By prime factorisation, we have:

$$\begin{array}{r|r} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

36 = 2 × 2 × 3 × 3;

45 = 3 × 3 × 5

2 appears maximum 2 times in one of the prime factorisations;

3 appears maximum 2 times in one of the prime factorisations;

5 appears maximum 1 time in one of the prime factorisations.

So, L.C.M. of 36 and 45 = 2 × 2 × 3 × 3 × 5 = 180

Hence, the L.C.M. of 36 and 45 is 180.

By prime factorisation, we have:

$$\begin{array}{r|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

48 = 2 × 2 × 2 × 2 × 3;

72 = 2 × 2 × 2 × 3 × 3

2 appears maximum 4 times in one of the prime factorisations;

3 appears maximum 2 times in one of the prime factorisations.

So, L.C.M. of 48 and 72 = 2 × 2 × 2 × 2 × 3 × 3 = 144

Hence, the L.C.M. of 48 and 72 is 144.

By prime factorisation, we have:

$$\begin{array}{r|r} 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 3 & 105 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

63 = 3 × 3 × 7;

105 = 3 × 5 × 7

3 appears maximum 2 times in one of the prime factorisations;

5 appears maximum 1 time in one of the prime factorisations;

7 appears maximum 1 time in one of the prime factorisations.

So, L.C.M. of 63 and 105 = 3 × 3 × 5 × 7 = 315

Hence, the L.C.M. of 63 and 105 is 315.

By prime factorisation, we have:

$$\begin{array}{r|r} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

27 = 3 × 3 × 3;

54 = 2 × 3 × 3 × 3;

90 = 2 × 3 × 3 × 5

2 appears maximum 1 time in one of the prime factorisations;

3 appears maximum 3 times in one of the prime factorisations;

5 appears maximum 1 time in one of the prime factorisations.

So, L.C.M. of 27, 54 and 90 = 2 × 3 × 3 × 3 × 5 = 270

Hence, the L.C.M. of 27, 54 and 90 is 270.

By prime factorisation, we have:

$$\begin{array}{r|r} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \qquad \text{and} \quad \begin{array}{r|r} 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \qquad \text{and} \quad \begin{array}{r|r} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

36 = 2 × 2 × 3 × 3;

54 = 2 × 3 × 3 × 3;

81 = 3 × 3 × 3 × 3

2 appears maximum 2 times in one of the prime factorisations;

3 appears maximum 4 times in one of the prime factorisations.

So, L.C.M. of 36, 54 and 81 = 2 × 2 × 3 × 3 × 3 × 3 = 324

Hence, the L.C.M. of 36, 54 and 81 is 324.

By prime factorisation, we have:

$$\begin{array}{r|r} 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \qquad \text{and} \quad \begin{array}{r|r} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

60 = 2 × 2 × 3 × 5;

75 = 3 × 5 × 5;

80 = 2 × 2 × 2 × 2 × 5

90 = 2 × 3 × 3 × 5

2 appears maximum 4 times in one of the prime factorisations;

3 appears maximum 2 times in one of the prime factorisations;

5 appears maximum 2 times in one of the prime factorisations.

So, L.C.M. of 60, 75, 80 and 90 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 3600

Hence, the L.C.M. of 60, 75, 80 and 90 is 3600.

By prime factorisation, we have:

$$\begin{array}{r|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

48 = 2 × 2 × 2 × 2 × 3;

60 = 2 × 2 × 3 × 5;

72 = 2 × 2 × 2 × 3 × 3;

96 = 2 × 2 × 2 × 2 × 2 × 3

2 appears maximum 5 times in one of the prime factorisations;

3 appears maximum 2 times in one of the prime factorisations;

5 appears maximum 1 time in one of the prime factorisations.

So, L.C.M. of 48, 60, 72 and 96 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440

Hence, the L.C.M. of 48, 60, 72 and 96 is 1440.

By common division method, we get:

$$\begin{array}{r|rrrr} 2 & 8, & 10, & 12, & 16 \\ \hline 2 & 4, & 5, & 6, & 8 \\ \hline 2 & 2, & 5, & 3, & 4 \\ \hline & 1, & 5, & 3, & 2 \\ \end{array}$$

The L.C.M. of 8, 10, 12, 16 = 2 × 2 × 2 × 5 × 3 × 2 = 240

Hence, L.C.M. of 8, 10, 12, 16 = 240.

By common division method, we get:

$$\begin{array}{r|rrrr} 2 & 12, & 15, & 18, & 24 \\ \hline 2 & 6, & 15, & 9, & 12 \\ \hline 3 & 3, & 15, & 9, & 6 \\ \hline & 1, & 5, & 3, & 2 \\ \end{array}$$

The L.C.M. of 12, 15, 18, 24 = 2 × 2 × 3 × 5 × 3 × 2 = 360

Hence, L.C.M. of 12, 15, 18, 24 = 360.

By common division method, we get:

$$\begin{array}{r|rrrr} 2 & 15, & 18, & 20, & 27 \\ \hline 3 & 15, & 9, & 10, & 27 \\ \hline 3 & 5, & 3, & 10, & 9 \\ \hline 5 & 5, & 1, & 10, & 3 \\ \hline & 1, & 1, & 2, & 3 \\ \end{array}$$

The L.C.M. of 15, 18, 20, 27 = 2 × 3 × 3 × 5 × 2 × 3 = 540

Hence, L.C.M. of 15, 18, 20, 27 = 540.

By common division method, we get:

$$\begin{array}{r|rrrr} 2 & 20, & 25, & 30, & 45 \\ \hline 3 & 10, & 25, & 15, & 45 \\ \hline 5 & 10, & 25, & 5, & 15 \\ \hline & 2, & 5, & 1, & 3 \\ \end{array}$$

The L.C.M. of 20, 25, 30, 45 = 2 × 3 × 5 × 2 × 5 × 3 = 900

Hence, L.C.M. of 20, 25, 30, 45 = 900.

By common division method, we get:

$$\begin{array}{r|rrrr} 3 & 21, & 45, & 63, & 81 \\ \hline 3 & 7, & 15, & 21, & 27 \\ \hline 7 & 7, & 5, & 7, & 9 \\ \hline & 1, & 5, & 1, & 9 \\ \end{array}$$

The L.C.M. of 21, 45, 63, 81 = 3 × 3 × 7 × 5 × 9 = 2835

Hence, L.C.M. of 21, 45, 63, 81 = 2835.

By common division method, we get:

$$\begin{array}{r|rrrrr} 2 & 16, & 18, & 24, & 32, & 36 \\ \hline 2 & 8, & 9, & 12, & 16, & 18 \\ \hline 2 & 4, & 9, & 6, & 8, & 9 \\ \hline 2 & 2, & 9, & 3, & 4, & 9 \\ \hline 3 & 1, & 9, & 3, & 2, & 9 \\ \hline 3 & 1, & 3, & 1, & 2, & 3 \\ \hline & 1, & 1, & 1, & 2, & 1 \\ \end{array}$$

The L.C.M. of 16, 18, 24, 32, 36 = 2 × 2 × 2 × 2 × 3 × 3 × 2 = 288

Hence, L.C.M. of 16, 18, 24, 32, 36 = 288.

By common division method, we get:

$$\begin{array}{r|rrr} 2 & 120, & 210, & 225 \\ \hline 3 & 60, & 105, & 225 \\ \hline 5 & 20, & 35, & 75 \\ \hline & 4, & 7, & 15 \\ \end{array}$$

The L.C.M. of 120, 210, 225 = 2 × 3 × 5 × 4 × 7 × 15 = 12600

Hence, L.C.M. of 120, 210, 225 = 12600.

By common division method, we get:

$$\begin{array}{r|rrr} 2 & 175, & 182, & 350 \\ \hline 5 & 175, & 91, & 175 \\ \hline 5 & 35, & 91, & 35 \\ \hline 7 & 7, & 91, & 7 \\ \hline & 1, & 13, & 1 \\ \end{array}$$

The L.C.M. of 175, 182, 350 = 2 × 5 × 5 × 7 × 13 = 4550

Hence, L.C.M. of 175, 182, 350 = 4550.

It is given that,

H.C.F. = 23

L.C.M. = 276

Given number = 92

We know that,

The other number = $\frac{\text{H.C.F.} \times \text{L.C.M.}}{\text{given number}}$

= $\frac{23 \times 276}{92} = \frac{6348}{92}$ = 69

Hence, the other number is 69.

It is given that,

H.C.F. = 144

L.C.M. = 2880

Given number = 576

We know that,

The other number = $\frac{\text{H.C.F.} \times \text{L.C.M.}}{\text{given number}}$

= $\frac{144 \times 2880}{576} = \frac{414720}{576}$ = 720

Hence, the other number is 720.

It is given that,

H.C.F. = 15

Product of two numbers = 2925

We know that,

L.C.M. = $\frac{\text{Product of the two numbers}}{\text{H.C.F.}}$

= $\frac{2925}{15}$

= 195

Hence, the L.C.M. is 195.

It is given that,

L.C.M. = 150

Product of two numbers = 3750

We know that,

H.C.F. = $\frac{\text{Product of the two numbers}}{\text{L.C.M.}}$

= $\frac{3750}{150}$

= 25

Hence, the H.C.F. is 25.

The H.C.F. of two numbers perfectly divides their L.C.M. without leaving any remainder.

We are given that, H.C.F. is 12 and L.C.M. is 64.

On dividing the L.C.M. by the H.C.F.,

$$\begin{array}{l} \phantom{22 ))}{5} \\ 12\overline{\smash{\big)}64} \\ \phantom{22}\phantom{)}\mathllap{-,}\underline{60} \\ \phantom{{22 )}}04 \\ \end{array}$$

The remainder is 4.

Thus, 12 does not divide 64 exactly.

Hence, there cannot be two numbers having an H.C.F. of 12 and an L.C.M. of 64.

The least number exactly divisible by the given numbers will be their L.C.M. So, we need to find the L.C.M. of 15, 18 and 24.

$$\begin{array}{r|rrr} 2 & 15, & 18, & 24 \\ \hline 3 & 15, & 9, & 12 \\ \hline & 5, & 3, & 4 \\ \end{array}$$

The L.C.M. of 15, 18, 24 = 2 × 3 × 5 × 3 × 4 = 360

Hence, the least number exactly divisible by 15, 18, and 24 is 360.

First, we find the L.C.M. of 15, 20, 25, and 30. By common division method, we get:

$$\begin{array}{r|rrrr} 2 & 15, & 20, & 25, & 30 \\ \hline 3 & 15, & 10, & 25, & 15 \\ \hline 5 & 5, & 10, & 25, & 5 \\ \hline & 1, & 2, & 5, & 1 \\ \end{array}$$

The L.C.M. of 15, 20, 25 and 30 = 2 × 3 × 5 × 2 × 5 = 300

Required number = L.C.M. of (15, 20, 25, 30) + 5 = 300 + 5 = 305

Hence, the required smallest number is 305.

The bells will toll together again after an interval of time equal to the L.C.M. of their individual tolling intervals (2, 4, 6, 8, 10, and 12 minutes).

By common division method, we get:

$$\begin{array}{r|rrrrrr} 2 & 2, & 4, & 6, & 8, & 10, & 12 \\ \hline 2 & 1, & 2, & 3, & 4, & 5, & 6 \\ \hline 3 & 1, & 1, & 3, & 2, & 5, & 3 \\ \hline & 1, & 1, & 1, & 2, & 5, & 1 \\ \end{array}$$

The L.C.M. of 2, 4, 6, 8, 10, and 12 = 2 × 2 × 3 × 2 × 5 = 120 minutes = 2 hours

Hence, the bells will toll together again after 2 hours.

The devices will beep together again after an interval of time equal to the L.C.M. of 15 and 20.

By common division method, we get:

$$\begin{array}{r|rr} 5 & 15, & 20 \\ \hline & 3, & 4 \\ \end{array}$$

The L.C.M. of 15 and 20 = 5 × 3 × 4 = 60

Therefore, the devices will beep together again after 60 minutes. Since 60 minutes equals 1 hour, we add 1 hour to their initial beep time.

Next beep time = 10 a.m. + 1 hour = 11 a.m.

Hence, the devices will make the next beep together at 11 a.m.

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

536146 $\longrightarrow$ Last three digits are 146. But 146 is not divisible by 8.

789214 $\longrightarrow$ Last three digits are 214. But 214 is not divisible by 8.

955626 $\longrightarrow$ Last three digits are 626. But 626 is not divisible by 8.

813256 $\longrightarrow$ Last three digits are 256. 256 is divisible by 8.

Since the number formed by the last three digits of 813256 is divisible by 8, the number 813256 is also divisible by 8.

Hence, Option 4 is the correct option.

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

We need to find the sum of the digits of each given number.

314526 $\longrightarrow$ 3 + 1 + 4 + 5 + 2 + 6 = 21

578142 $\longrightarrow$ 5 + 7 + 8 + 1 + 4 + 2 = 27

964315 $\longrightarrow$ 9 + 6 + 4 + 3 + 1 + 5 = 28

876421 $\longrightarrow$ 8 + 7 + 6 + 4 + 2 + 1 = 28

Among the sums, 27 is divisible by 9. Therefore 578142 is also divisible by 9.

Hence, Option 2 is the correct option.

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

Calculating the difference for each number,

3333333 $\longrightarrow$ (3 + 3 + 3 + 3) - (3 + 3 + 3) = 12 - 9 = 3

66666 $\longrightarrow$ (6 + 6 + 6) - (6 + 6) = 18 - 12 = 6

999999 $\longrightarrow$ (9 + 9 + 9) - (9 + 9 + 9) = 27 - 27 = 0

Since the difference for 999999 is 0, the number 999999 is divisible by 11.

Hence, Option 3 is the correct option.

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

345672 $\longrightarrow$ Last digit is 2 (divisible by 2) and the sum of digits 3 + 4 + 5 + 6 + 7 + 2 = 27 (divisible by 3).

154636 $\longrightarrow$ Last digit is 6 (divisible by 2) but the sum of digits 1 + 5 + 4 + 6 + 3 + 6 = 25 (not divisible by 3).

236794 $\longrightarrow$ Last digit is 4 (divisible by 2) but the sum of digits 2 + 3 + 6 + 7 + 9 + 4 = 31 (not divisible by 3).

Since 345672 is divisible by both 2 and 3, it is divisible by 6.

Hence, Option 1 is the correct option.

A prime number is a number that has only two factors; 1 and itself.

81 $\longrightarrow$ Factors are 1, 3, 9, 27 and 81. It has more than two factors.

87 $\longrightarrow$ Factors are 1, 3, 29 and 87. It has more than two factors.

91 $\longrightarrow$ Factors are 1, 7, 13 and 91. It has more than two factors.

97 $\longrightarrow$ Factors are 1 and 97. It has exactly two factors.

Since 97 has exactly two factors, it is a prime number.

Hence, Option 4 is the correct option.

A composite number has more than two factors.

23 $\longrightarrow$ Factors are 1 and 23. It has exactly two factors.

29 $\longrightarrow$ Factors are 1 and 29. It has exactly two factors.

87 $\longrightarrow$ Factors are 1, 3, 29 and 87. It has more than two factors.

97 $\longrightarrow$ Factors are 1 and 97. It has exactly two factors.

Since 87 has more than two factors, it is a composite number.

Hence, Option 3 is the correct option.

Two numbers are co-primes if their H.C.F. is 1.

Using the prime factorisation method, we need to find the H.C.F. of each pair of numbers:

H.C.F. of 13 and 91:

$$\begin{array}{r|r} 13 & 13 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 7 & 91 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$$

So,

13 = 1 × 13;

91 = 7 × 13

∴ H.C.F. of 13 and 91 is 13.

H.C.F. of 23 and 46 :

$$\begin{array}{r|r} 23 & 23 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 46 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$$

So,

23 = 1 × 23;

46 = 2 × 23

∴ H.C.F. of 23 and 46 is 23.

H.C.F. of 19 and 20 :

$$\begin{array}{r|r} 19 & 19 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

So,

19 = 1 × 19 ;

20 = 2 × 2 × 5

∴ H.C.F. of 19 and 20 is 1.

H.C.F. of 63 and 84 :

$$\begin{array}{r|r} 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 84 \\ \hline 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

So,

63 = 3 × 3 × 7 ;

84 = 2 × 2 × 3 × 7

∴ H.C.F. of 63 and 84 = 3 × 7 = 21.

As the H.C.F. of 19 and 20 is 1, 19 and 20 are co-prime.

Hence, Option 3 is the correct option.

By prime factorisation method, we get:

$$\begin{array}{r|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

So,

72 = 2 × 2 × 2 × 3 × 3;

90 = 2 × 3 × 3 × 5 ;

96 = 2 × 2 × 2 × 2 × 2 × 3

The common factors are 2 and 3.

∴ H.C.F. = 2 × 3= 6

Hence, Option 3 is the correct option.

Using the long division method, first, we will find the H.C.F. of 147 and 210.

$$\begin{array}{l} 147\overline{\smash{\big)}210 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{147} \\ \phantom{1444)}\overline{63\smash{\big)}147 \smash{\big(}2} \\ \phantom{144 4444}\phantom{)}{126} \\ \phantom{1444 5678)}\overline{21\smash{\big)}63 \smash{\big(}3} \\ \phantom{space 144 5678 }\underline{63\phantom{1}} \\ \phantom{space 1444 567)}0\ \end{array}$$

∴ H.C.F. of 147 and 210 is 21.

Next, we find the H.C.F. of 21 and 294.

$$\begin{array}{l} 21\overline{\smash{\big)}294 \smash{\big(}}14 \\ \phantom{14}\phantom{)}\underline{21} \\ \phantom{14}\phantom{1)}{84} \\ \phantom{14}\phantom{2)}\underline{84} \\ \phantom{ 1141)}0\ \end{array}$$

The H.C.F. of 21 and 294 is 21.

Hence, Option 4 is the correct option.

To reduce a fraction to its lowest terms, we divide both numerator and denominator by their H.C.F.

Using the long division method to find the H.C.F. :

$$\begin{array}{l} 289\overline{\smash{\big)}391 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{289} \\ \phantom{1444}\overline{102\smash{\big)}289 \smash{\big(}2} \\ \phantom{144 4444}\phantom{)}{204} \\ \phantom{1444 5678)}\overline{85\smash{\big)}102 \smash{\big(}1} \\ \phantom{144 444478911}\phantom{)}{85} \\ \phantom{1444 56787891)}\overline{17\smash{\big)}85 \smash{\big(}5} \\ \phantom{space 144 56785677 }\underline{85\phantom{1}} \\ \phantom{space 1444 567 5678)}0\ \end{array}$$

The H.C.F. of 289 and 391 is 17.

Dividing both numerator and denominator of $\frac{289}{391}$ by 17,

$\frac{289}{391} = \frac{289 \div 17}{391 \div 17} = \frac{17}{23}$

Hence, Option 4 is the correct option.

By common division method, we get:

$$\begin{array}{r|rrrr} 2 & 12, & 15, & 20, & 27 \\ \hline 2 & 6, & 15, & 10, & 27 \\ \hline 3 & 3, & 15, & 5, & 27 \\ \hline 5 & 1, & 5, & 5, & 9 \\ \hline & 1, & 1, & 1, & 9 \\ \end{array}$$

The L.C.M. of 12, 15, 20, 27 = 2 × 2 × 3 × 5 × 9 = 540

Hence, Option 4 is the correct option.

If $a$ and $b$ are co-primes, their H.C.F. is 1.

For any two numbers $a$ and $b$,

L.C.M. = $\frac{a \times b}{\text{H.C.F.}} = \frac{a \times b}{1} = ab$

Hence, Option 2 is the correct option.

It is given that,

H.C.F. = 23

L.C.M. = 276

Given number = 92

We know that,

The other number = $\frac{\text{H.C.F.} \times \text{L.C.M.}}{\text{given number}}$

= $\frac{23 \times 276}{92} = \frac{6348}{92}$ = 69

Hence, the other number is 69.

Hence, Option 1 is the correct option.

It is given that,

H.C.F. = 16

Product of two numbers = 2160

We know that,

L.C.M. = $\frac{\text{product of the two numbers}}{\text{H.C.F.}}$

= $\frac{2160}{16}$ = 135

Hence, the L.C.M. is 135.

Hence, Option 3 is the correct option.

First, we find the L.C.M. of 4, 5, and 6:

$$\begin{array}{r|rrr} 2 & 4, & 5, & 6 \\ \hline 2 & 2, & 5, & 3 \\ \hline & 1, & 5, & 3 \\ \end{array}$$

The L.C.M. of 4, 5, and 6 = 2 × 2 × 5 × 3 = 60

The smallest 4-digit number is 1000. Dividing 1000 by 60 gives a remainder of 40.

So the Required number = 1000 + (60 - 40) = 1020.

Hence, Option 3 is the correct option.

The least number divisible by the given numbers is their L.C.M.

By common division method:

$$\begin{array}{r|rrrrr} 2 & 15, & 20, & 24, & 32, & 36 \\ \hline 2 & 15, & 10, & 12, & 16, & 18 \\ \hline 2 & 15, & 5, & 6, & 8, & 9 \\ \hline 3 & 15, & 5, & 3, & 4, & 9 \\ \hline 5 & 5, & 5, & 1, & 4, & 3 \\ \hline & 1, & 1, & 1, & 4, & 3 \\ \end{array}$$

The L.C.M. of 15, 20, 24, 32, 36 = 2 × 2 × 2 × 3 × 5 × 4 × 3 = 1440

Hence, Option 2 is the correct option.

By prime factorisation, we have:

$$\begin{array}{r|r} 3 & 87 \\ \hline 29 & 29 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|l} 5 & 145 \\ \hline 29 & 29 \\ \hline & 1 \end{array}$$

87 = 3 × 29;

145 = 5 × 29

So, L.C.M. of 87 and 145 = 3 × 5 × 29 = 1305

Hence, Option 3 is the correct option.

To reduce a fraction to its lowest terms, we divide both numerator and denominator by their H.C.F.

Using the long division method to find the H.C.F. :

$$\begin{array}{l} 561\overline{\smash{\big)}748 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{561} \\ \phantom{1444}\overline{187\smash{\big)}561 \smash{\big(}3} \\ \phantom{space 1442 }\underline{561\phantom{1}} \\ \phantom{space 1444 )}0\ \end{array}$$

The H.C.F. of 561 and 748 is 187.

Dividing both numerator and denominator of $\frac{561}{748}$ by 187,

$\frac{561}{748} = \frac{561 \div 187}{748 \div 187} = \frac{3}{4}$

Hence, Option 1 is the correct option.

As per the hint provided,

H.C.F. of $\frac{3}{16}, \frac{5}{12}$ and $\frac{7}{8}$ = $\frac{\text {H.C.F. of } 3, 5, 7 }{ \text{L.C.M. of } 16, 12, 8}$

Since 3, 5, and 7 are co-prime, their H.C.F. is 1.

To find the L.C.M. of 16, 12, 8:

By common division method:

$$\begin{array}{r|rrr} 2 & 8, & 12, & 16 \\ \hline 2 & 4, & 6, & 8 \\ \hline 2 & 2, & 3, & 4 \\ \hline & 1, & 3, & 2 \\ \end{array}$$

The L.C.M. of 8, 12, 16 = 2 × 2 × 2 × 3 × 2 = 48

H.C.F. of $\frac{3}{16}, \frac{5}{12}$ and $\frac{7}{8}$ = $\frac{\text {H.C.F. of } 3, 5, 7 }{ \text{L.C.M. of } 16, 12, 8}$ = $\frac{1}{48}$

Hence, Option 2 is the correct option.

As per the hint provided,

L.C.M. of $\frac{1}{3} , \frac{5}{6}, \frac{5}{9},$ and $\frac{10}{27}$ = $\frac{\text {L.C.M. of } 1, 5, 5, 10 }{ \text{H.C.F. of } 3, 6, 9, 27}$

L.C.M. of 1, 5, 5, 10:

$$\begin{array}{r|rrrr} 5 & 1, & 5, & 5, & 10 \\ \hline & 1, & 1, & 1, & 2 \\ \end{array}$$

The L.C.M. of 1, 5, 5, 10 = 5 × 2 = 10

H.C.F. of 3, 6, 9, 27 :

The prime factorizations are :

3 = 1 × 3

6 = 2 × 3

9 = 3 × 3

27 = 3 × 3 × 3

So the H.C.F. of 3, 6, 9, 27 is 3.

L.C.M. of $\frac{1}{3} , \frac{5}{6}, \frac{5}{9},$ and $\frac{10}{27}$ = $\frac{\text {L.C.M. of } 1, 5, 5, 10 }{ \text{H.C.F. of } 3, 6, 9, 27}$ = $\frac{10}{3}$

Hence, Option 4 is the correct option.

(i) We have,

20 = 1 × 20

20 = 2 × 10

20 = 4 × 5

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

(ii) L.C.M. of two numbers = $\frac{\text{Product of the given numbers}}{\text{H.C.F. of two numbers}}$

(iii) If H.C.F. of two numbers is 1, then these numbers are called co-primes.

(iv) Yes, 65781 is divisible by 9. Because, the sum of the digits of 65781 is,

6 + 5 + 7 + 8 + 1 = 27

Since the sum of its digits 27 is divisible by 9, 65781 is also divisible by 9.

(v) The smallest three prime numbers are 2, 3 and 5.

2 × 3 × 5 = 30,

So the smallest number having three different prime factors is 30.

(i) The statement is F. 2 is the smallest prime; 1 is neither prime nor composite.

(ii) The statement is T. A prime number is divisible only by 1 and itself.

(iii) The statement is T. If the H.C.F. of two numbers is 1, then the two numbers have no common factors other than 1, which defines co-prime numbers.

(iv) The statement is T. Any number multiplied by 1 gives itself.

(v) The statement is T. If $a$ and $b$ are co-primes, their H.C.F. is 1. For any two numbers $a$ and $b$, L.C.M. = $\frac{a \times b}{\text{H.C.F.}} = \frac{a \times b}{1} = ab$.

(vi) The statement is T. L.C.M. of 2, 3 and 5 = 2 × 3 × 5 = 30, and the smallest 4-digit multiple of 30 = 30 × 34 = 1020.

Two composite numbers can be co-prime if they do not share any common prime factors. For example, 8 and 9 are both composite, but they are co-prime. So, the Assertion (A) is true.

The Reason (R) is true and it provides the definitive rule that allows for the Assertion to occur. As, the only requirement is the absence of a shared prime factor, the numbers themselves are not required to be prime.

Hence, Both A and R are true and R is the correct explanation of A.

If a number is divisible by 2 and 3, it is divisible by their least common multiple, which is 6. It is not necessarily divisible by 12. A counterexample is 6 (divisible by 2 and 3, but not 12) or 18. So the Assertion (A) is false.

All the numbers divisible by 6 are multiples of 6. Since 6 is an even number, all its integer multiples are also even. So the Reason (R) is true.

Hence, A is false but R is true.

If one number is a multiple of the other, the larger number is the L.C.M. For example, the L.C.M. of 4 and 12 is 12. So the Assertion (A) is true.

For two given numbers, product of the numbers = product of their H.C.F. and L.C.M. So the Reason (R) is a valid mathematical property.

Both statements are true, but the Reason (R) does not explain the Assertion. The Assertion is true due to the concept of multiples and factors, not because of the product relationship between the H.C.F. and L.C.M.

Hence, Both A and R are true but R is not the correct explanation of A.

A number is divisible by 9 only if the sum of its digits is divisible by 9. The digit sum is 18, which is divisible by 9, so 1980, 2079 and 1089 satisfy this (the digit sum of 3036 is 12, so 3036 is rejected).

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

For the smallest option, 1089: (1 + 8) - (0 + 9) = 9 - 9 = 0, which is divisible by 11. Also 1089 = 9 × 121 = 11 × 99, so it is divisible by both 9 and 11.

Since, 1089 is the smallest value satisfying all the conditions, thus the smallest possible value of N = 1089.

Hence, Option 4 is the correct option.

The product of any two primes is greater than each prime and is divisible by at least one of those primes besides 1 and itself; therefore it is composite.

Hence, Option 1 is the correct option.

A number divisible by each of 2, 3, 4, 5, 6 and 7 must be a multiple of their L.C.M.

The L.C.M. of 2, 3, 4, 5, 6 and 7 is 420.

On dividing 10000 by 420, we get 23 with a remainder, so the nearest multiples of 420 are 420 × 23 = 9660 (just below 10000) and 420 × 24 = 10080 (just above 10000).

Since these are the multiples of 420 closest to 10000, thus the two required numbers are 9660 and 10080.

Hence, Option 1 is the correct option.

We know that,

The other number = $\frac{\text{H.C.F.} \times \text{L.C.M.}}{\text{given number}}$ = $\frac{13 \times 1989}{117} = \frac{25857}{117}$ = 221

Since the other number is 221, thus the sum of both the numbers = 117 + 221 = 338.

Hence, Option 3 is the correct option.

Checking each statement:

1. 1 is not a prime number; the smallest prime number is 2. So this statement is false.

2. 8 and 9 are co-prime, yet both are composite. So at least one of them need not be prime, and this statement is false.

3. 2 is a prime number that is even. So a prime number need not be odd, and this statement is false.

4. Two consecutive odd numbers differ by 2; if both are prime, they form a twin-prime pair. So this statement is true.

Since only the fourth statement holds, thus the true statement is that two consecutive odd prime numbers are always twin primes.

Hence, Option 4 is the correct option.

The sum of the digits 0, 4 and 5 is 0 + 4 + 5 = 9, which is divisible by 3. So every 3-digit number formed using these digits is divisible by 3, and hence is composite.

(For example, 405 is odd, 450 is even and 504 is composite, so the number can be odd, even or composite.)

Since every such number is always divisible by 3, thus the number formed will never be a prime number.

Hence, Option 1 is the correct option.

The 2nd multiple of 9 is 9 × 2 = 18 and the 6th multiple of 6 is 6 × 6 = 36.

So, P = L.C.M. of 18 and 36 = 36.

Also, 60 = 2 × 2 × 3 × 5 and 84 = 2 × 2 × 3 × 7, so Q = H.C.F. of 60 and 84 = 2 × 2 × 3 = 12.

Since P = 36 and Q = 12, thus the L.C.M. of P and Q = L.C.M. of 36 and 12 = 36.

Hence, Option 4 is the correct option.

Among any three consecutive numbers, at least one is divisible by 2 and exactly one is divisible by 3. So their product is always divisible by 2 × 3 = 6.

It need not be divisible by 4, 12 or 24; for example, 1 × 2 × 3 = 6 is divisible by 6 but not by 4, 12 or 24.

Since the product always contains the factors 2 and 3, thus the product of three consecutive numbers is always divisible by 6.

Hence, Option 2 is the correct option.

The least number that can be divided exactly into groups of 2, 3, 5 and 6 is the L.C.M. of these numbers.

The L.C.M. of 2, 3, 5 and 6 = 30.

Since the least such number is the L.C.M. of the group sizes, thus the least number of students the class can have = 30.

Hence, Option 2 is the correct option.

For 373a to be divisible by 9, the sum of its digits 3 + 7 + 3 + a = 13 + a must be divisible by 9. This gives a = 5.

For 473b to be divisible by 11, the difference between the sum of digits at odd places and even places, (b + 7) - (3 + 4) = b, must be 0 or a multiple of 11. This gives b = 0.

For 371c to be divisible by 6, it must be divisible by both 2 and 3. For divisibility by 2, c must be even; for divisibility by 3, the digit sum 3 + 7 + 1 + c = 11 + c must be divisible by 3. The even digit satisfying this is c = 4.

Since a = 5, b = 0 and c = 4, thus a + b + c = 5 + 0 + 4 = 9.

Hence, Option 4 is the correct option.

Statement-1 is the correct definition of a perfect number. For example, the factors of 6 are 1, 2, 3 and 6, and their sum 1 + 2 + 3 + 6 = 12 = 2 × 6. So, statement-1 is true.

The product of two prime numbers is not always odd; for example, 2 × 3 = 6 is even. So, statement-2 is false.

∴ Statement-1 is true but Statement-2 is false.

Hence, Option 2 is the correct option.