Playing With Numbers

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

28 - 3 × 8 ÷ 6

= 28 - 3 × $\frac{8}{6}$

= 28 - $\frac{24}{6}$

= 28 - 4

= 24

Hence, 28 - 3 × 8 ÷ 6 = 24

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

[(4 × 2) - (4 ÷ 2)] + 8

= [8 - $\frac{4}{2}$] + 8

= [8 - 2] + 8

= 6 + 8

= 14

Hence, [(4 × 2) - (4 ÷ 2)] + 8 = 14

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

15 × 12 ÷ (5 - 2)

= 15 × 12 ÷ 3

= 15 × $\frac{12}{3}$

= 15 × 4

= 60

Hence, 15 × 12 ÷ (5 - 2) = 60

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

32 + 48 ÷ 12 - 3 × 7

= 32 + $\frac{48}{12}$ - 3 × 7

= 32 + 4 - 21

= 36 - 21

= 15

Hence, 32 + 48 ÷ 12 - 3 × 7 = 15

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

16 - 4 of 5 ÷ 10 × 2

= 16 - 20 ÷ 10 × 2

= 16 - $\frac{20}{10}$ × 2

= 16 - 2 × 2

= 16 - 4

= 12

Hence, 16 - 4 of 5 ÷ 10 × 2 = 12

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

24 of 36 ÷ 9 - 13 × 4

= 864 ÷ 9 - 13 × 4

= $\frac{864}{9}$ - 13 × 4

= 96 - 52

= 44

Hence, 24 of 36 ÷ 9 - 13 × 4 = 44

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

19 - [18 - {12 - (7 - 5)}]

= 19 - [18 - {12 - 2}]

= 19 - [18 - 10]

= 19 - 8

= 11

Hence, 19 - [18 - {12 - (7 - 5)}] = 11

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

17 - [14 - {40 + 8 ÷ (7 - $\overline{6 - 3}$)}]

= 17 - [14 - {40 + 8 ÷ (7 - 3)}]

= 17 - [14 - {40 + 8 ÷ 4}]

= 17 - [14 - {40 + $\frac{8}{4}$}]

= 17 - [14 - {40 + 2}]

= 17 - [14 - 42]

= 17 - (-28)

= 17 + 28

= 45

Hence, 17 - [14 - {40 + 8 ÷ (7 - $\overline{6 - 3}$)}] = 45

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

25 - [12 - {5 + 18 ÷ (4 - $\overline{5 - 3}$)}]

= 25 - [12 - {5 + 18 ÷ (4 - 2)}]

= 25 - [12 - {5 + 18 ÷ 2}]

= 25 - [12 - {5 + $\frac{18}{2}$}]

= 25 - [12 - {5 + 9}]

= 25 - [12 - 14]

= 25 - (-2)

= 25 + 2

= 27

Hence, 25 - [12 - {5 + 18 ÷ (4 - $\overline{5 - 3}$)}] = 27

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

15 - [16 - {12 + 21 ÷ (9 - 2)}]

= 15 - [16 - {12 + 21 ÷ 7}]

= 15 - [16 - {12 + $\frac{21}{7}$}]

= 15 - [16 - {12 + 3}]

= 15 - [16 - 15]

= 15 - 1

= 14

Hence, 15 - [16 - {12 + 21 ÷ (9 - 2)}] = 14

(i) Factors of 15 = 1, 3, 5 and 15

(ii) Factors of 55 = 1, 5, 11 and 55

(iii) Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48

(iv) Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36

A prime number is a natural number that is divisible by 1 and itself only.

(i) Prime numbers less than 25 = 2, 3, 5, 7, 11, 13, 17, 19 and 23

(ii) Prime numbers between 15 and 35 = 17, 19, 23, 29 and 31

(iii) Prime numbers between 8 and 76 = 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71 and 73

A prime number is a natural number that is divisible by 1 and itself only.

(i) Prime numbers from 5 to 45 = 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 and 43

(ii) Prime numbers from 2 to 32 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31

(iii) Prime numbers from 8 to 48 = 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47

(i) 16 = 2 × 2 × 2 × 2

∴ The prime factor of 16 is 2.

(ii) 35 = 5 × 7

∴ The prime factors of 35 are 5 and 7.

(iii) 49 = 7 × 7

∴ The prime factor of 49 is 7.

Factors of 25, i.e. F₂₅ = 1, 5 and 25

F₂₀ = 1, 2, 4, 5, 10 and 20

Now, Factors that are common to F₂₅ and F₂₀,

= 1 and 5

From the common factors, the highest common factor = 5

⇒ H.C.F. of 25 and 20 = 5

F₈ = 1, 2, 4 and 8

F₁₂ = 1, 2, 3, 4, 6 and 12

F₁₈ = 1, 2, 3, 6, 9 and 18

Now factors common to F₈, F₁₂ and F₁₈,

= 1 and 2

From the common factors, the highest common factor = 2

⇒ H.C.F. of 8, 12 and 18 = 2

F₂₄ = 1, 2, 3, 4, 6, 8, 12 and 24

F₃₆ = 1, 2, 3, 4, 6, 9, 12, 18 and 36

F₄₅ = 1, 3, 5, 9, 15 and 45

F₆₀ = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60

Factors common to F₂₄, F₃₆, F₄₅ and F₆₀,

= 1 and 3

From the common factors, the highest common factor = 3

⇒ H.C.F. of 24, 36, 45 and 60 = 3

Splitting each number into its prime factors :

40 = 2 × 2 × 2 × 5

60 = 2 × 2 × 3 × 5

80 = 2 × 2 × 2 × 2 × 5

The prime factors common to the given numbers are 2, 2 and 5.

⇒ H.C.F. of 40, 60 and 80 = 2 × 2 × 5 = 20

Splitting each number into its prime factors :

48 = 2 × 2 × 2 × 2 × 3

84 = 2 × 2 × 3 × 7

88 = 2 × 2 × 2 × 11

The prime factors common to the given numbers are 2 and 2.

⇒ H.C.F. of 48, 84 and 88 = 2 × 2 = 4

Splitting each number into its prime factors :

12 = 2 × 2 × 3

16 = 2 × 2 × 2 × 2

28 = 2 × 2 × 7

The prime factors common to the given numbers are 2 and 2.

⇒ H.C.F. of 12, 16 and 28 = 2 × 2 = 4

By division method,

$$\begin{array}{ll} 16\overline{\smash{\big)}24\smash{\big(}} 1 & \text{[Dividing the bigger number 24 by the smaller number 16]} \\ \phantom{)}\phantom{)}\underline{-16} & \\ \phantom{++} 8 \overline{\smash{\big)} 16\smash{\big(}} 2 & \text{[Dividing the divisor 16 by the remainder 8]} \\ \phantom{++)}\phantom{)}\underline{-16} & \\ \phantom{+++(} \phantom{} \times & \text{[No remainder is left, so the last divisor 8 is the H.C.F.]} \\ \end{array}$$

Hence, H.C.F. of 16 and 24 = 8

Let us first find the H.C.F. of 7 and 14.

$$\begin{array}{ll} 7\overline{\smash{\big)}14\smash{\big(}} 2 & \text{[Dividing the bigger number 14 by the smaller number 7]} \\ \phantom{}\phantom{.} \underline{-14} & \\ \phantom{+} \phantom{}\phantom{.} \times & \text{[No remainder is left, so the last divisor 7 is the H.C.F.]} \\ \end{array}$$

H.C.F. of 7 and 14 = 7

Since, the third number is 24 and the H.C.F. obtained above is 7, find the H.C.F. of 24 and 7.

$$\begin{array}{ll} 7\overline{\smash{\big)}24\smash{\big(}} 3 & \text{[Dividing 24 by the H.C.F. 7 obtained above]} \\ \phantom{}\phantom{.}\underline{-21} & \\ \phantom{+)} \phantom{0} 3 \overline{\smash{\big)} 7\smash{\big(}} 2 & \text{[Dividing the divisor 7 by the remainder 3]} \\ \phantom{+)}\phantom{.0}\underline{-6} & \\ \phantom{++} \phantom{.0} 1 \overline{\smash{\big)} 3\smash{\big(}} 3 & \text{[Dividing the divisor 3 by the remainder 1]} \\ \phantom{++)}\phantom{)}\phantom{.0}\underline{-3} & \\ \phantom{+++(} \phantom{0} \times & \text{[No remainder is left, so the last divisor 1 is the H.C.F.]} \\ \end{array}$$

Hence, H.C.F. of 7, 14 and 24 = 1

Let us first find the H.C.F. of 32 and 56.

$$\begin{array}{ll} 32\overline{\smash{\big)}56\smash{\big(}} 1 & \text{[Dividing the bigger number 56 by the smaller number 32]} \\ \phantom{)}\phantom{)}\underline{-32} & \\ \phantom{+()} 24 \overline{\smash{\big)} 32\smash{\big(}} 1 & \text{[Dividing the divisor 32 by the remainder 24]} \\ \phantom{++)}\phantom{)}\underline{-24} & \\ \phantom{+++()} 8 \overline{\smash{\big)} 24\smash{\big(}} 3 & \text{[Dividing the divisor 24 by the remainder 8]} \\ \phantom{+++)}\phantom{)}\underline{-24} & \\ \phantom{++++(} \phantom{} \times & \text{[No remainder is left, so the last divisor 8 is the H.C.F.]} \\ \end{array}$$

H.C.F. of 32 and 56 = 8

Since, the third number is 46 and the H.C.F. obtained above is 8, find the H.C.F. of 46 and 8.

$$\begin{array}{ll} 8\overline{\smash{\big)}46\smash{\big(}} 5 & \text{[Dividing 46 by the H.C.F. 8 obtained above]} \\ \phantom{}\phantom{)}\underline{-40} & \\ \phantom{++} 6 \overline{\smash{\big)} 8\smash{\big(}} 1 & \text{[Dividing the divisor 8 by the remainder 6]} \\ \phantom{++)}\phantom{)}\underline{-6} & \\ \phantom{+++)} 2 \overline{\smash{\big)} 6\smash{\big(}} 3 & \text{[Dividing the divisor 6 by the remainder 2]} \\ \phantom{+++)}\phantom{)}\underline{-6} & \\ \phantom{++++(} \phantom{} \times & \text{[No remainder is left, so the last divisor 2 is the H.C.F.]} \\ \end{array}$$

Hence, H.C.F. of 32, 56 and 46 = 2

Splitting each number into its prime factors :

45 = 3 × 3 × 5

75 = 3 × 5 × 5

135 = 3 × 3 × 3 × 5

The prime factors common to the given numbers are 3 and 5.

⇒ H.C.F. of 45, 75 and 135 = 3 × 5 = 15

Splitting each number into its prime factors :

66 = 2 × 3 × 11

33 = 3 × 11

132 = 2 × 2 × 3 × 11

The prime factors common to the given numbers are 3 and 11.

⇒ H.C.F. of 66, 33 and 132 = 3 × 11 = 33

Splitting each number into its prime factors :

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

60 = 2 × 2 × 3 × 5

132 = 2 × 2 × 3 × 11

The prime factors common to the given numbers are 2, 2 and 3.

⇒ H.C.F. of 24, 36, 60 and 132 = 2 × 2 × 3 = 12

The greatest number that divides each of the given numbers completely is their H.C.F.

Splitting each number into its prime factors :

180 = 2 × 2 × 3 × 3 × 5

225 = 3 × 3 × 5 × 5

315 = 3 × 3 × 5 × 7

The prime factors common to the given numbers are 3, 3 and 5.

⇒ H.C.F. = 3 × 3 × 5 = 45

Hence, the greatest number that divides 180, 225 and 315 completely is 45.

Two numbers are said to be co-prime if their H.C.F. is 1.

Splitting each number into its prime factors :

45 = 3 × 3 × 5

56 = 2 × 2 × 2 × 7

Since, there is no prime factor common to 45 and 56,

⇒ H.C.F. of 45 and 56 = 1

Hence, 45 and 56 are co-prime numbers.

Multiples of 8 = 8, 16, 24, 32, 40, 48, ......

Multiples of 12 = 12, 24, 36, 48, ......

Multiples of 24 = 24, 48, 72, ......

The least common multiple = 24

Hence, L.C.M. of 8, 12 and 24 = 24

Multiples of 10 = 10, 20, 30, 40, 50, 60, 70, ......

Multiples of 15 = 15, 30, 45, 60, 75, ......

Multiples of 20 = 20, 40, 60, 80, ......

The least common multiple = 60

Hence, L.C.M. of 10, 15 and 20 = 60

Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, ......

Multiples of 6 = 6, 12, 18, 24, 30, 36, ......

Multiples of 9 = 9, 18, 27, 36, 45, ......

Multiples of 12 = 12, 24, 36, 48, ......

The least common multiple = 36

Hence, L.C.M. of 3, 6, 9 and 12 = 36

Prime factor method :

18 = 2 × 3 × 3 = 2 × 3²

24 = 2 × 2 × 2 × 3 = 2³ × 3

96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

L.C.M. = Product of all the prime factors with the highest power of each

= 2⁵ × 3² = 32 × 9 = 288

Common division method :

$$\begin{array}{l|rrr} 2 & 18, & 24, & 96 \\ \hline 2 & 9, & 12, & 48 \\ \hline 2 & 9, & 6, & 24 \\ \hline 2 & 9, & 3, & 12 \\ \hline 2 & 9, & 3, & 6 \\ \hline 3 & 9, & 3, & 3 \\ \hline 3 & 3, & 1, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

⇒ L.C.M. of 18, 24 and 96 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

Prime factor method :

14 = 2 × 7

21 = 3 × 7

98 = 2 × 7 × 7 = 2 × 7²

L.C.M. = 2 × 3 × 7² = 2 × 3 × 49 = 294

Common division method :

$$\begin{array}{l|rrr} 2 & 14, & 21, & 98 \\ \hline 3 & 7, & 21, & 49 \\ \hline 7 & 7, & 7, & 49 \\ \hline 7 & 1, & 1, & 7 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

⇒ L.C.M. of 14, 21 and 98 = 2 × 3 × 7 × 7 = 294

Prime factor method :

34 = 2 × 17

85 = 5 × 17

51 = 3 × 17

L.C.M. = 2 × 3 × 5 × 17 = 510

Common division method :

$$\begin{array}{l|rrr} 2 & 34, & 85, & 51 \\ \hline 3 & 17, & 85, & 51 \\ \hline 5 & 17, & 85, & 17 \\ \hline 17 & 17, & 17, & 17 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

⇒ L.C.M. of 34, 85 and 51 = 2 × 3 × 5 × 17 = 510

The least number which when divided by the given numbers leaves no remainder is their L.C.M.

$$\begin{array}{l|rrrr} 2 & 15, & 25, & 40, & 50 \\ \hline 2 & 15, & 25, & 20, & 25 \\ \hline 2 & 15, & 25, & 10, & 25 \\ \hline 3 & 15, & 25, & 5, & 25 \\ \hline 5 & 5, & 25, & 5, & 25 \\ \hline 5 & 1, & 5, & 1, & 5 \\ \hline & 1, & 1, & 1, & 1 \\ \end{array}$$

L.C.M. = 2 × 2 × 2 × 3 × 5 × 5 = 600

Hence, the required least number is 600.

First we find the L.C.M. of 18, 36 and 48.

$$\begin{array}{l|rrr} 2 & 18, & 36, & 48 \\ \hline 2 & 9, & 18, & 24 \\ \hline 2 & 9, & 9, & 12 \\ \hline 2 & 9, & 9, & 6 \\ \hline 3 & 9, & 9, & 3 \\ \hline 3 & 3, & 3, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

L.C.M. = 2 × 2 × 2 × 2 × 3 × 3 = 144

The smallest number divisible by 18, 36 and 48 is 144.

∴ The smallest number which leaves a remainder of 7 = 144 + 7 = 151

Hence, the required number is 151.

First we find the L.C.M. of 16, 24, 40 and 45.

$$\begin{array}{l|rrrr} 2 & 16, & 24, & 40, & 45 \\ \hline 2 & 8, & 12, & 20, & 45 \\ \hline 2 & 4, & 6, & 10, & 45 \\ \hline 2 & 2, & 3, & 5, & 45 \\ \hline 3 & 1, & 3, & 5, & 45 \\ \hline 3 & 1, & 1, & 5, & 15 \\ \hline 5 & 1, & 1, & 5, & 5 \\ \hline & 1, & 1, & 1, & 1 \\ \end{array}$$

L.C.M. = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

The smallest number completely divisible by 16, 24, 40 and 45 is 720.

∴ The required number = 720 - 11 = 709

Hence, the required number is 709.

Splitting each number into its prime factors :

24 = 2 × 2 × 2 × 3

30 = 2 × 3 × 5

H.C.F. = product of common prime factors = 2 × 3 = 6

L.C.M. = 2 × 2 × 2 × 3 × 5 = 120

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

$\frac{\text{L.C.M.}}{\text{H.C.F.}} = \frac{120}{6} = 20$

Since, the remainder is 0,

Hence, L.C.M. = 120, H.C.F. = 6 and the L.C.M. is completely divisible by the H.C.F.

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ First number × Second number = H.C.F. × L.C.M.

⇒ 150 × Second number = 50 × 300

⇒ Second number = $\frac{50 \times 300}{150}$

⇒ Second number = $\frac{15000}{150}$

⇒ Second number = 100

Hence, the other number is 100.

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ 432 = H.C.F. × 72

⇒ H.C.F. = $\frac{432}{72}$

⇒ H.C.F. = 6

Hence, the H.C.F. of the two numbers is 6.

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ 19200 = 40 × L.C.M.

⇒ L.C.M. = $\frac{19200}{40}$

⇒ L.C.M. = 480

Hence, the L.C.M. of the two numbers is 480.

The smallest number which when divided by the given numbers leaves no remainder is their L.C.M.

$$\begin{array}{l|rrrrr} 2 & 12, & 15, & 18, & 24, & 36 \\ \hline 2 & 6, & 15, & 9, & 12, & 18 \\ \hline 2 & 3, & 15, & 9, & 6, & 9 \\ \hline 3 & 3, & 15, & 9, & 3, & 9 \\ \hline 3 & 1, & 5, & 3, & 1, & 3 \\ \hline 5 & 1, & 5, & 1, & 1, & 1 \\ \hline & 1, & 1, & 1, & 1, & 1 \\ \end{array}$$

L.C.M. = 2 × 2 × 2 × 3 × 3 × 5 = 360

Hence, the required smallest number is 360.

According to the divisibility of 2, if the unit digit of the number is zero or an even number then the number is divisible by 2.

(i) 352 — Last digit is 2 (even number).

(ii) 523 — Last digit is 3 (odd number).

(iii) 496 — Last digit is 6 (even number).

Hence, 352 and 496 are divisible by 2.

According to the divisibility of 4, if the two digit number formed by its ten's digit and unit digit is divisible by 4 then the number is divisible by 4.

(i) 222 — Last 2 digits are 22 and 22 is not divisible by 4.

(ii) 532 — Last 2 digits are 32 and 32 is divisible by 4.

(iii) 678 — Last 2 digits are 78 and 78 is not divisible by 4.

Hence, 532 is divisible by 4.

According to the divisibility of 8, if the number formed by its hundred's, ten's and unit digits is divisible by 8 then the number is divisible by 8.

(i) 324 — Last 3 digits are 324 and 324 is not divisible by 8.

(ii) 2536 — Last 3 digits are 536 and 536 is divisible by 8.

(iii) 92760 — Last 3 digits are 760 and 760 is divisible by 8.

Hence, 2536 and 92760 are divisible by 8.

According to the divisibility of 3, if the sum of the digits of the number is divisible by 3 then the number is divisible by 3.

(i) 221 = 2 + 2 + 1 = 5. And 5 is not divisible by 3.

(ii) 543 = 5 + 4 + 3 = 12. And 12 is divisible by 3.

(iii) 28492 = 2 + 8 + 4 + 9 + 2 = 25. And 25 is not divisible by 3.

Hence, 543 is divisible by 3.

According to the divisibility of 9, if the sum of the digits of the number is divisible by 9 then the number is divisible by 9.

(i) 1332 = 1 + 3 + 3 + 2 = 9. And 9 is divisible by 9.

(ii) 53247 = 5 + 3 + 2 + 4 + 7 = 21. And 21 is not divisible by 9.

(iii) 4968 = 4 + 9 + 6 + 8 = 27. And 27 is divisible by 9.

Hence, 1332 and 4968 are divisible by 9.

According to the divisibility of 6, if the number is divisible by 2 and 3 both then the number is divisible by 6.

(i) 324

Last digit is 4, hence 324 is divisible by 2.

3 + 2 + 4 = 9, 9 is divisible by 3. So, 324 is divisible by 3.

Hence, 324 is divisible by 6.

(ii) 2010

Last digit is 0, hence 2010 is divisible by 2.

2 + 0 + 1 + 0 = 3, 3 is divisible by 3. So, 2010 is divisible by 3.

Hence, 2010 is divisible by 6.

(iii) 33278

Last digit is 8, hence 33278 is divisible by 2.

3 + 3 + 2 + 7 + 8 = 23, 23 is not divisible by 3. So, 33278 is not divisible by 3.

Hence, 33278 is not divisible by 6.

Hence, 324 and 2010 are divisible by 6.

According to the divisibility of 5, if the unit digit is 0 or 5 then the number is divisible by 5.

(i) 5080 — Last digit is 0.

(ii) 66666 — Last digit is 6.

(iii) 755 — Last digit is 5.

Hence, 5080 and 755 are divisible by 5.

According to the divisibility of 10, if the unit digit is 0 then the number is divisible by 10.

(i) 9990 — Last digit is 0.

(ii) 0 — is divisible by 10.

(iii) 847 — Last digit is 7.

Hence, 9990 and 0 are divisible by 10.

According to the divisibility of 11, if the difference between the sum of its digits in odd places and the sum of its digits in even places is either 0 or divisible by 11 then the number is divisible by 11.

(i) 5918 — the sum of its digits in odd places = 8 + 9 = 17 the sum of its digits in even places = 1 + 5 = 6 The difference = 17 - 6 = 11 (divisible by 11).

(ii) 68717 — the sum of its digits in odd places = 7 + 7 + 6 = 20 the sum of its digits in even places = 1 + 8 = 9 The difference = 20 - 9 = 11 (divisible by 11).

(iii) 3882 — the sum of its digits in odd places = 2 + 8 = 10 the sum of its digits in even places = 8 + 3 = 11 The difference = 10 - 11 = -1 (not divisible by 11).

Hence, 5918 and 68717 are divisible by 11.

According to the divisibility of 15, if the number is divisible by 3 and 5 both then the number is divisible by 15.

(i) 960

Last digit is 0, hence 960 is divisible by 5.

9 + 6 + 0 = 15, 15 is divisible by 3. So, 960 is divisible by 3.

Hence, 960 is divisible by 15.

(ii) 8295

Last digit is 5, hence 8295 is divisible by 5.

8 + 2 + 9 + 5 = 24, 24 is divisible by 3. So, 8295 is divisible by 3.

Hence, 8295 is divisible by 15.

(iii) 10243

Last digit is 3, hence 10243 is not divisible by 5.

Hence, 10243 is not divisible by 15.

Hence, 960 and 8295 are divisible by 15.

The smallest number that is completely divisible by 28 and 42 is their L.C.M.

$$\begin{array}{l|rr} 2 & 28, & 42 \\ \hline 2 & 14, & 21 \\ \hline 3 & 7, & 21 \\ \hline 7 & 7, & 7 \\ \hline & 1, & 1 \end{array}$$

L.C.M. = 2 × 2 × 3 × 7 = 84

Hence, the smallest number completely divisible by 28 and 42 is 84.

The largest number that can divide 28 and 42 completely is their H.C.F.

28 = 2 × 2 × 7

42 = 2 × 3 × 7

The prime factors common to both numbers are 2 and 7.

⇒ H.C.F. = 2 × 7 = 14

Hence, the largest number that can divide 28 and 42 completely is 14.

Let the two numbers each divisible by 8 be 16 and 24.

(i) Sum of the numbers = 16 + 24 = 40

40 ÷ 8 = 5

Since, 40 is completely divisible by 8, the sum is also divisible by 8.

Yes, the sum is divisible by 8.

(ii) Difference of the numbers = 24 - 16 = 8

8 ÷ 8 = 1

Since, 8 is completely divisible by 8, the difference is also divisible by 8.

Yes, the difference is divisible by 8.

(i) The H.C.F. of two consecutive numbers is 1. For example, H.C.F. of 7 and 8 = 1.

(ii) The H.C.F. of two consecutive even numbers is 2. For example, H.C.F. of 8 and 10 = 2.

(iii) The H.C.F. of two consecutive odd numbers is 1. For example, H.C.F. of 7 and 9 = 1.

First we find the L.C.M. of 6, 8 and 12.

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

L.C.M. = 2 × 2 × 2 × 3 = 24

The multiples of 24 are 24, 48, 72, 96, 120, ......

The smallest 3-digit multiple of 24 is 120.

Hence, the smallest 3-digit number exactly divisible by 6, 8 and 12 is 120.

First we find the L.C.M. of 6, 8 and 12.

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

The L.C.M. of 6, 8 and 12 = 24

The greatest 3-digit number is 999.

On dividing 999 by 24, we get :

$$\begin{array}{l} 24\overline{\smash{\big)}999\smash{\big(}} 41 \\ \phantom{)}\phantom{)}\underline{-96} \\ \phantom{++} 39 \\ \phantom{+)}\underline{-24} \\ \phantom{++} 15 \\ \end{array}$$

Remainder = 15

∴ The greatest 3-digit number divisible by 24 = 999 - 15 = 984

Hence, the greatest 3-digit number exactly divisible by 6, 8 and 12 is 984.

$$\begin{array}{l|rr} 2 & 140, & 168 \\ \hline 2 & 70, & 84 \\ \hline 2 & 35, & 42 \\ \hline 3 & 35, & 21 \\ \hline 5 & 35, & 7 \\ \hline 7 & 7, & 7 \\ \hline & 1, & 1 \\ \end{array}$$

L.C.M. = 2 × 2 × 2 × 3 × 5 × 7 = 840

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ 140 × 168 = H.C.F. × 840

⇒ H.C.F. = $\frac{140 \times 168}{840}$

⇒ H.C.F. = $\frac{23520}{840}$

⇒ H.C.F. = 28

Hence, L.C.M. = 840 and H.C.F. = 28.

Splitting each number into its prime factors :

108 = 2 × 2 × 3 × 3 × 3

450 = 2 × 3 × 3 × 5 × 5

The prime factors common to both numbers are 2, 3 and 3.

⇒ H.C.F. = 2 × 3 × 3 = 18

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ 108 × 450 = 18 × L.C.M.

⇒ L.C.M. = $\frac{108 \times 450}{18}$

⇒ L.C.M. = $\frac{48600}{18}$

⇒ L.C.M. = 2700

Hence, H.C.F. = 18 and L.C.M. = 2700.

Let the two numbers be 12 and 18.

Splitting each number into its prime factors :

12 = 2 × 2 × 3

18 = 2 × 3 × 3

H.C.F. = product of common prime factors = 2 × 3 = 6

L.C.M. = 2 × 2 × 3 × 3 = 36

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

$\frac{\text{L.C.M.}}{\text{H.C.F.}} = \frac{36}{6} = 6$

Since, the remainder is 0,

Hence, the L.C.M. (36) is completely divisible by the H.C.F. (6).

Splitting each number into its prime factors :

16 = 2 × 2 × 2 × 2

32 = 2 × 2 × 2 × 2 × 2

96 = 2 × 2 × 2 × 2 × 2 × 3

H.C.F. = product of common prime factors = 2 × 2 × 2 × 2 = 16

L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 = 96

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

$\frac{\text{L.C.M.}}{\text{H.C.F.}} = \frac{96}{16} = 6$

Since, the remainder is 0,

Hence, L.C.M. = 96, H.C.F. = 16 and the L.C.M. of 16, 32 and 96 is divisible by their H.C.F.

A number divisible by both 5 and 8 must be divisible by their L.C.M.

Since 5 and 8 are co-prime, L.C.M. of 5 and 8 = 5 × 8 = 40.

Hence, option 3 is the correct option.

Every number (other than 0) has a finite number of factors but an infinite number of multiples.

Hence, option 3 is the correct option.

By using BODMAS, we will solve in this order : Bracket, of, Division, Multiplication, Addition, Subtraction.

18 - (15 ÷ $\overline{12 - 9}$)

= 18 - (15 ÷ 3)

= 18 - $\frac{15}{3}$

= 18 - 5

= 13

Hence, option 1 is the correct option.

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

Factors of 75 = 1, 3, 5, 15, 25 and 75

Common factors of 60 and 75 = 1, 3, 5 and 15

The number of common factors = 4

Hence, option 3 is the correct option.

Only the number 1 has exactly one factor, i.e. itself.

So, only 1 natural number has exactly one factor.

Hence, option 2 is the correct option.

According to the order of operations (BODMAS), the operation inside the brackets should be performed first.

So, 16 ÷ 4 should be performed first.

Hence, option 2 is the correct option.

12 and 17 have no common factor other than 1, so they are co-prime.

⇒ H.C.F. of 12 and 17 = 1

⇒ L.C.M. of 12 and 17 = 12 × 17

Hence, option 4 is the correct option.

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ Product of two numbers = 4 × 56 = 56 × 4

Hence, option 3 is the correct option.

2 = 2

6 = 2 × 3

H.C.F. = 2

L.C.M. = 2 × 3 = 6

Hence, option 3 is the correct option.

7 and 15 have no common factor other than 1, so they are co-prime.

⇒ H.C.F. of 7 and 15 = 1

⇒ L.C.M. of 7 and 15 = 7 × 15 = 105

Hence, option 1 is the correct option.

For 4048, the sum of digits in odd places = 8 + 0 = 8 and the sum of digits in even places = 4 + 4 = 8.

The difference = (8 + 0) - (4 + 4) = 8 - 8 = 0.

Since the difference is 0, 4048 is divisible by 11.

So, statement 1 is true.

Also, (8 + 0) - (4 + 4) = 8 - 8 = 0.

So, statement 2 is true.

Hence, option 1 is the correct option.

6 ÷ 8 and 12 ÷ 8 do not give whole numbers, but 216 ÷ 8 = 27.

So, out of 6, 12 and 216, only 216 is a multiple of 8.

So, statement 1 is true.

Every non-zero number has an infinite number of multiples, not a finite number.

So, statement 2 is false.

Hence, option 3 is the correct option.

A number is divisible by 22 if it is divisible by both 2 and 11.

1848 has 8 (an even number) at its unit place, so it is divisible by 2.

For 1848, the sum of digits in odd places = 8 + 8 = 16 and the sum of digits in even places = 4 + 1 = 5.

The difference = (8 + 8) - (4 + 1) = 16 - 5 = 11, which is divisible by 11.

So, 1848 is divisible by 11, and hence by 22.

So, assertion (A) is true.

Also, (8 + 8) - (4 + 1) = 11, which is divisible by 11.

So, reason (R) is true.

Hence, option 3 is the correct option.

10 = 2 × 5 and 11 = 11.

Since 10 and 11 have no common factor other than 1, their greatest common divisor (H.C.F.) is 1.

So, 10 and 11 are co-prime numbers.

So, assertion (A) is true.

Two numbers are co-prime if their greatest common divisor is 1.

So, reason (R) is true.

Hence, option 3 is the correct option.