Playing with Numbers

(i) 89 = 10 × 8 + 9

(ii) 207 = 100 × 2 + 10 × 0 + 7

(iii) 369 = 100 × 3 + 10 × 6 + 9

Given number is 34.

Number obtained by reversing the digits is 43.

Sum = 34 + 43 = 77 = 11 × 7

(i) According to the property, when ab + ba is divided by 11, the quotient = a + b.

Hence, when 34 + 43 is divided by 11, the quotient = 3 + 4 = 7.

(ii) According to the property, when ab + ba is divided by a + b, the quotient = 11.

Hence, when 34 + 43 is divided by 7 (3 + 4 = 7), the quotient = 11.

Given number is 73.

Number obtained by reversing the digits is 37.

Difference = 73 - 37 = 36 = 9 × 4

(i) According to the property, when ab - ba is divided by 9, the quotient = a - b.

Hence, when 73 - 37 is divided by 9, the quotient = 7 - 3 = 4.

(ii) According to the property, when ab - ba is divided by a - b, the quotient = 9.

Hence, when 73 - 37 is divided by 4 (7 - 3 = 4), the quotient = 9.

⇒ abc + bca + cab

⇒ (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)

⇒ (100a + 10a + a) + (10b + 100b + b) + (c + 10c + 100c)

⇒ 111a + 111b + 111c

⇒ 111(a + b + c)

⇒ 3 × 37 × (a + b + c)

(i) When the sum is divided by 111, the quotient = (a + b + c).

(ii) When the sum is divided by (a + b + c), the quotient = 111.

(iii) Since 111(a + b + c) = 37 × 3(a + b + c), when the sum is divided by 37, the quotient = 3(a + b + c).

(iv) Since 111(a + b + c) = 3 × 37(a + b + c), when the sum is divided by 3, the quotient = 37(a + b + c).

Given number is 843.

Number obtained by reversing the digits is 348.

∴ Difference = 843 - 348 = 495 = 99 × 5

(i) When the difference is divided by 99, the quotient = 5.

(ii) When the difference is divided by 5, the quotient = 99.

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Given, sum of digits = 11

⇒ a + b = 11 .....(1)

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the number obtained by reversing the digits is 9 less than the original number.

⇒ (10a + b) - (10b + a) = 9

⇒ 9a - 9b = 9

⇒ 9(a - b) = 9

⇒ a - b = 1 .....(2)

Adding (1) and (2), we get

2a = 12

⇒ a = 6

Putting a = 6 in (1), we get b = 11 - 6 = 5

∴ 2-digit number = 10 × 6 + 5 = 65

Hence, the required 2-digit number is 65.

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the difference of the 2-digit number and the number obtained by reversing the digits is 36.

⇒ (10a + b) - (10b + a) = 36

⇒ 9a - 9b = 36

⇒ 9(a - b) = 36

⇒ a - b = 4

Hence, the difference between the digits of the 2-digit number is 4.

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the sum of the 2-digit number and the number obtained by reversing the digits is 55.

⇒ (10a + b) + (10b + a) = 55

⇒ 11a + 11b = 55

⇒ 11(a + b) = 55

⇒ a + b = 5

Hence, the sum of the digits of the 2-digit number is 5.

Let the 3-digit number be a0c i.e. 100a + 10 × 0 + c = 100a + c, where the middle digit is 0.

Given, sum of the other two digits = 11

⇒ a + c = 11 .....(1)

Number obtained by reversing the digits = c0a i.e. 100c + a

Given, the number obtained by reversing the digits exceeds the original number by 495.

⇒ (100c + a) - (100a + c) = 495

⇒ 99c - 99a = 495

⇒ 99(c - a) = 495

⇒ c - a = 5 .....(2)

Adding equations (1) and (2), we get:

⇒ (a + c) + (c - a) = 11 + 5

⇒ 2c = 16

⇒ c = 8

Putting c = 8 in (1), we get a = 11 - 8 = 3

∴ 3-digit number = 100 × 3 + 10 × 0 + 8 = 308

Hence, the required number is 308.

Given, unit's digit, ten's digit and hundred's digit are in the ratio 1 : 2 : 3.

Let unit's digit = x, ten's digit = 2x and hundred's digit = 3x.

∴ 3-digit number = 100 × 3x + 10 × 2x + x = 300x + 20x + x = 321x

Number obtained by reversing the digits = 100 × x + 10 × 2x + 3x = 100x + 20x + 3x = 123x

Given, the difference of the original number and the number obtained by reversing the digits is 594.

⇒ 321x - 123x = 594

⇒ 198x = 594

⇒ x = $\frac{594}{198}$

⇒ x = 3

∴ Unit's digit = 3, ten's digit = 2 × 3 = 6 and hundred's digit = 3 × 3 = 9

∴ 3-digit number = 100 × 9 + 10 × 6 + 3 = 963

Hence, the required number is 963.

Let hundred's digit = a.

Given, unit's digit is one more than the hundred's digit and ten's digit is one less than the hundred's digit.

∴ Ten's digit = a - 1 and unit's digit = a + 1

According to the property, the sum of a 3-digit number and the numbers obtained by changing the order of digits cyclically = 111 × (sum of digits).

Given, this sum is 2664.

⇒ 111 × (sum of digits) = 2664

⇒ Sum of digits = $\frac{2664}{111}$

⇒ Sum of digits = 24

⇒ a + (a - 1) + (a + 1) = 24

⇒ 3a = 24

⇒ a = 8

∴ Hundred's digit = 8, ten's digit = 8 - 1 = 7 and unit's digit = 8 + 1 = 9

∴ 3-digit number = 100 × 8 + 10 × 7 + 9 = 879

Hence, the required number is 879.

Given,

$$\begin{matrix} & 4 & \text{A} \\ + & 3 & 5 \\ \hline & \text{B} & 2 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, A + 5 is a number whose ones digit is 2.

⇒ A + 5 = 2, A + 5 = 12, A + 5 = 22; and so on.

⇒ A = 2 - 5, A = 12 - 5, A = 22 - 5; and so on.

⇒ A = -3, A = 7, A = 17; and so on.

Since, A is a digit.

⇒ A = 7.

$$\begin{matrix} & \overset{1}{4} & \text{7} \\ + & 3 & 5 \\ \hline & \text{B} & 2 \\ \hline \end{matrix}$$

Now, we will find the value of B.

1 + 4 + 3 = B

⇒ B = 8

$$\begin{matrix} & \overset{1}{4} & \text{7} \\ + & 3 & 5 \\ \hline & \text{8} & 2 \\ \hline \end{matrix}$$

Hence, A = 7 and B = 8.

Given,

$$\begin{matrix} & 5 & \text{A} \\ + & 7 & 9 \\ \hline \text{C} & \text{B} & 3 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, A + 9 is a number whose ones digit is 3.

⇒ A + 9 = 3, A + 9 = 13, A + 9 = 23; and so on.

⇒ A = 3 - 9, A = 13 - 9, A = 23 - 9; and so on.

⇒ A = -6, A = 4, A = 14; and so on.

Since, A is a digit. ∴ A = 4.

$$\begin{matrix} & \overset{1}{5} & \text{4} \\ + & 7 & 9 \\ \hline \text{C} & \text{B} & 3 \\ \hline \end{matrix}$$

Secondly, we will find the values of B and C.

1 + 5 + 7 = CB

⇒ CB = 13

⇒ B = 3 and C = 1

$$\begin{matrix} & \overset{1}{5} & \text{4} \\ + & 7 & 9 \\ \hline \text{1} & \text{3} & 3 \\ \hline \end{matrix}$$

Hence, A = 4, B = 3 and C = 1.

Given,

$$\begin{matrix} & 4 & 2 & \text{A} \\ + & 2 & \text{A} & 5 \\ \hline & \text{A} & 0 & 2 \\ \hline \end{matrix}$$

Clearly, A + 5 is a number whose ones digit is 2.

⇒ A + 5 = 2, A + 5 = 12, A + 5 = 22; and so on.

⇒ A = 2 - 5, A = 12 - 5, A = 22 - 5; and so on.

⇒ A = -3, A = 7, A = 17; and so on.

Since, A is a digit. ∴ A = 7.

Checking with A = 7:

Ones column: 7 + 5 = 12, write 2 and 1 is carried over to ten's column.

Ten's column: 2 + 7 + 1 = 10, write 0 and 1 is carried over to hundred's column.

Hundred's column: 4 + 2 + 1 = 7 = A, which is correct.

$$\begin{matrix} & \overset{1}{4} & \overset{1}{2} & \text{7} \\ + & 2 & \text{7} & 5 \\ \hline & \text{7} & 0 & 2 \\ \hline \end{matrix}$$

Hence, A = 7.

Given,

$$\begin{matrix} & \text{A} & \text{A} \\ + & \text{A} & \text{A} \\ \hline \text{B} & \text{A} & 8 \\ \hline \end{matrix}$$

Clearly, A + A is a number whose ones digit is 8.

⇒ 2A = 8 or 2A = 18

⇒ A = 4 or A = 9

Let's take A = 4:

44 + 44 = 88, which is a 2-digit number. But the sum BA8 is a 3-digit number. Hence, A ≠ 4.

Let's take A = 9:

99 + 99 = 198, which is of the form BA8 with B = 1 and A = 9.

$$\begin{matrix} & \overset{1}{\text{9}} & \text{9} \\ + & \text{9} & \text{9} \\ \hline \text{1} & \text{9} & 8 \\ \hline \end{matrix}$$

Hence, A = 9 and B = 1.

Given,

$$\begin{matrix} & 1 & 8 & \text{A} \\ + & \text{B} & \text{A} & 7 \\ \hline & \text{C} & \text{B} & 2 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, A + 7 is a number whose ones digit is 2.

⇒ A + 7 = 2, A + 7 = 12, A + 7 = 22; and so on.

⇒ A = 2 - 7, A = 12 - 7, A = 22 - 7; and so on.

⇒ A = -5, A = 5, A = 15; and so on.

Since, A is a digit. ∴ A = 5 and 1 is carried over to ten's column.

$$\begin{matrix} & 1 & \overset{1}{8} & \text{5} \\ + & \text{B} & \text{5} & 7 \\ \hline & \text{C} & \text{B} & 2 \\ \hline \end{matrix}$$

Secondly, we will find the value of B.

Ten's column: 8 + 5 + 1 = 14, write 4 and 1 is carried over to hundred's column.

⇒ B = 4

Lastly, we will find the value of C.

Hundred's column: 1 + B + 1 = 1 + 4 + 1 = 6

⇒ C = 6

$$\begin{matrix} & \overset{1}{1} & \overset{1}{8} & \text{5} \\ + & \text{4} & \text{5} & 7 \\ \hline & \text{6} & \text{4} & 2 \\ \hline \end{matrix}$$

Hence, A = 5, B = 4 and C = 6.

Given,

$$\begin{matrix} & \text{A} & 2 & 1 & \text{B} \\ + & 1 & \text{C} & \text{A} & \text{B} \\ \hline & \text{B} & 4 & 9 & 6 \\ \hline \end{matrix}$$

Firstly, we will find the value of B.

Clearly, B + B is a number whose ones digit is 6.

⇒ 2B = 6 or 2B = 16

⇒ B = 3 or B = 8

Let's take B = 3:

Ones column: 3 + 3 = 6, no carry over.

Ten's column: 1 + A is a number whose ones digit is 9 ⇒ A = 8.

Thousand's column: A + 1 = B ⇒ 8 + 1 = 9 ≠ 3. Hence, B ≠ 3.

Let's take B = 8:

Ones column: 8 + 8 = 16, write 6 and 1 is carried over to ten's column.

Ten's column: 1 + A + 1 is a number whose ones digit is 9 ⇒ A + 2 = 9 ⇒ A = 7, no carry over.

Hundred's column: 2 + C is a number whose ones digit is 4 ⇒ C = 2, no carry over.

Thousand's column: A + 1 = 7 + 1 = 8 = B, which is correct.

$$\begin{matrix} & \text{7} & 2 & \overset{1}{1} & \text{8} \\ + & 1 & \text{2} & \text{7} & \text{8} \\ \hline & \text{8} & 4 & 9 & 6 \\ \hline \end{matrix}$$

Hence, A = 7, B = 8 and C = 2.

Given,

$$\begin{matrix} & \text{B} & 3 & 4 & 5 \\ + & \text{C} & 9 & \text{B} & \text{A} \\ \hline & 8 & \text{B} & \text{A} & 2 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, 5 + A is a number whose ones digit is 2.

⇒ 5 + A = 2, 5 + A = 12, 5 + A = 22; and so on.

⇒ A = 2 - 5, A = 12 - 5, A = 22 - 5; and so on.

⇒ A = -3, A = 7, A = 17; and so on.

Since, A is a digit. ∴ A = 7 and 1 is carried over to ten's column.

Secondly, we will find the value of B.

Ten's column: 4 + B + 1 is a number whose ones digit is A i.e. 7.

⇒ B + 5 = 7 or B + 5 = 17

⇒ B = 2 or B = 12

Since, B is a digit. ∴ B = 2.

Hundred's column: 3 + 9 = 12, whose ones digit is 2 = B, which is correct, and 1 is carried over to thousand's column.

Lastly, we will find the value of C.

Thousand's column: B + C + 1 = 8

⇒ 2 + C + 1 = 8

⇒ C = 5

$$\begin{matrix} & \overset{1}{\text{2}} & 3 & \overset{1}{4} & 5 \\ + & \text{5} & 9 & \text{2} & \text{7} \\ \hline & 8 & \text{2} & \text{7} & 2 \\ \hline \end{matrix}$$

Hence, A = 7, B = 2 and C = 5.

Given,

$$\begin{matrix} & 4 & 7 & 6 & \text{A} \\ - & \text{D} & 2 & \text{B} & 9 \\ \hline & 1 & \text{C} & 8 & 9 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, A - 9 is a number whose ones digit is 9.

As it is not possible to subtract 9 from any digit A to get 9, borrowing 1 from ten's digit gives 10 + A - 9.

⇒ 10 + A - 9 = 9

⇒ A + 1 = 9

⇒ A = 8

Since, A is a digit. ∴ A = 8 and 1 is borrowed from ten's digit.

Secondly, we will find the value of B.

Ten's column: After borrowing, 6 - 1 = 5 and 5 - B is a number whose ones digit is 8.

As it is not possible to subtract B from 5 to get 8, borrowing 1 from hundred's digit gives 15 - B.

⇒ 15 - B = 8

⇒ B = 15 - 8

⇒ B = 7

Since, B is a digit. ∴ B = 7 and 1 is borrowed from hundred's digit.

Thirdly, we will find the value of C.

Hundred's column: After borrowing, 7 - 1 = 6 and 6 - 2 = 4

⇒ C = 4

Lastly, we will find the value of D.

Thousand's column: 4 - D = 1

⇒ D = 4 - 1

⇒ D = 3

$$\begin{matrix} & 4 & \overset{6}{\cancel{7}} & \overset{15}{\cancel{6}} & \overset{18}{\cancel{8}} \\ - & \text{3} & 2 & \text{7} & 9 \\ \hline & 1 & \text{4} & 8 & 9 \\ \hline \end{matrix}$$

Hence, A = 8, B = 7, C = 4 and D = 3.

Given,

$$\begin{matrix} & 2 & \text{A} \\ \times & 3 & \text{A} \\ \hline \text{B} & 7 & \text{A} \\ \hline \end{matrix}$$

Clearly, A × A is a number whose ones digit is A.

0 × 0 = 0

1 × 1 = 1

5 × 5 = 25

6 × 6 = 36

So, possible values of A = 0, 1, 5 or 6.

Let's take A = 0: 20 × 30 = 600, which is not of the form B7A.

Let's take A = 1: 21 × 31 = 651, which is not of the form B7A.

Let's take A = 5: 25 × 35 = 875, which is of the form B7A with B = 8 and A = 5.

Let's take A = 6: 26 × 36 = 936, which is not of the form B7A.

$$\begin{matrix} & 2 & \text{5} \\ \times & 3 & \text{5} \\ \hline \text{8} & 7 & \text{5} \\ \hline \end{matrix}$$

Hence, A = 5 and B = 8.

Given,

$$\begin{matrix} & \text{A} & \text{B} \\ \times & \text{A} & \text{B} \\ \hline 6 & \text{A} & \text{B} \\ \hline \end{matrix}$$

Clearly, AB × AB = 6AB, i.e. the square of the 2-digit number AB is a 3-digit number lying between 600 and 699.

Now,

24² = 576

25² = 625

26² = 676

27² = 729

So, AB can be 25 or 26.

Let's take AB = 25:

25 × 25 = 625, which is of the form 6AB with A = 2 and B = 5.

Let's take AB = 26:

26 × 26 = 676. Here, 6AB with A = 2 and B = 6 gives 626 ≠ 676. Hence, AB ≠ 26.

$$\begin{matrix} & \text{2} & \text{5} \\ \times & \text{2} & \text{5} \\ \hline 6 & \text{2} & \text{5} \\ \hline \end{matrix}$$

Hence, A = 2 and B = 5.

Given,

$$\begin{matrix} & \text{A} & \text{A} \\ \times & 4 & \text{A} \\ \hline 9 & \text{A} & 4 \\ \hline \end{matrix}$$

Clearly, A × A is a number whose ones digit is 4.

2 × 2 = 4

8 × 8 = 64

So, possible values of A = 2 or 8.

Let's take A = 2: 22 × 42 = 924, which is of the form 9A4 with A = 2.

Let's take A = 8: 88 × 48 = 4224, which is a 4-digit number, but 9A4 is a 3-digit number. Hence, A ≠ 8.

$$\begin{matrix} & \text{2} & \text{2} \\ \times & 4 & \text{2} \\ \hline 9 & \text{2} & 4 \\ \hline \end{matrix}$$

Hence, A = 2.

Given,

$$\begin{array}{l} \text{A}\overline{\smash{\big)}3\text{BC}\smash{\big(}} 5\text{D} \\ \phantom{}\phantom{)}\underline{-30} \\ \phantom{++} 5\text{C} \\ \phantom{+}\phantom{)}\underline{-5\text{C}} \\ \phantom{++} \times \\ \end{array}$$

In the first step of division, A × 5 = 30

⇒ A = 30 ÷ 5

⇒ A = 6

Now, 3B - 30 = 5

⇒ 3B = 35

⇒ B = 5

On bringing down C, we get 5C.

Since the remainder is 0, A × D = 5C

⇒ 6 × D = 5C

Now, 6 × 9 = 54

⇒ D = 9 and C = 4

Checking:

$$\begin{array}{l} 6\overline{\smash{\big)}354\smash{\big(}} 59 \\ \phantom{}\phantom{)}\underline{-30} \\ \phantom{++} 54 \\ \phantom{+}\phantom{)}\underline{-54} \\ \phantom{++} \times \\ \end{array}$$

Hence, A = 6, B = 5, C = 4 and D = 9.

Place 4 at the top vertex, 5 at the bottom left vertex and 6 at the bottom right vertex. Place 3 at the middle of the left side, 2 at the middle of the right side and 1 at the middle of the bottom side.

Then,

Left side: 4 + 3 + 5 = 12

Right side: 4 + 2 + 6 = 12

Bottom side: 5 + 1 + 6 = 12

Hence, each side of the magic triangle adds up to 12.

Using variables to represent the missing numbers and determining their values:

$$\begin{array}{|c|c|c|c|} \hline 3 & 14 & x & 0 \\ \hline 8 & r & 6 & p \\ \hline 4 & q & s & 7 \\ \hline y & t & 1 & 12 \\ \hline \end{array}$$

First complete the first row.

3 + 14 + x + 0 = 30 ⇒ x = 13

Now complete the first column.

3 + 8 + 4 + y = 30 ⇒ y = 15

Complete the fourth column.

0 + p + 7 + 12 = 30 ⇒ p = 11

Complete the diagonal starting from 0.

0 + 6 + q + 15 = 30 ⇒ q = 9

Complete the second row.

8 + r + 6 + 11 = 30 ⇒ r = 5

Complete the diagonal starting from 3.

3 + 5 + s + 12 = 30 ⇒ s = 10

Complete the fourth row.

15 + t + 1 + 12 = 30 ⇒ t = 2

Now, third row: 4 + 9 + 10 + 7 = 30, second column: 14 + 5 + 9 + 2 = 30 and third column: 13 + 6 + 10 + 1 = 30, which are all correct.

Hence, the magic square can be completed as:

$$\begin{array}{|c|c|c|c|} \hline 3 & 14 & 13 & 0 \\ \hline 8 & 5 & 6 & 11 \\ \hline 4 & 9 & 10 & 7 \\ \hline 15 & 2 & 1 & 12 \\ \hline \end{array}$$

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

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

(i) 87035 — Last digit is 5. So, it is divisible by 5 but not by 10.

(ii) 75060 — Last digit is 0. So, it is divisible by both 5 and 10.

(iii) 9685 — Last digit is 5. So, it is divisible by 5 but not by 10.

(iv) 10730 — Last digit is 0. So, it is divisible by both 5 and 10.

Hence, the numbers (i), (ii), (iii) and (iv) are divisible by 5 and the numbers (ii) and (iv) are divisible by 10.

According to the divisibility of 2, if the unit's digit of the number is 0, 2, 4, 6 or 8 then the number is divisible by 2.

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

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

(i) 67894 — Last digit is 4, so it is divisible by 2. Number formed by last two digits = 94, which is not divisible by 4. So, it is neither divisible by 4 nor by 8.

(ii) 5673244 — Last digit is 4, so it is divisible by 2. Number formed by last two digits = 44 = 4 × 11, so it is divisible by 4. Number formed by last three digits = 244, which is not divisible by 8. So, it is not divisible by 8.

(iii) 9685048 — Last digit is 8, so it is divisible by 2. Number formed by last two digits = 48 = 4 × 12, so it is divisible by 4. Number formed by last three digits = 048 = 8 × 6, so it is divisible by 8.

(iv) 6533142 — Last digit is 2, so it is divisible by 2. Number formed by last two digits = 42, which is not divisible by 4. So, it is neither divisible by 4 nor by 8.

(v) 75379 — Last digit is 9, which is an odd number. So, it is not divisible by 2, 4 or 8.

Hence, the numbers (i), (ii), (iii) and (iv) are divisible by 2, the numbers (ii) and (iii) are divisible by 4 and the number (iii) is 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.

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) 45639 — Sum of digits = 4 + 5 + 6 + 3 + 9 = 27, which is divisible by both 3 and 9.

(ii) 301248 — Sum of digits = 3 + 0 + 1 + 2 + 4 + 8 = 18, which is divisible by both 3 and 9.

(iii) 567081 — Sum of digits = 5 + 6 + 7 + 0 + 8 + 1 = 27, which is divisible by both 3 and 9.

(iv) 345903 — Sum of digits = 3 + 4 + 5 + 9 + 0 + 3 = 24, which is divisible by 3 but not by 9.

(v) 345046 — Sum of digits = 3 + 4 + 5 + 0 + 4 + 6 = 22, which is neither divisible by 3 nor by 9.

Hence, the numbers (i), (ii), (iii) and (iv) are divisible by 3 and the numbers (i), (ii) and (iii) are divisible by 9.

According to the divisibility of 11, if the difference of the sum of digits at the odd places (starting from unit's place) and the sum of the digits at even places is either 0 or divisible by 11, then the number is divisible by 11.

(i) 10835

Sum of digits at odd places = 5 + 8 + 1 = 14

Sum of digits at even places = 3 + 0 = 3

Their difference = 14 - 3 = 11, which is divisible by 11.

∴ 10835 is divisible by 11.

(ii) 380237

Sum of digits at odd places = 7 + 2 + 8 = 17

Sum of digits at even places = 3 + 0 + 3 = 6

Their difference = 17 - 6 = 11, which is divisible by 11.

∴ 380237 is divisible by 11.

(iii) 504670

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

Sum of digits at even places = 7 + 4 + 5 = 16

Their difference = 16 - 6 = 10, which is neither 0 nor divisible by 11.

∴ 504670 is not divisible by 11.

(iv) 28248

Sum of digits at odd places = 8 + 2 + 2 = 12

Sum of digits at even places = 4 + 8 = 12

Their difference = 12 - 12 = 0

∴ 28248 is divisible by 11.

Hence, the numbers (i), (ii) and (iv) are divisible by 11.

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

(i) 15414

Last digit is 4, so it is divisible by 2.

Sum of digits = 1 + 5 + 4 + 1 + 4 = 15, which is divisible by 3, so it is divisible by 3.

∴ 15414 is divisible by 6.

(ii) 213888

Last digit is 8, so it is divisible by 2.

Sum of digits = 2 + 1 + 3 + 8 + 8 + 8 = 30, which is divisible by 3, so it is divisible by 3.

∴ 213888 is divisible by 6.

(iii) 469876

Last digit is 6, so it is divisible by 2.

Sum of digits = 4 + 6 + 9 + 8 + 7 + 6 = 40, which is not divisible by 3, so it is not divisible by 3.

∴ 469876 is not divisible by 6.

Hence, the numbers (i) and (ii) are divisible by 6.

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) ∵ 34x is a multiple of 3

⇒ 3 + 4 + x is divisible by 3

⇒ 7 + x is divisible by 3

⇒ 7 + x = 0, 3, 6, 9, 12, 15, 18, ...............

⇒ x = -7, -4, -1, 2, 5, 8, 11, ...............

Since x is a digit.

Hence, x = 2, 5 or 8.

(ii) ∵ 74x5284 is a multiple of 3

⇒ 7 + 4 + x + 5 + 2 + 8 + 4 is divisible by 3

⇒ 30 + x is divisible by 3

⇒ 30 + x = 30, 33, 36, 39, 42, ...............

⇒ x = 0, 3, 6, 9, 12, ...............

Since x is a digit.

Hence, x = 0, 3, 6 or 9.

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.

∵ 42z3 is a multiple of 9

⇒ 4 + 2 + z + 3 is divisible by 9

⇒ 9 + z is divisible by 9

⇒ 9 + z = 9, 18, 27, ...............

⇒ z = 0, 9, 18, ...............

Since z is a digit. So, z = 0 or 9.

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) ∵ 49*2207 is divisible by 9

⇒ 4 + 9 + * + 2 + 2 + 0 + 7 is divisible by 9

⇒ 24 + * is divisible by 9

⇒ 24 + * = 27, 36, ...............

⇒ * = 3, 12, ...............

Since * is a digit,

∴ * is to be replaced by 3.

The number formed is 4932207, which is divisible by 9.

Hence, * is to be replaced by 3.

(ii) ∵ 5938*623 is divisible by 9

⇒ 5 + 9 + 3 + 8 + * + 6 + 2 + 3 is divisible by 9

⇒ 36 + * is divisible by 9

⇒ 36 + * = 36, 45, 54, ...............

⇒ * = 0, 9, 18, ...............

Since * is a digit, ∴ * is to be replaced by 0 or 9.

The numbers formed are 59380623 and 59389623, which are divisible by 9.

Hence, * is to be replaced by 0 or 9.

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

(i) Since the unit's digit of 97*542 is 2, so the number formed is divisible by 2.

Therefore, the number formed will be divisible by 6 if it is divisible by 3 i.e. the sum of its digits is divisible by 3.

⇒ 9 + 7 + * + 5 + 4 + 2 is divisible by 3

⇒ 27 + * is divisible by 3

⇒ 27 + * = 27, 30, 33, 36, 39, ...............

⇒ * = 0, 3, 6, 9, 12, ...............

Since * is a digit.

Hence, * is to be replaced by 0, 3, 6 or 9.

(ii) Since the unit's digit of 709*94 is 4, so the number formed is divisible by 2.

Therefore, the number formed will be divisible by 6 if it is divisible by 3 i.e. the sum of its digits is divisible by 3.

⇒ 7 + 0 + 9 + * + 9 + 4 is divisible by 3

⇒ 29 + * is divisible by 3

⇒ 29 + * = 30, 33, 36, 39, ...............

⇒ * = 1, 4, 7, 10, ...............

Since * is a digit. Hence, * is to be replaced by 1, 4 or 7.

According to the divisibility of 11, if the difference of the sum of digits at the odd places (starting from unit's place) and the sum of the digits at even places is either 0 or divisible by 11, then the number is divisible by 11.

(i) For the number 64*2456,

Sum of digits at odd places = 6 + 4 + * + 6 = 16 + *

Sum of digits at even places = 5 + 2 + 4 = 11

Their difference = 16 + * - 11 = 5 + *, which should be either 0 or divisible by 11.

⇒ 5 + * = 11, 22, ...............

⇒ * = 6, 17, ...............

Since * is a digit.

Hence, * is to be replaced by 6.

(ii) For the number 86*6194,

Sum of digits at odd places = 4 + 1 + * + 8 = 13 + *

Sum of digits at even places = 9 + 6 + 6 = 21

Their difference = 21 - (13 + *) = 8 - *, which should be either 0 or divisible by 11.

⇒ 8 - * = 0, 11, 22, ...............

⇒ * = 8, -3, -14, ...............

Since * is a digit.

Hence, * is to be replaced by 8.

(i) The sum of a 2-digit number and number obtained by reversing the digits is always divisible by 11 and sum of the digits.

(ii) The difference of a 2-digit number and number obtained by reversing the digits is always divisible by 9 and difference of the digits.

(iii) The difference of a 3-digit number abc (a > c) and number obtained by reversing the digits is always divisible by 99 and (a - c).

(iv) The next number of the series 0, 1, 1, 2, 3, 5, 8, 13, ..... is 21.

(v) General form of a 2 digit number 57 is 10 × 5 + 7.

(vi) Usual form of number 100 × 7 + 10 × 4 + 1 is 741.

(vii) If the division N ÷ 5 leaves remainder 4 and the division N ÷ 2 leaves remainder 1, then unit's digit of N must be 9.

(viii) If 213x52 is divisible by 9, then digit x is 5.

(ix) The number ab - ba where a, b are digits and a > b is divisible by 9.

(i) False

If the division N ÷ 5 leaves a remainder 4, then the unit's digit of N may be 4 or 9.

(ii) True

If the division N ÷ 2 leaves a remainder 1, then N is an odd number, so its unit's digit may be 1, 3, 5, 7 or 9.

(iii) True

If a number is divisible by any number m, then it will also be divisible by each of the factors of m.

(iv) False

For example, 5 is divisible by the sum 2 + 3 = 5, but 5 is neither divisible by 2 nor by 3.

(v) True

Sum of digits of 3651x = 3 + 6 + 5 + 1 + x = 15 + x, which is divisible by 9 when 15 + x = 18 i.e. x = 3.

(vi) False

Sum of digits of 42x7 = 4 + 2 + x + 7 = 13 + x, which is divisible by 3 when 13 + x = 15, 18, 21 i.e. x = 2, 5 or 8. So, x cannot be 4.

(vii) False

Sum of digits of 152875 = 1 + 5 + 2 + 8 + 7 + 5 = 28, which is not divisible by 9.

(viii) True

Number formed by last three digits of 389072 = 072 = 8 × 9, which is divisible by 8.

(ix) True

Last digit of 27402 is 2, so it is divisible by 2. Sum of digits = 2 + 7 + 4 + 0 + 2 = 15, which is divisible by 3, so it is divisible by 3. Hence, 27402 is divisible by 6.

(x) False

For the number 359249, sum of digits at odd places = 9 + 2 + 5 = 16 and sum of digits at even places = 4 + 9 + 3 = 16. Their difference = 16 - 16 = 0. Hence, 359249 is divisible by 11.

In a 3-digit number xyz, hundred's digit is x, ten's digit is y and unit's digit is z.

∴ xyz = 100 × x + 10 × y + z = 100x + 10y + z

Hence, option 2 is the correct option.

100a + b + 10c = 100a + 10c + b

Here, hundred's digit is a, ten's digit is c and unit's digit is b.

∴ Usual form = acb

Hence, option 4 is the correct option.

ab + ba = (10a + b) + (10b + a) = 11a + 11b = 11(a + b)

On dividing by (a + b), we get

Quotient = 11(a + b)/(a + b) = 11

Hence, option 3 is the correct option.

abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111a + 111b + 111c = 111(a + b + c)

On dividing by 111, we get

Quotient = 111(a + b + c)/111 = (a + b + c)

Hence, option 3 is the correct option.

A + A + A = B1

⇒ 3A is a 2-digit number whose ones digit is 1.

Now, 3 × 7 = 21

⇒ A = 7 and B1 = 21

⇒ B = 2

Hence, option 4 is the correct option.

According to the divisibility of 2, if the unit's digit of the number is 0, 2, 4, 6 or 8 then the number is divisible by 2.

The unit's digit of 437821 is 1, which is an odd number.

∴ 437821 is not divisible by 2.

Hence, option 2 is the correct option.

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

The unit's digit of 32075 is 5.

∴ 32075 is not divisible by 10.

Hence, option 4 is the correct option.

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

Number formed by last two digits of 98764 = 64 = 4 × 16, which is divisible by 4.

Number formed by last two digits of 98746 = 46, number formed by last two digits of 98674 = 74 and number formed by last two digits of 98647 = 47, which are not divisible by 4.

∴ 98764 is divisible by 4.

Hence, option 1 is the correct option.

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

Number formed by last three digits of 32456 = 456 = 8 × 57, which is divisible by 8.

Number formed by last three digits of 32466, 32476 and 32486 are 466, 476 and 486 respectively, which are not divisible by 8.

∴ 32456 is divisible by 8.

Hence, option 4 is the correct option.

According to the divisibility of 11, if the difference of the sum of digits at the odd places (starting from unit's place) and the sum of the digits at even places is either 0 or divisible by 11, then the number is divisible by 11.

For 725824,

Sum of digits at odd places = 4 + 8 + 2 = 14

Sum of digits at even places = 2 + 5 + 7 = 14

Their difference = 14 - 14 = 0

∴ 725824 is divisible by 11.

Hence, option 1 is the correct option.

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.

Sum of digits of 24354 = 2 + 4 + 3 + 5 + 4 = 18, which is divisible by 9.

Sum of digits of 24453 = 2 + 4 + 4 + 5 + 3 = 18, which is divisible by 9.

Sum of digits of 24534 = 2 + 4 + 5 + 3 + 4 = 18, which is divisible by 9.

Sum of digits of 24564 = 2 + 4 + 5 + 6 + 4 = 21, which is not divisible by 9.

∴ 24564 is not divisible by 9.

Hence, option 4 is the correct option.

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.

∵ 467x8 is divisible by 3

⇒ 4 + 6 + 7 + x + 8 is divisible by 3

⇒ 25 + x is divisible by 3

⇒ 25 + x = 27, 30, 33, ...............

⇒ x = 2, 5, 8, ...............

Out of the given options, x = 2.

Hence, option 2 is the correct option.

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.

∵ 36x52y8 is divisible by 9

⇒ 3 + 6 + x + 5 + 2 + y + 8 is divisible by 9

⇒ 24 + x + y is divisible by 9

⇒ 24 + x + y = 27, 36, ...............

⇒ x + y = 3, 12, ...............

Out of the given options, x + y = 3.

Hence, option 2 is the correct option.

Since 6 = 2 × 3, so 2 is a factor of 6. Therefore, every number divisible by 6 is also divisible by 2.

So, statement I is true.

A number divisible by 2 need not be divisible by 6. For example, 4 is divisible by 2 but not by 6.

So, statement II is false.

Hence, option 1 is the correct option.

An even number need not be divisible by 4. For example, 6 is an even number but it is not divisible by 4.

So, statement I is false.

Since 4 = 2 × 2, so every number divisible by 4 is also divisible by 2 i.e. it is an even number.

So, statement II is true.

Hence, option 2 is the correct option.

∵ 456x78 is divisible by 3

⇒ 4 + 5 + 6 + x + 7 + 8 is divisible by 3

⇒ 30 + x is divisible by 3

⇒ 30 + x = 30, 33, 36, 39, ...............

⇒ x = 0, 3, 6, 9, ...............

Since x is a digit, so x = 0, 3, 6 or 9.

So, statement I is true.

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

So, statement II is true.

Hence, option 3 is the correct option.

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Given, sum of digits = 7

⇒ a + b = 7 .....(1)

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the difference of the 2-digit number and the number obtained by reversing the digits is 9.

⇒ (10a + b) - (10b + a) = 9

⇒ 9a - 9b = 9

⇒ 9(a - b) = 9

⇒ a - b = 1 .....(2)

Adding (1) and (2), we get

2a = 8

⇒ a = 4

Putting a = 4 in (1), we get b = 7 - 4 = 3

∴ 2-digit number = 10 × 4 + 3 = 43

If the number obtained by reversing the digits is greater than the original number, then

b - a = 1, which on solving with (1) gives a = 3 and b = 4.

∴ 2-digit number = 10 × 3 + 4 = 34

Hence, the required number is 43 or 34.

Let the 3-digit number be abc i.e. 100a + 10b + c, where hundred's digit is a, ten's digit is b and unit's digit is c.

Given, a - c = 5

Number obtained by reversing the digits = cba i.e. 100c + 10b + a

Difference = (100a + 10b + c) - (100c + 10b + a)

= 100a + 10b + c - 100c - 10b - a

= 99a - 99c

= 99(a - c)

On dividing by 9, we get

Quotient = $\frac{99(a - c)}{9}$ = 11(a - c) = 11 × 5 = 55

Hence, the required quotient is 55.

The numbers 675 and 756 are obtained by changing the order of digits of 567 cyclically.

According to the property, the sum of a 3-digit number abc and the numbers obtained by changing the order of digits cyclically = 111(a + b + c) = 3 × 37 × (a + b + c).

Here, a + b + c = 5 + 6 + 7 = 18

∴ 567 + 675 + 756 = 111 × 18

(i) When the sum is divided by 111, the quotient = 5 + 6 + 7 = 18.

(ii) When the sum is divided by 18 (5 + 6 + 7 = 18), the quotient = 111.

(iii) Since 111 × 18 = 37 × 3 × 18 = 37 × 54, when the sum is divided by 37, the quotient = 3 × 18 = 54.

(iv) Since 111 × 18 = 3 × 37 × 18 = 3 × 666, when the sum is divided by 3, the quotient = 37 × 18 = 666.

Given,

$$\begin{matrix} & \text{A} & \text{B} & 2 \\ + & & 2 & \text{A} \\ \hline & \text{A} & \text{A} & 8 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, 2 + A is a number whose ones digit is 8.

⇒ 2 + A = 8, 2 + A = 18; and so on.

⇒ A = 8 - 2, A = 18 - 2; and so on.

⇒ A = 6, A = 16; and so on.

Since, A is a digit. ∴ A = 6 and there is no carry over.

Secondly, we will find the value of B.

Ten's column: B + 2 is a number whose ones digit is A i.e. 6.

⇒ B + 2 = 6 or B + 2 = 16

⇒ B = 4 or B = 14

Since, B is a digit. ∴ B = 4 and there is no carry over.

Hundred's column: A = A, which is correct.

$$\begin{matrix} & \text{6} & \text{4} & 2 \\ + & & 2 & \text{6} \\ \hline & \text{6} & \text{6} & 8 \\ \hline \end{matrix}$$

Hence, A = 6 and B = 4.

Given,

$$\begin{matrix} & & 4 & \text{A} & \text{B} \\ + & & \text{A} & \text{B} & 7 \\ \hline & 1 & 0 & 2 & 3 \\ \hline \end{matrix}$$

Firstly, we will find the value of B.

Clearly, B + 7 is a number whose ones digit is 3.

⇒ B + 7 = 3, B + 7 = 13, B + 7 = 23; and so on.

⇒ B = 3 - 7, B = 13 - 7, B = 23 - 7; and so on.

⇒ B = -4, B = 6, B = 16; and so on.

Since, B is a digit. ∴ B = 6 and 1 is carried over to ten's column.

Secondly, we will find the value of A.

Ten's column: A + B + 1 = A + 6 + 1 = A + 7 is a number whose ones digit is 2.

⇒ A + 7 = 2, A + 7 = 12, A + 7 = 22; and so on.

⇒ A = 2 - 7, A = 12 - 7, A = 22 - 7; and so on.

⇒ A = -5, A = 5, A = 15; and so on.

Since, A is a digit. ∴ A = 5 and 1 is carried over to hundred's column.

Hundred's column: 4 + A + 1 = 4 + 5 + 1 = 10, write 0 and 1 is carried over to thousand's column, which is correct.

$$\begin{matrix} & & \overset{1}{4} & \overset{1}{\text{5}} & \text{6} \\ + & & \text{5} & \text{6} & 7 \\ \hline & 1 & 0 & 2 & 3 \\ \hline \end{matrix}$$

Hence, A = 5 and B = 6.

Given,

$$\begin{matrix} & \text{A} & \text{A} & \text{B} \\ \times & & & \text{B} \\ \hline & 8 & 8 & \text{A} \\ \hline \end{matrix}$$

Clearly, B × B is a number whose ones digit is A.

Also, AAB × B = 88A, a 3-digit number lying between 880 and 889.

Let's take B = 1: Then A = 1 (as 1 × 1 = 1) and 111 × 1 = 111, which is not of the form 88A. Hence, B ≠ 1.

Let's take B = 2: Then A = 4 (as 2 × 2 = 4) and 442 × 2 = 884, which is of the form 88A with A = 4.

Let's take B = 3: Then A = 9 (as 3 × 3 = 9) and 993 × 3 = 2979, which is a 4-digit number. Hence, B ≠ 3.

Similarly, for B = 4, 5, 6, 7, 8 or 9, the product AAB × B becomes a 4-digit number. Hence, these values are not possible.

$$\begin{matrix} & \text{4} & \text{4} & \text{2} \\ \times & & & \text{2} \\ \hline & 8 & 8 & \text{4} \\ \hline \end{matrix}$$

Hence, A = 4 and B = 2.

According to the divisibility of 11, if the difference of the sum of digits at the odd places (starting from unit's place) and the sum of the digits at even places is either 0 or divisible by 11, then the number is divisible by 11.

For the number 923x783,

Sum of digits at odd places = 3 + 7 + 3 + 9 = 22

Sum of digits at even places = 8 + x + 2 = 10 + x

Their difference = 22 - (10 + x) = 12 - x, which should be either 0 or divisible by 11.

⇒ 12 - x = 0, 11, 22, ...............

⇒ x = 12, 1, -10, ...............

Since x is a digit, therefore x = 1.

Hence, the value of digit x is 1.

Last digit is 3, which is an odd number, so it is not divisible by 2.

Sum of digits = 7 + 6 + 5 + 4 + 3 = 25, which is neither divisible by 3 nor by 9, so it is not divisible by 3 or 9.

Sum of digits at odd places = 3 + 5 + 7 = 15 and sum of digits at even places = 4 + 6 = 10. Their difference = 15 - 10 = 5, which is neither 0 nor divisible by 11, so it is not divisible by 11.

Hence, 76543 is not divisible by 2, 3, 9 or 11.

Last digit is 2, so it is divisible by 2.

Sum of digits = 6 + 5 + 4 + 3 + 2 = 20, which is neither divisible by 3 nor by 9, so it is not divisible by 3 or 9.

Sum of digits at odd places = 2 + 4 + 6 = 12 and sum of digits at even places = 3 + 5 = 8. Their difference = 12 - 8 = 4, which is neither 0 nor divisible by 11, so it is not divisible by 11.

Hence, 65432 is divisible by 2 only.

Last digit is 6, so it is divisible by 2.

Sum of digits = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 6 = 48, which is divisible by 3 but not by 9, so it is divisible by 3 but not by 9.

Sum of digits at odd places = 6 + 4 + 6 + 8 = 24 and sum of digits at even places = 3 + 5 + 7 + 9 = 24. Their difference = 24 - 24 = 0, so it is divisible by 11.

Hence, 98765436 is divisible by 2, 3 and 11 but not by 9.

Last digit is 7, which is an odd number, so it is not divisible by 2.

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

Sum of digits at odd places = 7 + 5 + 3 = 15 and sum of digits at even places = 6 + 4 + 2 = 12. Their difference = 15 - 12 = 3, which is neither 0 nor divisible by 11, so it is not divisible by 11.

Hence, 234567 is divisible by 3 and 9 but not by 2 or 11.

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

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

Last digit of 23565 is 5. So, it is divisible by 5 but not by 10.

Hence, 23565 is divisible by 5 but not by 10.

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

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

Last digit of 45270 is 0. So, it is divisible by both 5 and 10.

Hence, 45270 is divisible by both 5 and 10.

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

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

Number formed by last two digits of 47596 = 96 = 4 × 24, so it is divisible by 4. Number formed by last three digits = 596, which is not divisible by 8. So, it is not divisible by 8.

Hence, 47596 is divisible by 4 but not by 8.

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

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

Number formed by last two digits of 593024 = 24 = 4 × 6, so it is divisible by 4. Number formed by last three digits = 024 = 8 × 3, so it is divisible by 8.

Hence, 593024 is divisible by both 4 and 8.

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the difference of the 2-digit number and the number obtained by reversing the digits is 45.

⇒ (10a + b) - (10b + a) = 45

⇒ 9a - 9b = 45

⇒ 9(a - b) = 45

⇒ a - b = 5

The possible pairs of digits with a - b = 5 are:

a = 6, b = 1; a = 7, b = 2; a = 8, b = 3; a = 9, b = 4

(When a = 5 and b = 0, the number obtained by reversing the digits of 50 i.e. 05 is not a 2-digit number, so it is not considered.)

∴ The possible numbers are 61, 72, 83, 94 and the numbers obtained by reversing their digits are 16, 27, 38, 49.

Hence, all possible 2-digit numbers are 16, 61; 27, 72; 38, 83; 49, 94.