Playing with Numbers

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

So, when 68 + 86 is divided by 14 (6 + 8 = 14), the quotient = 11.

Hence, option 3 is the correct option.

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

So, when 75 - 57 is divided by 9, the quotient = 7 - 5 = 2.

Hence, option 2 is the correct option.

$$\begin{matrix} & 1 & \text{A} \\ + & \text{A} & 3 \\ \hline 1 & 0 & 1 \\ \hline \end{matrix}$$

A + 3 must give a number whose ones digit is 1.

⇒ A + 3 = 1, A + 3 = 11, A + 3 = 21, ...............

⇒ A = 1 - 3, A = 11 - 3, A = 21 - 3, ...............

⇒ A = -2, A = 8, A = 18, ...............

⇒ A = 8, as A is a digit.

$$\begin{matrix} & \overset{1}{1} & \text{8} \\ + & \text{8} & 3 \\ \hline 1 & 0 & 1 \\ \hline \end{matrix}$$

Hence, option 4 is the correct option.

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

So, when 53 - 35 is divided by 2 (5 - 3 = 2), the quotient = 9

Hence, option 2 is the correct option.

$$\begin{matrix} & 1 & 2 & \text{A} \\ - & & \text{A} & 9 \\ \hline & & 4 & 8 \\ \hline \end{matrix}$$

A - 9 must give a number whose ones digit is 8.

⇒ A - 9 = 8, A - 9 = 18, A - 9 = 28, ...............

⇒ A = 8 + 9, A = 18 + 9, A = 28 + 9, ...............

⇒ A = 17, A = 27, A = 37, ...............

⇒ A = 7, as A is a digit.

Hence, option 2 is the correct option.

(a) According to the property, when ab + ba is divided by 11, the quotient = a + b.
So, when 94 + 49 is divided by 11, the quotient = 9 + 4 = 13

(b) According to the property, when ab + ba is divided by a + b, the quotient = 11.
So, when 94 + 49 is divided by 13 (9 + 4 = 13), the quotient = 11.

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

So, when 94 - 49 is divided by 5 (9 - 4 = 5), the quotient = 9

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

So, when 94 - 49 is divided by 9, the quotient = (9 - 4) = 5.

$$\begin{aligned}abc = bac\\[1em] ⇒ a \times b \times c = b \times a \times c\\[1em] ⇒ a \times a \times c = a \times a \times c\\[1em] ⇒ a^2 \times c = a^2 \times c\\[1em] ⇒ \cancel {a^2} \times c = \cancel {a^2} \times c\\[1em] ⇒ c = c\end{aligned}$$

∴ If a = b, then abc = bac

Lets take the 3-digit number as abc

∴ Number obtained on interchanging its ones and hundreds digits will be cba

abc - cba

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

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

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

= (99a ) + (- 99c)

= 99 (a - c)

= 99(a - c)

Here a = 6, b = 9 and c = 4

(i) On dividing by 33:

$$\begin{aligned}\frac{99(a - c)}{33} \\[1em] = 3(a - c)\end{aligned}$$

Substituting values of a and c we get,

3(6 - 4) = 3 x 2 = 6

∴ When the difference between 694 and number obtained on interchanging its ones and hundreds digits is divided by 33, its quotient will be 6.

(ii) On dividing by 99:

$$\begin{aligned}\frac{99(a - c)}{99} \\[1em] = (a - c)\end{aligned}$$

Substituting values of a and c we get,

6 - 4 = 2

∴ When the difference between 694 and number obtained on interchanging its ones and hundreds digits is divided by 99, its quotient will be 2.

(iii) On dividing by 2:

$\frac{99(a - c)}{2}$

Substituting values of a and c we get,

$$\begin{aligned}\frac{99(6 - 4)}{2} \\[1em] = \frac{99 \times 2}{2} \\[1em] = 99\end{aligned}$$

∴ When the difference between 694 and number obtained on interchanging its ones and hundreds digits is divided by 2, its quotient will be 99.

(i) According to the property 3, when abc + bca + cab is divided by 111, the quotient = a + b + c.

Here the number is 842 and when it is arranged in cyclic order it becomes 428 and 284.

When 842 + 428 + 284 is divided by 111, the quotient = 8 + 4 + 2 = 14.

(ii) According to the property 3, when abc + bca + cab is divided by a + b + c, the quotient = 111 .

Here the number is 842 and when it arrange in cyclic order it becomes 428 and 284.

When 842 + 428 + 284 is divided by 8 + 4 + 2 (14), the quotient = 111.

(iii) Here the number is 842 and when it arrange in cyclic order it becomes 428 and 284.

$$\begin{aligned}\frac{842 + 428 + 284}{37}\\[1em] = \frac{1554}{37}\\[1em] = 42\end{aligned}$$

When 842 + 428 + 284 is divided by 37, the quotient = 42.

If a * b = a + 2b + 5

Putting a = 4 and b = 5, we get

4 * 5 = 4 + 2 x 5 + 5

⇒ 4 * 5 = 4 + 10 + 5

⇒ 4 * 5 = 19

Hence, 4 * 5 = 19

If a * b = 2a – b – 2

Putting a = 8 and b = 7, we get

8 * 7 = 2 x 8 – 7 – 2

⇒ 8 * 7 = 16 – 7 – 2

⇒ 8 * 7 = 7

Hence, 8 * 7 = 7

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

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.

As A is a digit, so A = 7.

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

Now, finding the value of B.

1 + 3 + 2 = B

6 = 5

Since, A is a digit. ∴ A = 7, B = 6.

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

Firstly, we will find the value of A.

Clearly, 8 + A is a number whose ones digit is 3.

⇒ 8 + A = 3, 8 + A = 13, 8 + A = 23; and so on.

⇒ A = 3 - 8 , A = 13 - 8, A = 23 - 8; and so on.

⇒ A = -5 , A = 5, A = 15; and so on.

Since, A is a digit. ∴ A = 5

$$\begin{matrix} & \overset{1}{9} & 8 \\ + & 4 & \text{5} \\ \hline \text{C} & \text{B} & 3 \\ \hline \end{matrix}$$ Secondly, we will find the value of B

1 + 9 + 4 = B

14 = B

Since, B is a digit. ∴ B = 4 and hence, C = 1

$$\begin{matrix} & 9 & 8 \\ + & 4 & \text{5} \\ \hline \text{1} & \text{4} & 3 \\ \hline \end{matrix}$$ A = 5, B = 4 and C = 1

$$\begin{matrix} & \text{A} & 1 \\ + & 1 & \text{B} \\ \hline & \text{B} & 0 \\ \hline \end{matrix}$$

Firstly we will find the value of B.

Clearly, 1 + B is a number whose ones digit is 0.

⇒ 1 + B = 0, 1 + B = 10, 1 + B = 20; and so on.

⇒ B = 0 - 1, B = 10 - 1, B = 20 - 1; and so on.

⇒ B = -1, B = 9, B = 19; and so on.

Since, B is a digit, ∴ B = 9.

Secondly we will find the value of A.

$$\begin{matrix} & \overset{1}{\text{A}} & 1 \\ + & 1 & \text{9} \\ \hline & \text{9} & 0 \\ \hline \end{matrix}$$

Clearly, 1 is carried over on tens place. A + 1 + 1 = 9.

⇒ A + 2 = 9 , A + 2 = 19 , A + 2 = 29; and so on.

⇒ A = 9 - 2 , A = 19 - 2 , A = 29 - 2; and so on.

⇒ A = 7 , B = 17 ,B = 27; and so on.

Since, A is a digit.∴ A = 7.

$$\begin{matrix} & \text{7} & 1 \\ + & 1 & \text{9} \\ \hline & \text{9} & 0 \\ \hline \end{matrix}$$ A = 7 and B = 9

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

Firstly, we will find the value of B.

Clearly, B + 1 is a number whose ones digit is 8.

⇒ B + 1 = 8, B + 1 = 18, B + 1 = 28; and so on.

⇒ B = 8 - 1, B = 18 - 1, B = 28 - 1; and so on.

⇒ B = 7, B = 17, B = 27; and so on.

Since, B is a digit. ∴ B = 7.

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

Secondly, we will find the value of A.

Clearly, A + 7 is a number whose ones digit is 1.

⇒ A + 7 = 1, A + 7 = 11, A + 7 = 21; and so on.

⇒ A = 1 - 7, A = 11 - 7, A = 21 - 7; and so on.

⇒ A = -6 , A = 4 , A = 14; and so on.

Since, A is a digit. ∴ A = 4.

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

A = 4 and B = 7.

$$\begin{matrix} & 1 & 2 & \text{A} \\ + & 6 & \text{A} & \text{B} \\ \hline & \text{A} & 0 & 9 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, 2 + A is a number whose ones digit is 0.

⇒ 2 + A = 0, 2 + A = 10, 2 + A = 20; and so on.

⇒ A = 0 - 2, A = 10 - 2, A = 20 - 2; and so on.

⇒ A = -2, A = 8, A = 18; and so on.

Since, A is a digit. ∴ A = 8.

$$\begin{matrix} & \overset{1}{1} & 2 & \text{8} \\ + & 6 & \text{8} & \text{B} \\ \hline & \text{8} & 0 & 9 \\ \hline \end{matrix}$$

Secondly, we will find the value of B.

Clearly, 8 + B is a number whose ones digit is 9.

⇒ 8 + B = 9 , 8 + B = 19 , 8 + B = 29; and so on.

⇒ B = 9 - 8 , B = 19 - 8 ,B = 29 - 8; and so on.

⇒ B = 1 , B = 11 , B = 21; and so on.

Since, B is a digit. ∴ B = 1.

$$\begin{matrix} & \overset{1}{1} & 2 & \text{8} \\ + & 6 & \text{8} & \text{1} \\ \hline & \text{8} & 0 & 9 \\ \hline \end{matrix}$$

A = 8 and B = 1.

$$\begin{matrix} & 1 & \text{A} \\ & \times & \text{A} \\ \hline & 9 & \text{A} \\ \hline \end{matrix}$$

Clearly, we will find the value of A.

A x A = A

⇒ A² = A

Find a number whose square gives the same number on unit place.

1 x 1 = 1

5 x 5 = 25

6 x 6 = 36

Lets take A = 1

$$\begin{matrix} & 1 & \text{1} \\ & \times & \text{1} \\ \hline & 1 & \text{1} \\ \hline \end{matrix}$$

Hence, 1 does not satisfy the equation.

Lets take A = 5

$$\begin{matrix} & \overset{2}{1} & \text{5} \\ & \times & \text{5} \\ \hline & 7 & \text{5} \\ \hline \end{matrix}$$

5 also does not satisfy the multiplication.

Lets take A = 6

$$\begin{matrix} & \overset{3}{1} & \text{A} \\ & \times & \text{A} \\ \hline & 9 & \text{A} \\ \hline \end{matrix}$$

Hence, A = 6.

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

Firstly, we will find the value of B.

Hence, B x 6 = B

A number multiplied with 6 gives same number in its unit place.

$$\begin{aligned}2 \times 6 = 12\\[1em] 4 \times 6 = 24\\[1em] 6 \times 6 = 36\\[1em] 8 \times 6 = 48\\[1em]\end{aligned}$$

Hence, B = 2 or 4 or 6 or 8

Now find the value of A

Lets take B = 2

Then

$$\begin{matrix} & & \overset{1}{\text{A}} & \text{2} \\ & & \times & 6 \\ \hline & \text{2} & \text{2} & \text{2} \\ \hline \end{matrix}$$

1 carried on A means A x 6 + 1 = 22. Since no value of A satisfy the equation. Hence B ≠ 2.

Lets take B = 4

Then

$$\begin{matrix} & & \overset{2}{\text{A}} & \text{4} \\ & & \times & 6 \\ \hline & \text{4} & \text{4} & \text{4} \\ \hline \end{matrix}$$

2 carried on A means A x 6 + 2 = 44 ⇒ A x 6 = 42 ⇒ A = 7.

Lets take B = 6

Then

$$\begin{matrix} & & \overset{3}{\text{A}} & \text{6} \\ & & \times & 6 \\ \hline & \text{6} & \text{6} & \text{6} \\ \hline \end{matrix}$$

3 carried on A means A x 6 + 3 = 66. Since no value of A satisfy the equation. Hence B ≠ 6.

Lets take B = 8

Then

$$\begin{matrix} & & \overset{4}{\text{A}} & \text{8} \\ & & \times & 6 \\ \hline & \text{8} & \text{8} & \text{8} \\ \hline \end{matrix}$$

4 carried on A means A x 6 + 4 = 88. Hence A = 14. But A is single digit number. So, B ≠ 8

$$\begin{matrix} & & \overset{2}{\text{A}} & \text{4} \\ & & \times & 6 \\ \hline & \text{4} & \text{4} & \text{4} \\ \hline \end{matrix}$$

Hence, A = 7 and B = 4

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

Firstly, we will find the value of B.

$$\begin{aligned}B \times 3 = B\\[1em] 0 \times 3 = 0\\[1em] 5 \times 3 = 15\\[1em]\end{aligned}$$

So, 0 and 5 are possible values of B.

Now finding the value of A.

Lets take B = 0 then A x 3 = A

$$\begin{matrix} & & \text{A} & \text{0} \\ & & \times & 3 \\ \hline & \text{C} & \text{A} & \text{0} \\ \hline \end{matrix}$$

0 x 3 = 0

5 x 3 = 15

As we need 2-digit number in answer. So A = 5 and C = 1.

Lets take B = 5

$$\begin{matrix} & & \overset{1}{\text{A}} & \text{5} \\ & & \times & 3 \\ \hline & \text{C} & \text{A} & \text{5} \\ \hline \end{matrix}$$

A x 3 + 1 = A

⇒ A x 3 = A - 1

There is no value for A which satisfies the equation. Hence B ≠ 5

$$\begin{matrix} & & \text{5} & \text{0} \\ & & \times & 3 \\ \hline & \text{1} & \text{5} & \text{0} \\ \hline \end{matrix}$$

Hence, A = 5, B = 0 and C = 1.

$$\begin{matrix} & & \text{A} & \text{B} \\ & & \times & 5 \\ \hline & \text{C} & \text{A} & \text{B} \\ \hline \end{matrix}$$

Firstly, find the value of B.

$$\begin{aligned}B \times 5 = B\\[1em] 0 \times 5 = 0\\[1em] 5 \times 5 = 25\\[1em]\end{aligned}$$

Possible values of B = 0 and 5.

Now, find the value of A

Lets take B = 0

$$\begin{matrix} & & \text{A} & \text{0} \\ & & \times & 5 \\ \hline & \text{C} & \text{A} & \text{0} \\ \hline \end{matrix}$$

$$\begin{aligned}A \times 5 = \text{CA}\\[1em] 5 \times 5 = 25\\[1em]\end{aligned}$$

Hence, A = 5 and C = 2

Now lets take B = 5

$$\begin{matrix} & & \overset{2}{\text{A}} & \text{5} \\ & & \times & 5 \\ \hline & \text{C} & \text{A} & \text{5} \\ \hline \end{matrix}$$

$A \times 5 + 2 = CA$

No value for A satisfies this equation. Hence, B ≠ 5

$$\begin{matrix} & & \text{5} & \text{0} \\ & & \times & 5 \\ \hline & \text{2} & \text{5} & \text{0} \\ \hline \end{matrix}$$

A = 5, B = 0 and C = 2

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

Clearly, we will find the value of A.

5 + A is a number whose ones digit is 3.

⇒ 5 + A = 3, 5 + A = 13 , 5 + A = 23; and so on.

⇒ A = 3 - 5, A = 13 - 5, A = 23 - 5; and so on.

⇒ A = -2, A = 8, A = 18; and so on.

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

Since, A is a digit. ∴ A = 8

$$\begin{matrix} & 6 & \text{A} & \text{B} & 5 \\ + & \text{D} & 5 & 8 & \text{C} \\ \hline & 9 & 3 & 5 & 1 \\ \hline \end{matrix}$$

Firstly, we will find the value of C.

5 + C is a number whose ones digit is 1.

⇒ 5 + C = 1, 5 + C = 11 , 5 + C = 21; and so on.

⇒ C = 1 - 5, C = 11 - 5, C = 21 - 5; and so on.

⇒ C = -4, C = 6, C = 16; and so on.

Since, C is a digit. ∴ C = 6

$$\begin{matrix} & 6 & \text{A} & \overset{1}{\text{B}} & 5 \\ + & \text{D} & 5 & 8 & \text{6} \\ \hline & 9 & 3 & 5 & 1 \\ \hline \end{matrix}$$

Secondly, we will find the value of B.

B + 8 + 1 is a number whose ones digit is 5.

⇒ B + 9 = 5 , B + 9 = 15 , B + 9 = 25; and so on.

⇒ B = 5 - 9 , B = 15 - 9, B = 25 - 9; and so on.

⇒ B = -4 , B = 6, B = 16; and so on.

Since, B is a digit. ∴ B = 6

$$\begin{matrix} & 6 & \overset{1}{\text{A}} & \overset{1}{\text{6}} & 5 \\ + & \text{D} & 5 & 8 & \text{6} \\ \hline & 9 & 3 & 5 & 1 \\ \hline \end{matrix}$$

Thirdly, we will find the value of A.

A + 5 + 1 is a number whose ones digit is 3.

⇒ A + 6 = 3 , A + 6 = 13 , A + 6 = 23; and so on.

⇒ A = 3 - 6 , A = 13 - 6, A = 23 - 6; and so on.

⇒ A = -3 , A = 7, A = 17; and so on.

Since, A is a digit. ∴ A = 7

$$\begin{matrix} & \overset{1}{6} & \overset{1}{\text{7}} & \overset{1}{\text{6}} & 5 \\ + & \text{D} & 5 & 8 & \text{6} \\ \hline & 9 & 3 & 5 & 1 \\ \hline \end{matrix}$$

Lastly, we will find the value of D.

6 + D + 1 is a number whose ones digit is 9.

⇒ D + 7 = 9 , D + 7 = 19 , D + 7 = 29; and so on.

⇒ D = 9 - 7 , D = 19 - 7, D = 29 - 7; and so on.

⇒ D = 2 , D = 12, D = 22; and so on.

Since, D is a digit. ∴ D = 2

$$\begin{matrix} & \overset{1}{6} & \overset{1}{\text{7}} & \overset{1}{\text{6}} & 5 \\ + & \text{2} & 5 & 8 & \text{6} \\ \hline & 9 & 3 & 5 & 1 \\ \hline \end{matrix}$$

A = 7, B = 6, C = 6 and D = 2.

As we know abc = 100 x a + 10 x b + c.

Similarly, cba = 100 x c + 10 x b + a.

Hence,

$$\begin{aligned}\frac{abc - cba}{11}\\[1em] = \frac{(100 \times a + 10 \times b + c) - (100 \times c + 10 \times b + a)}{11}\\[1em] = \frac{100a + 10b + c - 100c - 10b - a}{11}\\[1em] = \frac{(100a - a )+ (10b - 10b) + (c - 100c)}{11}\\[1em] = \frac{(99a ) + (- 99c)}{11}\\[1em] = \frac{99 (a - c)}{11}\\[1em] = 9(a - c)\end{aligned}$$

Hence, if abc – cba is divided by 11 then the quotient is 9(a-c).

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

1. 192 Last digit is 2 (even number).

2. 1660 Last digit is 0.

3. 1101 Last digit is 1 (odd number)

Hence, 192 and 1101 are divisible by 2.

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.

1. 261 = 2 + 6 + 1 = 9. And 9 is divisible by 3

2. 777 = 7 + 7 + 7 = 21. And 21 is divisible by 3

3. 6657 = 6 + 6 + 5 + 7 = 24. And 24 is divisible by 3

Hence, all the three numbers, 261, 777 and 6657 are divisible by 3.

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

1. 2292

Last digit is 2, hence 2292 is divisible by 2

2 + 2 + 9 + 2 = 15, 15 is divisible by 3. So, 2292 is divisible by 3

Hence, 2292 is divisible by 6.

2. 2612

Last digit is 2, hence 2612 is divisible by 2

2 + 6 + 1 + 2 = 11, 11 is not divisible by 3. So, 2612 is not divisible by 3

Hence, 2612 is not divisible by 6.

3. 5940

Last digit is 0, hence 5940 is divisible by 2

5 + 9 + 4 + 0 = 18, 18 is divisible by 3. So, 5940 is divisible by 3

Hence, 5940 is divisible by 6.

Hence, option 1 and 3 are 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.

1. 1863 = 1 + 8 + 6 + 3 = 18. And 18 is divisible by 9.

2. 2064 = 2 + 0 + 6 + 4 = 12. And 12 is not divisible by 9.

3. 6111 = 6 + 1 + 1 + 1 = 9. And 9 is divisible by 9.

Hence, 1863 and 6111 are divisible by 9.

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.

1. 360 — Last 2 digits are 60 and 60 is divisible by 4.

2. 4093 — Last 2 digits are 93 and 93 is not divisible by 4.

3. 5348 — Last 2 digits are 48 and 48 is divisible by 4.

Hence, 360 and 5348 are divisible by 4.

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

1. 3250 — Last digit is 0.

2. 5557 — Last digit is 7.

3. 39255 — Last digit is 5.

Hence, 3250 and 39255 are divisible by 5.

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

1. 5100 — Last digit is 0.

2. 4612 — Last digit is 2.

3. 3400 — Last digit is 0.

Hence, 5100 and 3400 are divisible by 10.

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

1. 2563 — the sum of its digits in odd places = 2 + 6 = 8
   the sum of its digits in even places = 5 + 3 = 8
   The difference = 8 - 8 = 0.

2. 8307 — the sum of its digits in odd places = 8 + 0 = 8
   the sum of its digits in even places = 3 + 7 = 10
   The difference = 10 - 8 = 2 (Not divisible by 11)

3. 95635 — the sum of its digits in odd places = 9 + 6 + 5 = 20
   the sum of its digits in even places = 5 + 3 = 8
   The difference = 20 - 8 = 12 (Not divisible by 11)

Hence, 2563 is divisible by 11.

According to the divisibility of 7, take the last digit of the number, double it then subtract the result from the rest of the number, if the resulting number is divisible by 7 then the number is divisible by 7.

The number is 21952.

The last digit is 2 and its double is 4.

Rest of the number = 2195

Subtract 4 from 2195 to get 2195 - 4 = 2191.

Take 2191 as the number.

The last digit is 1 and its double is 2.

Rest of the number = 219

Subtract 2 from 219 to get 219 - 2 = 217

Take 217 as the number.

The last digit is 7 and its double is 14.

Rest of the number = 21

Subtract 14 from 21 to get 21 - 14 = 7 and 7 is divisible by 7.

Hence, 21952 is divisible by 7.

According to the divisibility of 7, take the last digit of the number, double it then subtract the result from the rest of the number, if the resulting number is divisible by 7 then the number is divisible by 7.

The number is 6486.

The last digit is 6 and its double is 12.

Rest of the number = 648

Subtract 12 from 648 to get 648 - 12 = 636.

Take 636 as the number.

The last digit is 6 and its double is 12.

Rest of the number = 63

Subtract 12 from 63 to get 63 - 12 = 51.

51 is not divisible by 7.

Hence, 6486 is not divisible by 7.

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.

1x5 = 1 + x + 5 = 6 + x. And (6 + x) should be a multiple of 3.

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

x = -6, -3, 0, 3, 6, 9, 12,.....

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

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.

31x5 = 3 + 1 + x + 5 = 9 + x. And (9 + x) should be a multiple of 3.

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

x = -9, -6, -3, 0, 3, 6, 9,.....

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

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.

28x6 = 2 + 8 + x + 6 = 16 + x. And (16 + x) is should be a multiple of 3.

Hence, 16 + x = 0 ,3, 6, 9, 12, 15, 18, 21, 24, 27, 30......

x = -16, -13, -10, -7, -4, -1, 2, 5, 8, 11.....

Since, x is a digit. ∴ x = 2, 5 or 8

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

24x is a multiple of 2 and 3 both.

To make 24x divisible by 2 ⇒ x should be 0 or an even number.

So, x = 0, 2, 4, 6 or 8 .............(1)

To make 24x divisible by 3 ⇒ 2 + 4 + x is a multiple of 3.

⇒ 6 + x is a multiple of 3.

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

⇒ x = -6, -3, 0, 3, 6, 9, 12, ...............

Since x is a digit. So, x = 0, 3, 6, 9 ............(2)

From (1) and (2), x = 0 or 6.

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

3x26 should be a multiple of 2 and 3 both.

3x26 is divisible by 2 as it has even number 6 at its unit place.

To make 3x26 divisible by 3 ⇒ 3 + x + 2 + 6 should be a multiple of 3.

⇒ 11 + x should be a multiple of 3.

⇒ 11 + x = 0 , 3, 6, 9, 12, 15, 18......

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

Since x is a digit. So, x = 1, 4, or 7.

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.

42x8 is divisible by 4

⇒ x8 is divisible by 4

⇒ 10x + 8 is divisible by 4

⇒ 10x + 8 = 8, 28, 48, 68, 88, 108,...............

⇒ 10x = 0, 20, 40, 60, 80, 100,...............

⇒ x = 0, 2, 4, 6, 8, 10,...............

Since x is a digit. So, x = 0, 2, 4, 6 or 8.

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.

9142x is divisible by 4

⇒ 2x is divisible by 4

⇒ 20 + x is divisible by 4

⇒ 20 + x = 20, 24, 28, 32, 36, 40,...............

⇒ x = 0, 4, 8, 12, 16,...............

Since x is a digit. So, x = 0, 4 or 8.

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.

∵ 7x34 is divisible by 9

⇒ 7 + x + 3 + 4 is divisible by 9

⇒ 14 + x is divisible by 9

⇒ 14 + x = 0 , 9 , 18 , 27 , 36, ...............

⇒ x = -14 , -5 , 4 , 13 , 22, ...............

Since x is a digit. So, x = 4.

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.

∵ 5x555 is divisible by 9

⇒ 5 + x + 5 + 5 + 5 is divisible by 9

⇒ 20 + x is divisible by 9

⇒ 20 + x = 0 , 9 , 18 , 27 , 36, ...............

⇒ x = -20 , -11 , -2 , 7 , 16, ...............

Since x is a digit. So, x = 7.

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

∴ 3x2 is divisible by 11
⇒ Sum of the digits in even place = x

Sum of the digits in odd place = 3 + 2 = 5

Difference between the sum of the digits in odd places and the sum of the digits in even places = 5 - x

⇒ 5 - x must divisible by 11

⇒ 5 - x = 0, 11, 22, 33, ...............

⇒ x = 5, -6, -17, -28, ...............

Since x is a digit. So, x = 5.

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

∴ 5x2 is divisible by 11
⇒ Sum of the digits in even place = x

Sum of the digits in odd place = 5 + 2 = 7

Difference between the sum of the digits in odd places and the sum of the digits in even places = 7 - x

⇒ 7 - x must be divisible by 11

⇒ 7 - x = 0, 11, 22, 33, ...............

⇒ x = 7, -4, -15, -26, ...............

Since x is a digit. So, x = 7.

ab + ba

= (10 x a + b) + (10 x b + a)

= 10a + b + 10b + a

= (10a + a) + (b + 10b)

= 11a + 11b

= 11(a + b)

Hence, option 3 is the correct option.

ab - ba

= (10a + b) - (10b + a)

= 10a + b - 10b - a

= (10a - a) + (b - 10 x b)

= 9a - 9 x b

= 9(a - b)

Hence, option 2 is the correct option.

$$\begin{aligned}ab ÷ ba\\[1em] = \frac{ab}{ba}\\[1em] = \frac{a \times b}{b \times a}\\[1em] = \frac{a \times a}{a \times a}\\[1em] = \frac{a^2}{a^2}\\[1em] = \frac{\cancel{a^2}}{\cancel{a^2}}\\[1em] = 1\end{aligned}$$ Hence, option 4 is the correct option.

abc - cba

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

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

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

= 99a - 99 x c

= 99(a - c)

Hence, option 4 is the correct option.

Clearly, A + 1 is a number whose ones digit is 9.

A + 1 = 9, A + 1 = 19, A + 1 = 29; and so on

⇒ A = 9 - 1, A = 19 - 1, A = 29 - 1; and so on

⇒ A = 8, A = 18, A = 28; and so on

Since A is a digit. So A = 8

Hence, option 3 is the correct option.

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

So, when 59 + 95 is divided by 11, the quotient = (5 + 9) = 14.

Hence, option 1 is the correct option.

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

So, when 74 - 47 is divided by 9, the quotient = 7 - 4 = 3.

Hence, option 3 is the correct option.

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

10a + 4 is divisible by 2 as 4 (even number) is at its unit place.

10a + 4 is divisible by 3 when a + 4 is divisible by 3.

Then, a + 4 is divisible by 6.

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.

∵ 3z4 is divisible by 9

⇒ 3 + z + 4 is divisible by 9

⇒ 7 + z is divisible by 9

⇒ 7 + z = 0 , 9 , 18 , 27 , 36, ...............

⇒ z = -7 , 2 , 11 , 20, ...............

Since z is a digit. So, z = 2.

Hence, option 2 is the correct option.

Let the digit at ten's place be x and at unit's place be y.

The two-digit number = 10 × x + y = 10x + y

Number obtained by reversing its digits = 10 × y + x = 10y + x

Sum of the number and reversed numbers= (10x + y) + (10y + x)

= 10x + x + y + 10y

= 11x + 11y

= 11(x + y)

On dividing by 11, we get :

$\frac{\text{11(x + y)}}{11}$ = (x + y)

The quotient is equal to the sum of the digits.

So, statement 1 is true.

On dividing the sum of numbers by sum of the digits = $\frac{11(x + y)}{x + y}$ = 11

The quotient is equal to 11.

So, statement 2 is true.

Hence, option 1 is the correct option.

Given,

759 = 100 x 7 + 10 x 5 + 1 x 9

Solving R.H.S. of the above equation, we get :

⇒ 100 x 7 + 10 x 5 + 1 x 9

⇒ 700 + 50 + 9

⇒ 759

Since, L.H.S. = R.H.S.

So, assertion (A) is true.

In a three-digit number 100p + 10q + 1r,

q and r can be 0 but p cannot be equal to 0 for 3-digit number.

So, reason (R) is false.

Hence, option 3 is the correct option.

Divisibility Rule for 9 : A number is divisible by 9 if the sum of its digits is divisible by 9.

So, if 36p52q9 is divisible by 9, then the sum of digits of 36p52q9, will also be divisible by 9.

⇒ 3 + 6 + p + 5 + 2 + q + 9

⇒ 25 + p + q

⇒ 27 (If, p + q = 2)

Since, 27 is divisible by 3, so if 36p52q9 is divisible by 9, then p + q = 2.

So, assertion (A) is true.

We know that,

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

This is a correct rule for divisibility by 3.

So, reason (R) is true, but it does not explains assertion.

Hence, option 2 is the correct option.

Given a three-digit number : 542

The cyclic permutations of the digits are : 542, 425 and 254.

Now we compute the sum of these three numbers : 542 + 425 + 254 = 1221

1221 = 1 x 11 x 111

So, assertion (A) is true.

Let the given a three-digit number : 100p + 10q + r

The cyclic permutations of the digits are: 100p + 10q + r, 100q + 10r + p and 100r + 10p + q

Now we compute the sum of these three numbers: (100p + 10q + r) + (100q + 10r + p) + (100r + 10p + q)

= (100p + 10p + p) + (100q + 10q + q) + (100r + 10r + r)

= 111p + 111q + 111r

= 111(p + q + r)

Factors of 111(p + q + r) = 1 x 111 x (p + q + r)

According to reason (R), 111 and sum of the digits are the factors of the sum of a three - digit number and the two number obtained by changing the digit cyclically, but 11 is not.

So, reason (R) is false.

Hence, option 3 is the correct option.

We know that,

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

So, reason (R) is true.

Lets check 2574 is divisible by 11. Digits at even place (5 and 4), digits at odd place (2 and 7).

⇒ (5 + 4) - (2 + 7)

⇒ 9 - 9

⇒ 0.

Since, the difference is 0,

∴ 2574 is divisible by 11.

Lets check 7083 is divisible by 11. Digits at even place (0 and 3), digits at odd place (7 and 8).

⇒ (0 + 3) - (7 + 8)

⇒ 3 - 15

⇒ -12

Since, difference is neither 0 nor divisible by 11,

∴ 7083 is not divisible by 11.

So, assertion (A) is true and reason explain assertion.

Hence, option 1 is the correct option.

abc + cab + bca

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

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

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

= 111a + 111b + 111c

= 111(a + b + c)

= 3 x 37 x (a + b + c)

Hence, (abc + cab + bca) is divisible by 37

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.

∵ 24x is divisible by 9

⇒ 2 + 4 + x is divisible by 9

⇒ 6 + x is divisible by 9

⇒ 6 + x = 0 , 9 , 18 , 27 , 36, ...............

⇒ x = -6 , 3 , 12 , 21 , 30, ...............

Since x is a digit. So, x = 3.

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.

∵ 31y5 is divisible by 3

⇒ 3 + 1 + y + 5 is divisible by 3

⇒ 9 + y is divisible by 3

⇒ 9 + y = 0 , 3 , 6 , 9 , 12, ...............

⇒ y = -9 , -6 , -3 , 0 , 3, 6, 9, ...............

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

Hence, least value of digit y = 0.

Let the 3-digit number be abc.

abc = 100a + 10b + c

Given,

b = 3c ........(1)

a = 2c ........(2)

and

a + b + c = 12 ........(3)

Putting value of b and a from eqns (1) and (2) into eq (3) we get,

2c + 3c + c = 12

⇒ 6c = 12

⇒ c = $\frac{12}{6}$

⇒ c = 2

Putting value of c in eq 1,

b = 3c = 3 x 2 = 6

Putting value of c in eq 2,

a = 2c = 2 x 2 = 4

Hence, the number is 462.

$$\begin{matrix} & 7 & 3 & \text{A} \\ - & \text{B} & \text{C} & 9 \\ \hline & 3 & 4 & 8 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, A - 9 is a number whose ones digit is 8.

⇒ A - 9 = 8 , A - 9 = 18 , A - 9 = 28; and so on.

⇒ A = 8 + 9 , A = 18 + 9 , A = 28 + 9; and so on.

⇒ A = 17 , A = 27 , A = 37; and so on.

Since, A is a digit. ∴ A = 7.

Secondly, we will find the value of C.

$$\begin{matrix} & 7 & \overset{2}{\cancel{3}} & \overset{17}{\cancel{7}} \\ - & \text{B} & \text{C} & 9 \\ \hline & 3 & 4 & 8 \\ \hline \end{matrix}$$

Clearly, 2 - C is a number whose ones digit is 4.

As it is not possible to subtract any number from 2 that gives 4. Hence, borrowing one number from preceding digit gives 12 - C.

$$\begin{matrix} & \overset{6}{\cancel{7}} & \overset{12}{\cancel{3}} & \overset{17}{\cancel{7}} \\ - & \text{B} & \text{C} & 9 \\ \hline & 3 & 4 & 8 \\ \hline \end{matrix}$$

⇒ 12 - C = 4 , 12 - C = 14 , 12 - C = 24; and so on.

⇒ C = 12 - 4 , C = 12 - 14 , C = 12 - 24; and so on.

⇒ C = 8 , C = -2 , C = -12; and so on.

Since, C is a digit. ∴ C = 8.

Lastly, we will find the value of B.

Clearly, 6 - B is a number whose ones digit is 3.

⇒ 6 - B = 3 , 6 - B = 13 , 6 - B = 23; and so on.

⇒ B = 6 - 3 , B = 6 - 13 , B = 6 - 23; and so on.

⇒ B = 3 , B = -7 , B = -17; and so on.

Since, B is a digit. ∴ B = 3.

$$\begin{matrix} & 7 & 3 & \text{7} \\ - & \text{3} & \text{8} & 9 \\ \hline & 3 & 4 & 8 \\ \hline \end{matrix}$$

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

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

Firstly, we will find the value of B

Clearly, B + 1 is a number whose ones digit is 8.

⇒ B + 1 = 8 , B + 1 = 18 , B + 1 = 28; and so on.

⇒ B = 8 - 1 , B = 18 - 1, B = 28 - 1; and so on.

⇒ B = 7 , B = 17, B = 27; and so on.

Since, B is a digit. ∴ B = 7

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

Now, we find the value of A.

Clearly, A + 7 is a number whose ones digit is 1.

⇒ A + 7 = 1 , A + 7 = 11 , A + 7 = 21; and so on.

⇒ A = 1 - 7 , A = 11 - 7, A = 21 - 7; and so on.

⇒ A = -6 , A = 4, A = 14; and so on.

Since, A is a digit. ∴ A = 4

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

Hence, A = 4 and B = 7.

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

Lets take A as a 1-digit number

B + 7 = A .....(1)

A + 3 = 9 .....(2)

A = 9 - 3 = 6

Putting the value of A in equation (1), we get

B + 7 = 6

B = 6 - 7

B = -1

∴ As we know B is a digit. Hence B ≠ -1.

Lets take A as a 2-digit number with ones digit as 1.

We can represent A as 10 + A

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

B + 7 = 10 + A .....(3)

1 + A + 3 = 9 (one carry on ten's digit) .....(4)

A + 4 = 9

A = 9 - 4 = 5

Putting the value of A in equation 3.

B + 7 = 10 + 5

B + 7 = 15

B = 15 - 7 = 8

Hence, A = 5, B = 8

Now, lets take A as a 2-digit number with ones digit as 2.

We can represent A as 20 + A

$$\begin{matrix} & \overset{2}{\text{A}} & \text{B} \\ + & 3 & 7 \\ \hline & 9 & \text{A} \\ \hline \end{matrix}$$

B + 7 = 20 + A .....(5)

2 + A + 3 = 9 (two carry on ten's digit) .....(6)

A + 5 = 9

A = 9 - 5 = 4

Putting the value of A in equation (5),

B + 7 = 20 + 4

B + 7 = 24

B = 24 - 7 = 17

∴ As we know B is a digit. Hence B ≠ 17.

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

Hence, A = 5 , B = 8.

$$\begin{matrix} & 1 & \text{B} \\ & \times & \text{B} \\ \hline & 9 & \text{B} \\ \hline \end{matrix}$$

Clearly, B x B = B

1 x 1 = 1

5 x 5 = 25

6 x 6 = 36

So, possible answer for B = 1 or 5 or 6.

But 1 and 5 did not satisfy the multiplication.

$$\begin{matrix} & \overset{3}{1} & \text{6} \\ & \times & \text{6} \\ \hline & 9 & \text{6} \\ \hline \end{matrix}$$ Hence, B = 6.

$$\begin{matrix} & 8 & \text{B} & 4 \\ - & 5 & 1 & \text{B} \\ \hline & \text{B} & 2 & 1 \\ \hline \end{matrix}$$

Clearly, 4 - B is a number whose ones digit is 1.

⇒ 4 - B = 1 , 4 - B = 11 , 4 - B = 21; and so on.

⇒ B = 4 - 1 , B = 4 - 11, B = 4 - 21; and so on.

⇒ B = 3 , B = -7, B = -17; and so on.

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

Since, B is a digit. ∴ B = 3