Lesson: Playing with Numbers

In each of the questions 1 to 17, out of the four options, only one is correct.

Write the correct answer.

Question 1

Generalised form of a four-digit number abdc is:

(a) 1000a + 100b + 10c + d

(b) 1000a + 100c + 10b + d

(c) 1000a + 100b + 10d + c

(d) a b c d

Answer:

(c)

Solution:

A four-digit number abdc is written as 1000a + 100b + 10d + c

Question 2

Generalised form of a two-digit number xy is:

(a) x + y

(b) 10x + y

(c) 10x – y

(d) 10y + x

Answer:

(b)

Solution:

A two digit number xy is written as 10x + y

Question 3

The usual form of 1000a + 10b + c is:

(a) abc

(b) abco

(c) aobc

(d) aboc

Answer:

(c)

Solution:

Since 1000a + 10b + c = 1000a + 100 b + c so the usual form of

1000a + 10b + c is aobc

Question 4

Let abc be a three - digit number, where a and c are different.

Then abc – cba is not divisible by:

(a) 9

(b) 11

(c) 13

(d) 33

Answer:

(c)

Solution:

abc – cba = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 11 (a – c)

Which is always divisible by 11, 9, 33 and 99.

This will be divisible by 13 provided a – c is divisible by 13, which is

not possible.

Therefore, abc – cba is not divisible by 13.

Question 5

The sum of all the numbers formed by using the digits x, y and z of the number xyz

is divisible by:

(a) 11

(b) 33

(c) 37

(d) 64

Answer:

(c)

Solution:

Given,

xyz = 100x + 10y + z

yxz = 100y + 10x + z

yzx = 100y + 10z + x

zyx = 100z + 10y + x

zxy = 100z + 10x + y

Sum of these numbers = 222 (x + y + z) = 37 x + y + z),

which is divisible by 37.

The sum of all the numbers formed by using the digits x, y and z of the number xyz

Is divisible by 37.

Question 6

A four-digit number aabb is divisible by 55.

Then possible value(s) of b is/are:

(a) 0 and 2

(b) 2 and 5

(c) 0 and 5

(d) 7

Answer:

(c)

Solution:

aabb = 1000a + 100a + 10b + b = 11 (100a + b),

Which is always divisible by 11.

Since, the number is divisible by 55; it is divisible by both 11 and 5.

This means that b has to be either 5 or 0.

Question 7

Let abc be a three digit number.

Then abc + bca + cab is not divisible by.

(a) a + b + c

(b) 3

(c) 37

(d) 9

Answer:

(d)

Solution:

abc + bca + cab

Can be written as:

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

= 111a + 111b + 111c

=111(a + b + c) = 37 (a + b + c)

So, 111(a + b + c) is not divisible by 9.

This number will be divisible by 9, if it is further given that a + b + c is divisible by 3.

Therefore, abc + bca + cab is not divisible by 9.

Question 8

A four-digit number 4ab5 is divisible by 55. Then the value of b – a is:

(a) 0

(b) 1

(c) 4

(d) 5

Answer:

(b)

Solution:

The four digit numbers 4ab5 is divisible by 55; hence, it is divisible by both

11 and 5.

For all vales of a and b the number is divisible by 5as the digit at the units

Place is 5.

Since it divisible by 11, we must have b + 4 – a – 5 = b – a – 1

Divisible by 11, which means that (b – a) must be 1.

Question 9

If abc is a three digit number, then the number abc – a – b – c is

divisible by:

(a) 9

(b) 90

(c) 10

(d) 11

Answer:

(a)

Solution:

If the number is abc, then,

abc – a – b – c = 100a + 10b + c – a – b – c = 99a + 9b

= 9 (11a + b), Which is divisible by 9.

Question 10

A six-digit number is formed by repeating a three-digit number.

For example 256256, 678678,etc.

Any number of this form is divisible by:

(a) 7 only

(b) 11 only

(c) 13 only

(d) 1001

Answer:

(d)

Solution:

Let the three digit number be abc.

So,

abcabc = 100000a + 10000b + 1000c + 100a + 10b + c

= 1000100a + 10010b + 1001c

= 1001 (100a + 10b + 1)

Which is divisible by all of 7, 11, 13 and 1001.

The correct option is (d).

Hence ,

the 6 digit number formed by repeating a three-digit number is divisible by 1001.

Question 11

If the sum of the digits of a number is divisible by three, then the number is always

divisible by:

(a) 2

(b) 3

(c) 6

(d) 9

Answer:

(b)

Solution:

If the sum of the digits of a number is divisible by three, then the numberis always

divisible by 3.

For example, let us consider a three digit number.

abc = 100a + 10b + c = 99a + 9b + a + b + c If it is further given

that a + b + c is divisible by 3, then a + b + c = 3n, for some natural

number n and abc = 3 (33a + 3b + n), which is clearly divisible by 3.

Question 12

If x + y + z = 6 and z is an odd digit, then the three-digit number xyz is:

(a) An odd multiple of 3

(b) An odd multiple of 6

(c) An even multiple of 3

(d) An even multiple of 9

Answer:

(a)

Solution:

Since, z is an odd digit, z can be either be1 or 3 or 5, because if z becomes 7 or

9, x + y + z = 6 will be impossible. Further, if z = 1, then

x + y = 5, if z = 3, then x + y = 3, if z = 5, then x + y = 1,

The possible values of

(x, y) = (1, 4), (2, 3), (3, 2), (4,1), (1, 2), (2,1), (1, 0) .

Which will be an odd multiple of 3 for all possible values of x and y.

Question 13

If 5A + B3 = 65, then the values of A and B are:

(a) A = 2, B = 3

(b) A = 3, B = 2

(c) A = 2, B = 1

(d) A = 1, B = 2

Answer:

(c)

Solution:

We have,

There are two letters A and B whose values we have to find.

The addition of A and 3is giving a number whose ones digit is 5.

This is possible only when digit A is 2. In that case, the addition of 2 and 3will

give 5.

Clearly, A is 2 as 2 + 3 = 5.

In the next step,

Clearly, B is 1as 5 + 1 = 6.

So the puzzle can be solved as shown below.

Hence, A and B are 2 and1 respectively.

Question 14

If A3 + 8B = 150, then the value of A + B is:

(a) 13

(b) 12

(c) 17

(d) 15

Answer:

(a)

Solution:

We have,

There are two letters A and B whose values we have to find. The addition of 3 and B

is giving a number whose ones digit is 0.

This is possible only when digit B is 7. In that case, the addition of 3 and 7 will

give 10.

Clearly, B is 7 as 3 + 7 = 10.

In the next step,

This is possible only when the digit A is 6. In that case, the addition of 1,6 and 8

will give 15.

Clearly, A is 6 as 1 + 6 + 8 = 15.

So the puzzle can be solved as shown below.

Hence, A and B are 6 and 7 respectively.

So, A + B = 6 + 7 = 13.

Question 15

If 5A then the value of A is:

(a) 3

(b) 6

(c) 7

(d) 9

Answer:

(c)

Solution:

Given,

Since A is at the unit’s place in both the numbers 5A and A,

A number with one’s digit 9.

A is either 3 or 7.

Taking A as 3, we get 53 3 = 159;talking a as 7,we get 57 7 = 399.

Therefore, A = 7.

Question 16

If 6A B = A8B, then the value of A – B is.

(a) – 2

(b) 2

(c) – 3

(d) 3

Answer:

(a)

Solution:

We have,

This means that the product of A and B is giving a number with unit’s digit B.

There are many possible combinations of numbers to give such a result.

The possible values of B are 2,3,4,5, 6, and 8.

Taking B as 8, we get A = 6, this does not satisfy the second step i.e.

6 B + 4 ≠ A8.

Taking B as 6, we get A = 6, this does not satisfy the second step i.e.

6 B + 3 ≠ A8.

Taking B as 5, we get A = 1 or 3 or 5 or 7 or 9.

For all these values, the second step does not get satisfied.

In the same manner B = 2does not satisfy the second step and its solutions.

Now, taking B = 3, A is 1.

Clearly, A = 1and B = 3

Hence A and B are 1 and 3 respectively.

A – B = 1 – 3 = – 2

Question 17

Which of the following numbers is divisible by 99?

(a) 913462

(b) 114345

(c) 135792

(d) 3572406

Answer:

(b)

Solution:

For a number to be divisible by 99, it must be divisible by both 11 and 9.

A number is divisible by 11 if the difference between the sum of the digits at its

odd places and that of digits at its even places is either

0 or divisible by 11.

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

The given number is 913462.

Sum of its digits = 9 + 1 + 3 + 4 + 6 + 2 = 25, which is not divisible by 9.

Given number is114345.

Sum of its digits = 1 + 1 + 4 + 3 + 4 + 5 = 18, which is divisible by 9 and

the difference of sums of its digits at odd places and even places is 9 – 9 =0.

It is divisible by 11.

Therefore, 114345 is divisible by 99.

Given number is135792.

Sum of its digits = 1 + 3 + 5 + 7 + 9 + 2 = 27, which is divisible by 9.

But, difference of sums of digits at odd places and even places (15 – 12 = 13) is

not divisible by 11.

135792 is not divisible by 99.

Given number is 3572406.

Sum of its digits = 3 + 5 + 7 + 2 + 4 + 0 + 6 = 27, which is divisible by 9.

But, difference of sums of digits at odd places and even places (20 – 7 = 13) is

not divisible by 11.

3572406 is not divisible by 99.

114345 is divisible by 99.

In questions 18 to 33, fill in the blanks to make the statements true.

Question 18

3134673 is divisible by 3 and _____________.

Answer:

9

Solution:

Given number is 3134673.

Sum of its digits = 3 + 1 + 3 + 4 + 6 + 7 + 3 = 27.

27 is divisible by 3 and 9.

Question 19

20x3 is a multiple of 3 if the digit x is _____________or _____________or

_____________.

Answer:

1, 4, 7

Solution:

20x3 is multiple of 3 if

Is a multiple of 3

Is a multiple of 3

………… -------- (i)

But, x is a digit of the number 20x3. S0, x can take values 0, 1, 2, 3…9.

can take values 5, 6, 7, 8... 14. ------- (ii)

From (i) and (ii), we get

Hence x can take values 1 or, 4 or, 7.

Question 20

3x5 is divisible by 9 if the digit x is _____________.

Answer:

1

Solution:

3x5 is divisible by 9 if 3 + x + 5 is a multiple of 9 is a multiple of 9.

But, x is a digit. So, it can take values 0, 1, 2, 3…, 9.

And hence,

8 x

can take values 8, 9, 10, 11..., 17. ------- (ii)

From (i) and (ii), we get

Hence,

1x

Question 21

The sum of a two–digit number and the number obtained by reversing the digits is

always divisible by _____________.

Answer:

11

Solution:

The sum of a two–digit number and the number obtained by reversing the digits is

always divisible by 11 as.

Question 22

The difference of a two–digit number and the number obtained by reversing its

digits is always divisible by _____________.

Answer:

9

Solution:

The difference between a 2-digit number and the number obtained by reversing its

digits is always divisible by 9 as.

Question 23

The difference of three-digit number and the number obtained by putting the digits

in reverse order is always divisible by 9 and _____________.

Answer:

11

Solution:

Any 3-digit number can be represented as:

100A + 10B + C

The number, when reversed is:

100C + 10B + A

The difference:

(100A + 10B + C) – (100C + 10B + A) = 99A – 99C = 99(A – C)

Since the result is equal to 99 times some integer, the result must be divisible by

99, 9 and 11.

Question 24

If

2

8

B

AB

A

then A = _____________and B = _____________.

Answer:

A = 6,B = 3

Solution:

There are two letters A and B whose values we have to find.

The addition of B and B is giving a number whose ones digit is A.

Also, there is a second A.

The above sum will be possible only when digit B is 3.

In that case, the addition of 3and 3 will give A.

Clearly, A is 6 as 3 + 3 = 6.

In the next step, 2 + A = 8 as 2 + 6 = 8

So the puzzle can be solved as shown below.

23

63

86

Hence, A and B are 6 and 3 respectively.

Question 25

If

96

AB

B

then A = _____________ and B = _____________.

Answer:

A = 2,B = 4 or A = 1, B = 6

Solution:

We have,

96

AB

B

This means that the product of B with itself is either 6 or it has unit’s digit as 6.

The possible values of B are 4 and 6 whose product with itself is a number having 6

at unit’s place.

Taking B = 4 we have

4

4

96

A

Clearly, A = 2

24

4

96

Taking B = 6 we have

6

6

96

A

Clearly, A = 1in this case.

Hence A and B are 2 and 4 respectively or 1 and 6 respectively.

Question 26

If

1

49

B

B

B

then B = _____________.

Answer:

7

Solution:

We have,

1

49

B

B

B

There is a letter B whose value we have to find.

This means that the product of B and 1 isa number whose ones digits is B.

If

1B

is put in the given equation then 11 1 11 doesn’t satisfy the equation.

If

2B

is put in the given equation then 21 2 = 42 doesn’t satisfy the equation.

If

3B

put is in the given equation then 31 3 = 93 doesn’t satisfy the equation.

If

4B

is put in the given equation then 41 4 = 164 doesn’t satisfy the equation.

If

5B

is put in the given equation then 51 5 = 255 doesn’t satisfy the equation.

If

6B

is put in the given equation then 61 6 = 366 doesn’t satisfy the equation.

If

7B

is put in the given equation then 71 7 = 497 satisfies the equation.

71 7 = 497

Hence B = 7

Question 27

1 x 35 is divisible by 9 if x = _____________.

Answer:

09or

Solution:

1 x 35 is divisible by 9 if

1 + x + 3 + 5 is a multiple of 9

9 + x is a multiple of 9

9 + x = 0 or 9 or 18 ………… -------- (i)

But, x is a digit. So, it can take values 0, 1, 2, 3… 9.

And hence, 9 + x can take values 9, 10, 11... 18. ------- (ii)

From (i) and (ii), we get

9 + x = 9 or 18

x = 0 or 9

Hence, x = 0 or 9

Question 28

A four-digit number abcd is divisible by 11, if d + b =_____________ or

_____________.

Answer:

a + c or a + c + 11 or a + c – 11

Solution:

A number is divisible by 11 if the difference between the sum of the digits at its

odd places and that of digits at its even places is either 0 or divisible by 11.

A four-digit number abcd is divisible by 11, if d + b = a + c or a + c + 11 or a + c – 11

as we have to keep difference of d + b and a + c divisible by 11 keeping in mind

that the value of d + b or a + c cannot exceed 18.

Question 29

A number is divisible by 11 if the difference between the sum of digits at its odd

places and that of digits at the even places is either 0 or divisible by ___________.

Answer:

11

Solution:

Let’s write a 3-digit number abc as

100a + 10b + c = 99a + 11b + (a – b + c) = 11 (9a + b) + (a – b + c)

If, (a + c – b), i.e., the difference between the sum of the digits at its odd places

and that of digits at the even places is either divisible by 11 or 0, then the number

abc will be divisible by 11.

Similarly let’s write a 4-digit number abcd as

1000a + 100b + 10c + d

= (1001a + 99b + 11c) – (a – b + c – d)

= 11 (91a + 9b + c) + [(b + d) – (a + c)]

If, [(b + d) – (a + c)], i.e., the difference between the sum of the digits at its

odd places and that of digits at the even places is either divisible by 11 or 0, then

the number abc will be divisible by 11.

Therefore, a number is divisible by 11 if the difference between the sum of digits

At its odd places and that of digits at the even places is either 0 or divisible by 11.

Question 30

If a 3-digit number abc is divisible by 11, then _____________ is either 0 or

multiple of 11.

Answer:

(a + c) – b

Solution:

Let’s write the 3-digit number abc as.

If the number abc is divisible by 11, then (a + c – b), i.e., the difference between

the sum of the digits at its odd places and that of digits at the even places should

either be divisible by 11 or 0.

If a 3-digit number abc is divisible by 11, then (a + c) – b is either 0 or multiple

of 11.

Question 31

If A 3 = 1A then A=_____________.

Answer:

5

Solution:

We have,

A 3 = 1 A ---------------- (i)

A 3 is a number with ones digit A itself.

This is possible only when A 5.

If A 5 is put in the equation (i) then 5 satisfy the equation.

5

Hence A 5

Question 32

If B B = AB, then either A = 2, B = 5 or A = _____________

B

_____________.

Answer:

3, 6

Solution:

We have,

B B = AB ------------------------------ (i)

B can be either 5 or 6.

Putting the value of

2A

and

5B

in equation (i), we get.

Putting the value of

3A

and

6B

in equation (i), we get.

6 6 = 36

Hence, either A = 2, B = 5 or A = 3, B = 6.

Question 33

If the digit 1 is placed after a 2-digit number whose tens digit is t and ones digit is

u, the new number is _____________.

Answer:

t u1

Solution:

If the digit 1 is placed after a 2-digit number whose tens digit is t and ones digit is

u, the new number is tu1.

State whether the statements given in questions 34 to 44 are true (T) or false (F):

Question 34

A two-digit number ab is always divisible by 2 if b is an even number.

Answer:

True

Solution:

A two-digit number ab is always divisible by 2 if b is an even number.

If b is an even number, then b must be a multiple of 2.

There for b can be written as 2n where n = 0 to 4 and n is a natural number.

Therefore, ab = 10a + b = 10a + 2n = 2 (5a + n), which is divisible by 2.

Question 35

A three-digit number abc is divisible by 5 if c is an even number.

Answer:

False

Solution:

A three-digit number abc is divisible by 5 if c is an even number.

This statement is false,

e.g; let the three digit numbers be 224, 246, 268, 286…

In each case the last digit of each number, though, is an even number.

Yet the numbers are not divisible by 5.

Any number is divisible by 5 if its ones digit is either 0 or 5.

Question 36

A four-digit number abcd is divisible by 4 if ab is divisible by 4.

Answer:

False

Solution:

A four-digit number abcd is divisible by 4 if ab is divisible by 4.

This statement is false,

e.g; let four digit number be 4467.

Here ab is divisible by 4, but the number 4467 is not divisible by 4.

Any number is divisible by 4 if the number formed by last two digits is divisible

by 4.

Question 37

A three-digit number abc is divisible by 6 if c is an even number and a + b + c

is a multiple of 3.

Answer:

True

Solution:

A three-digit number abc is divisible by 6 if c is an even number and a + b + c

is a multiple of 3.if c is even, then c can be written as 2k for k = 0 to 4 and k being

a natural number.

We can write abc = 100a + 10b + c = 100a + 10b + 2k.

If c is an even number, then abc is divisible by 2. Further, a + b+ c is a multiple

of 3.

Therefore, the number is divisible by 3.

If a number is divisible by both 2 and 3, it is divisible by 6.

Hence the given Statement is true.

Question 38

Number of the form 3N + 2 will leave remainder 2 when divided by 3.

Answer:

True

Solution:

Number of the form 3N + 2 will leave remainder 2 when divided by 3 is true.

Question 39

Number 7N + 1will leave remainder 1 when divided by 7.

Answer:

True

Solution:

Number 7N + 1will leave remainder 1 when divided by 7.

Question 40

If a number a is divisible by b, then it must be divisible by each factor of b.

Answer:

True

Solution:

If a number a is divisible by b, then it must be divisible by each factor of b.

Question 41

If AB 4 = 192, then A + B = 7.

Answer:

False

Solution:

We have,

A + B =

192

4

= 48

Hence A = 4 and B = 8. So, A + B = 4 + 8 = 12.

So, the given statement is false.

Question 42

If AB + 7C = 102, where B ≠ 0, C ≠ 0, then A + B + C = 14.

Answer:

True

Solution:

We have,

7

1 0 2

AB

C

There are three letters A, B and C whose values we have to find.

The addition of B and C is giving a number whose ones digit is 2.

This is possible when digit B plus C is 12.

In that case, 1 will be the carry-forward to the next step.

In the next step,

So, A = 2

Therefore A + B + C = A + (B + C) = 2 + 12 =14

Question 43

If 213x 27 is divisible by 9, then the value of x is 0.

Answer:

False

Solution:

Since 213x 27 is divisible by 9 is a multiple of 9.

15 + x is a multiple of 9

15 + x = 0 or 9 or 18 ………… -------- (i)

But, x is a digit. So, it can take values 0, 1, 2, 3… 9.

And hence,

15 x

can take values 15, 16, 17, 18... 24. ------- (ii)

if

0,x

then 15 + x = 15 which is not divisible by 9.

Hence,

0.x

So the given statement is false.

Question 44

If N ÷ 5 leaves remainder 3 and N ÷ 2 leaves remainder 0, then N ÷ 10 leaves

remainder 4.

Answer:

False

Solution:

If N ÷ 5 leaves remainder 3 and N ÷ 2 leaves remainder 0, then

10N

leaves

remainder 4.

Let’s take N = 18.

Does not satisfy the statesmen.

So, the Statement is False.

Solve the following:

Question 45

Find the least value that must be given to number a so that the number 91876a2 is

divisible by 8.

Answer:

3

Solution:

If the last three digits of a whole number are divisible by 8, then the entire number

is divisible by 8.

For 91876a2 to be divisible by 8, 6 a2 is divisible by 8.

Or, 6 a2 = 600 +10a + 2 = 8(75) + 10a + 2 is divisible by 8.

Or,

10 2a

is divisible by 8.

For this, a has to be either 3 or 7.

The least value of a is 3.

Question 46

If

1

6

p

p

Q

where Q – P = 3, then find the values of P and Q.

Answer:

6 and 9

Solution:

We have,

1

6

p

p

Q

There are two letters p and Q whose values we have to find.

This means that the product of P with itself is a number whose ones digits is 6.

P can be either 6 or 4.

In the next step, for

6P

, we have.

Satisfies this as 3 + 6 1 9

Therefore Q = 9.

For

4,P

we have.

1 + P 1 = Q. Q = 5satisfies this as 1 + 4 1 = 5.

Therefore Q = 5.

But in this case Q – p is not 3.

So the puzzle can be solved as shown below.

16

6

96

Hence, the values of P and Q are 6 and 9 respectively.

Question 47

If 1AB + CCA = 697 and there is no carry–over in addition, find the value of

A + B + C.

Answer:

12

Solution:

We have,

1AB + CCA = 697 i.e.

1

6 9 7

AB

C C A

There are three letters A, B and C whose values we have to find.

Since there is no carry-over, addition in unit’s place, ten’s place or hundred’s place

can be done in any order. Let’s take the addition in the hundred’s place.

1 + C = 6

C = 5

Taking the value of

5C

for the addition in the ten’s place, we get

A + 5 =9

A = 4

Taking the value of

4A

for the addition in the unit’s place, we get

B = 3

Therefore, A + B + C = 5 + 4 + 3 = 12

Question 48

A five-digit number AABAA is divisible by 33.

Write all the numbers of this form.

Answer:

33033, 66066, 99099

Solution:

AABAA must be divisible by 11 and 3.

If a number is divisible by 11, then the difference between the sum of digits at its

odd places and that of digits at the even places is either 0 or divisible by 11.

If a number is divisible by 3, then the sum of its digits is divisible by 3.

Since, AABAA to be divisible by 11 therefore (A + B + A) – (A + A) = B

should be divisible by 11.

So B is bound to be 0.

Since, AABAA to be divisible by 3, therefore, A + A + B + A + A = 4A + B

must be divisible by 3.

Since B = 0 , 4A should be divisible by 3.

Therefore, A could be 3, 6 or 9.

Question 49

Find the value of the letters in each of the following questions.

AA

AA

X A Z

.

Answer:

A = 9, Z = 8, X = 1

Solution:

There are three letters A, X and Z whose values we have to find.

The addition of A and A is a number whose ones digit is Z .In the second step

AA

gives A, a number whose ones digit is A.

Keeping in mind the placement of r A’s, the above addition is possible only when

digit A is 9.

In that case, the addition of 9 and 9 will give 18 and thus, 1 will be the carry- over

for the next step.

In the next step, a number with ones digit A.

Clearly, A is 9 as 9 + 9 = 18 and 1 will be a carry- over for the next step.

In the next step,

1 x

So the puzzle can be solved as shown below.

Clearly, Z is 8.

Hence, A, X and Z are 9, 1 and 8 respectively.

Question 50

85

4

3

A

BC

Answer

A = 8, B = 1, C = 3

Solution:

There are three letters A, B and C whose values we have to find.

The addition of 5 and A is a number whose ones digit is 3.

This is possible only when digit A is 8.

In that case, the addition of A (8) and 5 will give 13 and thus, 1 will be the

carry - over for the next step.

In the next step,

1 + 8 + 4 = 13

So the puzzle can be solved as shown below.

85

4

3

A

BC

Clearly, C is 3 and B is 1

Hence, A, B and C are 8, 1 and 3 respectively.

Question 51

6

8

2

B

A

CA

Answer:

A = 6, B = 7, C = 1

Solution:

There are three letters A, B and C whose values we have to find.

The addition of 6 and A is a number whose ones digits is 2.

This is possible only when the digit A is 6.

In that case, the addition of 6 and A will give 12 and thus, 1 will be the

carry- over for the next step.

In the next step,

1 + B + 8 = a number with ones digit A.

Clearly, B is 7 as 1 + 7 + 8 = 16 and 1 will be a carry-over for the next step.

In the next step,

So the puzzle can be solved as shown below.

76

86

1 6 2

Hence, the values of A, B and C are 6, 7 and 1 respectively.

Question 52

1

82

BA

A B A

B

Answer:

A = 6, B = 9

Solution:

There are two letters A and B whose values we have to find.

The addition of A and A is a number whose ones digits is 2.

This is possible only when the digit A is 6. In that case, the addition of 6 and 6

will give 12 and thus, 1 will be the carry-over for the next step.

In the next step,

A number with unit’s digit B.

Clearly, B is 9 as 1 + 9 + 9 = 19 and 1 will be a carry-over for the next step.

In the next step,

1 + 1 + A = 8 Clearly, A is 6 as1 + 1 + 6 = 8

So the puzzle can be solved as shown below.

1 9 6

6 9 6

8 9 2

Hence, the values of A and B are 6 and 9 respectively.

Question 53

1 3 0

C B A

C B A

A

Answer:

A = 5, B = 6, C = 7

Solution:

There are three letters A, B and C whose values we have to find.

The addition of A and A is a number whose ones digits is 0.

This is possible only when digit A is 5.

In that case, the addition of 5 and 5 will give 10 and thus, 1 will be the carry-over

for the next step.

In the next step,

1 + B + B = a number with unit’s digit as 3.

Clearly, B is 6 as 1 + 6 + 6 = 13 and 1 will be the carry-over for the next step.

In the next step,

1 + C + C = a number with the unit’s digit A.

1 + C + C = a number with the unit’s digit 5.

Clearly, C is 7 as 1 + 7 + 7 = 15 and 1 will be the carry-over for the next step.

In the next step,

1 1

So the puzzle can be solved as shown below.

7 6 5

7 6 5

1 5 3 0

Hence, the values of A, B and C are 5, 6 and 7 respectively.

Question 54

38

B A A

B A A

A

Answer:

A = 9, B = 1

Solution:

There are two letters A and B whose values we have to find.

The addition of A and A is a number whose ones digits is 8.

This happens only for

4A

and

9.A

If

4A

, then the sum is4 + 4 = 8, in next step A + A = a number with ones

digit.

A then 4 + 4 = 8 which makes

8A

too, which is not possible.

We reject this possibility.

If

9.A

In that case, the addition of 9 and 9 will give 18 and thus, 1 will be the carry-over

for the next step.

In the next step,

1 + A + A = a number with unit’s digit A.

Clearly, A is 9 as 1 + 9 + 9 = 19 and 1 will be carry-over for the next step.

In the next step,

Clearly, B is 1 as 1 + 1 + 1 = 3

So the puzzle can be solved as shown below.

1 9 9

1 9 9

3 9 8

Hence, the values of A and B are 9 and 1 respectively.

Question 55

01

10

1 0 8

AB

AB

B

Answer

A = 8, B = 9

Solution:

There are two letters A and B whose values we have to find.

The addition of B and B is a number whose ones digits is 8.

This happens only for B = 4 and B = 9.

If

4,B

then the sum is 4 + 4 = 8, in next step 1 + A = a number with one’s

digit 0 then 1 + 9 = 10 which makes A = 9.

In next step 1 + 0 + 0 = 1 and then in next step A +1 must give B but 9 + 1

is not 4.

We reject this possibility.

The addition of 9 and 9 will give 18 and thus, 1 will be the carry-over for the

next step.

In the next step,

A number with unit’s digit as 0

Clearly, A is 8 as 1 + 1 + 8 = 10 and 1 will be carry-over for the next step.

In the next step,

1 + 0 + 0 = 1In the next step,

A + 1 = B as 8 + 1 – 9

So the puzzle can be solved as shown below.

8 0 1 9

1 0 8 9

9 1 0 8

Hence, the values of A and B are 8 and 9 respectively.

Question 56

6

68

AB

C

Answer:

A = 7, B = 8, C = 4

Solution:

We have,

6

68

AB

C

There are three letters A, B and C whose values we have to find.

This means that the product of B and 6 is a number whose ones digits is 8.

Since

8B

satisfies B 6 = a number with ones digit 8 as 8

Therefore

8.B

4 will be the carry-over for the next step.

In the next step the product of A and 6 must give a number with ones digit 6.

Since

7A

, satisfies Therefore A = 7

4 will be the carry-over for the next step.

In the next step.

C = 4

So the puzzle can be solved as shown below.

6

68

AB

C

Hence, the values of A, B and C are 7, 8 and 4 respectively.

Question 57

6

AB

AB

AB

Answer:

A = 2, B = 5

Solution:

Here, we have to find the values of A and B.

Since ones digit of

BB

is B. Therefore B = 0 or B =5

Now, AB AB = 6 AB ---------------------- (i)

The square of a two digit number is a three digit number.

So, A can take values 1, 2 and 3.

We find that A = 2, B = 5 satisfies equation (i).

i.e. 25 25 = 625.

A = 2 and B = 5

No other pair of values of A and B satisfy the equation (i)

Hence, A = 2 and B = 5

Question 58

AA

A

C A B

Answer:

9, 1, 8A B C

Solution:

We have,

AA

A

C A B

There are three letters A, B and C whose values we have to find.

This means that the product of A and A is a number whose ones digits is B fulfils

the requirement.

8 will be the carry-over for the next step. In the next step the product of A and A

gives a number whose ones digits is A.

Since

9A

satisfies 8 + A A A as 8 + 9 9 = 89.

8 will be the carry-over for the next step.

In the next step.

8C

So the puzzle can be solved as shown below.

99

9

8 9 1

Hence, the values of A, B and C are 9, 1 and 8 respectively.

Question 59

7

45

AB

B

Answer:

A = 7, B = 2

Solution:

There are two letters A and B whose values we have to find.

The subtraction of 7 from B is giving a number whose ones digits is 5.

This is possible only when digit B is 2. In that case, the subtraction of 7 from 12

will give 5 and thus, 1 will be the borrow from the next step.

A number with unit’s digit 5.

Clearly, 12 – 7 = 5, hence B is 2.

In the next step,

A – B – 1 = 4

Clearly, A is 7 as 7 – 2 – 1 = 4

So the puzzle can be solved as shown below.

72

27

45

Hence, the values of A and B are7and 2 respectively.

Question 60

8

5

4 8 8

A B C

A B C

D

Answer:

A = 7, B = 2, C = 3, D = 1

Solution:

There are four letters A, B, C and D whose values we have to find.

The subtraction of 5 from C is a number whose ones digits is 8.

This is possible only when digit C is 3.

In that case, the subtraction of 5 from C will give 8 and thus, 1 will be the borrow

from the next step.

Clearly, 13 – 5 = 8. C = 3.

In the next step,

A number with unit’s digit 8.

i.e. B – 4 = a number with unit’s digit 8.

The subtraction of 4from B is giving a number whose ones digits is 8.

This is possible only when digit B is 2.

In that case, the subtraction of 4from B will give 8 and thus, 1 will be the borrow

for the current step to the next step.

Clearly, 12 – 4 = 8, hence

2B

.

In the next step,

A – B – 1 = a number with unit’s digit4.

A – 2 –1 = a number with unit’s digit4.

The subtraction of 3 from A is giving a number whose ones digits is 4.

This is possible only when digit A is 7.

In that case, the subtraction of 2 from A will give 4 and thus,

A – 2 –1 = a number with unit’s digit4.

Clearly, A is 7 as 7 – 2 – 1 = 4

In the next step,

8 – 7 = 1

So the puzzle can be solved as shown below.

8 7 2 3

7 2 3 5

1 4 8 8

Hence, the values of A, B, C and D are 7, 2, 3and 1 respectively.

Question 61

If 2A7 ÷ A = 33, then find the value of A.

Answer:

9

Solution:

Given

2A7 ÷ A = 33,

(200 + 10A + 7) ÷ A =33

200 + 10A + 7 = 33A

Hence,

9A

Question 62

212 x 5 is a multiple of 3 and 11.

Find the value of x.

Answer:

8

Solution:

If a number is divisible by 11, then the difference between the sum of digits at its

odd places and that of digits at the even places is either 0 or divisible by 11.

If a number is divisible by 3, then the sum of its digits is divisible by 3.

It is given that 212 x 5 is multiple of 3 and 11

Is a multiple of 3 and 9 – x – 1 is divisible by 11.

10 + x is a multiple of 3 and 8 – x is divisible by11.

Hence x can take the value of 8.

Question 63

Find the value of k where 31k 2 is divisible by 6.

Answer:

k is either 0 or 3 or 6 or 9

Solution:

Since 31k 2 is divisible by 6

∴ 2y – 10 is a multiple of 3

= 6 + k is a multiple of 3

Hence, k is either 0 or 3or 6 or 9

Question 64

1y3y6 is divisible by 11. Find the value of y.

Answer:

5

Solution:

If a number is divisible by 11, then the difference between the sum of digits at its

odd places and that of digits at the even places is either 0 or divisible by 11.

Since 1y3y6 is divisible by 11.

∴ 2y – 10 is a multiple of 11.

This is possible only if 2y – 10 = 0

Or,

5y

Hence, y is 5

Question 65

756 x is a multiple of 11, find the value of x.

Answer:

8

Solution:

If a number is divisible by 11, then the difference between the sum of digits at its

odd places and that of digits at the even places is either 0 or divisible by 11.

It is given that 756 x is multiple of 11.

∴ 5 + x – 13 is a multiple of 11.

This is possible if 5 + x – 13 = 0

Or, x = 8

Hence x is 8

Question 66

A three-digit number 2a3 is added to the number 326 to give a three-digit number

5b9 which is divisible by 9.

Find the value of

–.ba

Answer:

2

Solution:

We have,

A three digit number 2a3 which is added to the number 326 giving a three digit

number 5b9 which is divisible by 9.

23

3 2 6

59

a

b

5 + b + 9 is divisible by 9, therefore b must be 4. Hence, a must be 2.

So, b – a = 4 – 2 = 2

Question 67

Let E = 3, B = 7 and A =4

Find the other digits in the sum

B A S E

B A L L

G A M E S

Answer:

S = 8, L = 5, M = 9, G = 1

Solution:

Putting the values of E, B and A in the sum

7 4 3

74

43

S

LL

G M S

There are four letters S, L, M and G whose values we have to find.

The addition of 3 and L is a number whose ones digits is S.

There are two possibilities; 3 + L = S or 3 + L = 10 + S

In the second step.

The addition of S and L is a number whose ones digits is 3.

Here again there are four possibilities;

S + L = 3 or S + L + 1 = 3 or S + L = 10 +3 or S + L + 1 = 10 + 3

Considering that each of the digits are from 0 to 9 and the maximum value of the

addition of two digits will be 18and keeping in mind the placement of L and S, the

above addition is possible only for.

5L

and S = 8. In the next step.

SL

a number with ones digit 3.

8 + 5 = 13 and 1 will be the carry-over for the next step.

In the next step,

1 4 4

a number with the unit’s digit M. M = 9 as 1 + 4 + 4 = 9.

In the next step,

7 7

a number with the unit’s digit 4 and as 7 + 7 = 14 and 1 will be a

carry-over for the next step.

In the next step,

1G

So the puzzle can be solved as shown below.

7 4 8 3

7 4 5 5

1 4 9 3 8

Hence, the values of S, L, M and G are 8, 5, 9 and 1 respectively.

Question 68

Let D = 3, L =7 and A =8.

Find the other digits in the sum

M A D

AS

A

B U L L

Answer:

S = 6, M = 9, B = 1, U = 0

Solution:

Putting the values of D, L and A in the given sum

83

8

8

77

M

S

BU

There are four letters S, M, B and U whose values we have to find.

The addition of 3, M and 8 is giving a number whose ones digits is 7.

Here 1will be the carry-over to the next step as

88

has been shown as 17.

This happens only for S = 6as 3 + 6 + 8 = 17 and 1 will be the carry-over for

the next step.

In the next step,

1 + A + A = a number with ones digit 7.

Clearly, A is 8 as 1 + 8 + 8 = 17 and 1 will be the carry-over for the next step.

In the next step,

1 + M = BU

So the puzzle can be solved as shown below.

983

86

8

1 0 7 7

Hence, the values of S, M, B and U are 6, 9, 1and 0 respectively

Question 69

If from a two-digit number, we subtract the number formed by reversing its digits

then the result so obtained is a perfect cube.

How many such numbers are possible?

Write all of them.

Answer:

96, 85, 74, 63, 52, 41, 30

Solution:

Case 1: If the tens digit is larger than the ones digit (that is,

ab

) then

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

= 9a – 9b = 9 (a – b).

Case 2: If the ones digit is larger than the tens digit (that is,

ba

) then,

(10b + a) – (10a + b) = 9 (b – a).

Since, the difference is a perfect cube, in the above two cases,

(b – a) or (a – b) = 3

Case 3: If

ab

, the difference = 0, which is not a perfect cube.

Hence, the numbers are 96, 85, 74, 63, 52, 41and30.

Question 70

Work out the following multiplication.

1 2 3 4 5 6 7 9

9

─────────

Use the result to answer the following questions.

(a) What will be12345679 45?

(b) What will be12345679 63?

(c) By what number should 12345679 be multiplied to get

888888888?

(d) By what number should 12345679 be multiplied to get

999999999?

Answer:

(a) 555555555

(b) 777777777

(c) 72

(d) 81

Solution:

Given,

1 2 3 4 5 6 7 9

9

─────────

The product of 9 9 = 81so, a number whose ones digit is 1 and 8 will the

carry-over for the next step.

In the next step

9 so, a number whose tens digit is 1and 7 will the

carry-over for the next step.

In the next step

9 6 + 7 = 54 + 7 = 61 so, a number whose hundredths digit is 1and 6 will

the carry-over for the next step.

In the next step

9 5 + 6 = 45 + 6 = 51 so, a number whose thousands digit is 1and 5 will

the carry-over for the next step.

In the next step

9 4 + 5 = 36 + 5 = 41 so, a number whose ten thousands digit is 1 and 4

will the carry-over for the next step.

In the next step

9 3 + 4= 27 + 4 = 31 so, a number whose lakhs digit is 1and 3 will the

carry-over for the next step.

In the next step

9 2 + 3 = 18 + 3 = 21 so, a number whose ten lakhs digit is 1and 2 will the

carry-over for the next step.

In the next step

9 1 + 2 = 9 + 2 =11 so, a number whose crores digit is 1 and 1 will

carry-over for the next step.

In the next step 1

Multiplication of 12345679 and 9 is 111111111

12345679 45 = 111111111 5 = 555555555

12345679 63 = 111111111 7 = 777777777

12345679 72 = 111111111 8 = 888888888

12345679 81 = 111111111 9 = 999999999

Question 71

Find the value of the letters in each of the following:

6

PQ

Q Q Q

(ii)

2

1

18

LM

LM

M

Answer:

(i) P = 7, Q = 4

(ii) M = 7, L = 4

Solution:

(i) We have,

6

PQ

Q Q Q

There are two letters P and Q whose values we have to find.

This means that the product of Q and 6 is a number whose ones digits is Q.

The possible values of Q are 2, 4, 6 and 8.

Putting these values, one by one we find that Q = 4 satisfies Q 6 = a number

with unit’s digit Q as 4 6 = 24.Therefore Q = 4.

2 will be the carry-over for the next step. In the next step,

Since P = 7 satisfies 2 + P 6 = a number with ones digit Q as 2 + 7 6 = 44.

4 will be the carry-over for the next step.

In the next step Q = 4

So the puzzle can be solved as shown below.

74

6

4 4 4

Hence, the values of P and Q are 7 and 4 respectively

(ii)

2

1

18

LM

LM

M

There are two letters L and M whose values we have to find.

The addition of M and 1 is giving a number whose ones digits is 8.

This happens only for

7.M

In that case, the addition of 7 and 1 will give 8 as 7 + 1 = 8.

In the next step L and M when added gives a number with unit’s digit1.

This happens only for

4L

.

In that case, the addition of 4 and 7 will give 11.

Clearly, L is 4 as 4 + 7 = 11 and 1 will be carry-over for the next step.

In the next step,

1 + 2 + L = M

Clearly, L is 4 as 1 + 2 + 4 = 7

So the puzzle can be solved as shown below.

2 4 7

4 7 1

7 1 8

Hence, the values of L and M are 4 and 7 respectively

Question 72

If 148101B095 is divisible by 33, find the value of B.

Answer:

4

Solution:

Since 148101B095 is divisible by 33, it is divisible by both 11 and 3.

If a number is divisible by 11, then the difference between the sum of digits at its

odd places and that of digits at the even places is either 0 or divisible by 11.

1 + 4 + 8 + 1 + 0 + 1 + B + 0 + 9 + 5 is a multiple of 3

29 + B is a multiple of 3. Also, 18 + B – 11 is divisible by 11

Hence, B is 4.

Question 73

If 123123A4 is divisible by 11, find the value of A.

Answer:

4

Solution:

If a number is divisible by 11, then the difference between the sum of digits at its

odd places and that of digits at the even places is either 0 or divisible by 11.

Since 123123A4 is divisible by 11,

6 + A – 10 = A – 4 is divisible by 11

This means A – 4 = 0 or A = 4

Hence, A is 4.

Question 74

If 56x32y is divisible by 18, find the least value of y.

Answer:

Least value of y is 0

Solution:

Since 56x32y is divisible by 18, it is divisible by both 9 and 2.

If a number is divisible by 9, then the sum of its digits is divisible by 9.

5 + 6 + x + 3 + 2 + y is a multiple of 9

16 + x + y is a multiple of 9. Also, y is even.

If y = 0 and x =2, 16 + x + y = 18 is divisible by 9

Hence, minimum value of y can be 0