Lesson: Playing with Numbers

Exercise 16.1

Find the values of the letters in each of the following and give reasons for the

steps involved.

Question 1

3

25

2

A

B

Answer:

7, 6AB

Solution:

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

The addition of A and 5 is giving a number whose ones digit is 2.

This is possible only when digit A is 7.

In that case, the addition of A (7) and 5 will give 12 and thus, 1 will be the carry-over

for the next step. In the next step,

1 3 2 6

Clearly, B is 6

So the puzzle can be solved as shown below.

37

25

62

Hence, A and B are 7 and 6 respectively

Question 2

4

98

3

A

CB

Answer:

5, 4, 1A B C

Solution:

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

The addition of A and 8 is giving a number whose ones digit is 3.

This is possible only when digit A is 5.

In that case, the addition of A and 8 will give 13 and thus, 1 will be the carry-over

for the next step. In the next step,

1 4 9 14

So the puzzle can be solved as shown below.

45

98

1 4 3

Clearly, B and C are 4 and 1 respectively.

Hence, A, B, and C are 5, 4, and 1 respectively.

Question 3

1

9

A

A

A

Answer:

6A

Solution:

There is just one letter A whose value we have to find.

The multiplication of A with A itself gives a number whose ones digit is A again.

This happens only when

1, 5, 6.A or

If

1,A

then the multiplication will be

11 1 11

.

However, here, the tens digit is given as 9.

Therefore,

1A

is not possible.

Similarly, if

5,A

then the multiplication will be

15 5 75.

Thus,

5A

is also not possible.

If we take

6,A

then

16 6 96.

Therefore, A should be 6.

So the puzzle can be solved as shown below.

16

6

96

Hence, the value of A is 6

Question 4

37

6

AB

A

Answer:

2, 5AB

Solution:

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

The addition of A and 3 is giving 6. There can be two cases.

(1) First step is not producing a carry-over

In that case, A comes to be 3 as

3 3 6.

Considering the first step in which the

addition of B and 7 is giving A (i.e., 3), B should be a number such that the unit

’s digit of this addition comes to be 3. It is possible only when

6B

. In this

case,

6 7 13.A

However, A is a single digit number. Hence, it is not

possible.

(2) First step is producing a carry-over

In that case, A comes to be 2 as

1 2 3 6.

considering the first step in

which the addition of B and 7 is giving A (i.e., 2), B should be a number such

that the unit’s digit of this addition comes to be 2. It is possible only when

5B

and

5 7 12.

So the puzzle can be solved as shown below.

25

37

62

Hence, the values of A and B are 2 and 5 respectively.

Question 5

3

AB

C A B

Answer:

5, 0, 1A B C

Solution:

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

The multiplication of 3 and B gives a number whose ones digit is B again.

Hence, B must be 0 or 5.

Let B be 5.

Multiplication of first the step

3 5 15

1 will be a carry-over for the next step.

We have,

3 1 A

a two digit number CA (which is, in fact,

10CA

).

This is not possible for any value of A. Hence, B must be 0 only. If

0,B

then there

will be no carry-over for the next step.

We should obtain,

3 ,A CA

which is in fact

10CA

.

That is, the ones digit of

3 A

should be A.

This is possible when either

5 0.A or

However, A cannot be 0.

Therefore, A must be 5 only.

So the puzzle can be solved as shown below.

50

3

1 5 0

Hence, the values of A, B, and C are 5, 0, and 1 respectively.

Question 6

5

AB

C A B

Answer:

There are 3 sets of possible values of A, B, and C.

(i) 5, 0, and 2 respectively

(ii) 2, 5, and 1 respectively

(iii) 7, 5, and 3 respectively

Solution:

There are 3 sets of possible values of A, B, and C.

(i) 5, 0, and 2 respectively

(ii) 2, 5, and 1 respectively

(iii) 7, 5, and 3 respectively

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

The multiplication of B and 5 is giving a number whose ones digit is B again.

This is possible when

5 0B or B

only.

In case of

5,B

the product,

5 5 5 25B

2 will be the carry-over for

the next step.

We have,

5 2 A CA

(which is, in fact,

10CA

). This is possible

for

2 7A or

The multiplication is as follows.

25

5

1 2 5

75

5

3 7 5

If

0,B

5 BB

0 5 0

There will not be any carry-over in this step.

In the next step,

5 A CA

(this is, in fact,

10CA

)

It can happen only when

5 0A or A

However, A cannot be 0 as AB is a two-digit number.

Hence, A can be 5 only.

So the puzzle can be solved as shown below.

50

5

2 5 0

Hence, there are 3 possible values of A, B, and C.

(i) 5, 0, and 2 respectively

(ii) 2, 5, and 1 respectively

(iii) 7, 5, and 3 respectively

Question 7

6

AB

B B B

Answer:

7, 4AB

Solution:

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

The multiplication of 6 and B gives a number whose ones digit is B again.

It is possible only when

0, 2, 4, 6, 8B or

If

0,B

then the product will be 0. Therefore, this value of B is not possible.

If

2,B

then

6 12B

and 1 will be a carry-over for the next step.

6 1 A BB

(which is

10BB

)

22

6 21A

and hence, any integer value of A is not possible.

If

6,B

then

6 36B

and 3 will be a carry-over for the next step.

6 3 A BB

(which is

10BB

)

66

6 63A

and hence, any integer value of A is not possible.

If

8,B

then

6 48B

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

6 4 A BB

(which is

10BB

)

88

6 84A

and hence,

14A

. However, A is a single digit number.

Therefore, this value of A is not possible.

If

4,B

then

6 24B

and 2 will be a carry-over for the next step.

6 2 A BB

(which is

10BB

)

44

6 42A

and hence,

7A

So the puzzle can be solved as shown below.

74

6

4 4 4

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

Question 8

1

1

0

A

B

B

Answer:

7, 9AB

Solution:

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

The addition of 1 and B is giving a number whose ones digit is 0.

This is possible only when digit B is 9.

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

for the next step.

In the next step,

1 1 AB

Clearly,

7 1 7 1 9 A is as B

So the puzzle can be solved as shown below.

71

19

90

Hence, the values of A and B are 7 and 9 respectively.

Question 9

2

1

18

AB

AB

B

Answer:

4, 7AB

Solution:

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

The addition of B and 1 is giving a number whose ones digit is 8.

This is possible only when digit B is 7.

In that case, the addition of B and 1 will give 8. In the next step,

AB

is giving

a number with one’s digit 1.

Clearly, A is 4.

4 7 11

and 1 will be a carry-over for the next step.

In the next step,

1 2 AB

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 A and B are 4 and 7 respectively.

Question 10

12

6

09

A

AB

A

Answer:

8, 1AB

Solution:

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

The addition of A and B is giving 9, i.e., a number whose ones digit is 9.

The sum can be 9 only as the sum of two single digit numbers cannot be 19.

Therefore, there will not be any carry-over in this step.

In the next step,

2 A

giving a number with one’s digit 0

It is possible only when

8A

2 8 10

and 1 will be the carry-over for the next step.

1 1 6 A

Clearly, A is 8.

We know that the addition of A and B is giving 9.

As A is 8,therefore, B is 1.

So the puzzle can be solved as shown below.

1 2 8

6 8 1

8 0 9

So, the values of A and B are 8 and 1 respectively.

Exercise16.2

Question 1

If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Answer:

1y

Solution:

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

Sum of the digits of

21 5 2 1 5 8 y y y

Hence,

8 y

should be a multiple of 9.

This is possible when

8 y

is any one of these numbers 0, 9, 18, 27, and so on …

However, since y is a single digit number, this sum can be 9 only.

So, y should be 1 only.

Question 2

If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

You will find that there are two possible values of z. Why is this so?

Answer:

0 9z or

Solution:

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

Sum of the digits of

31 5 3 1 5 9 z z z

Hence,

9 z

should be a multiple of 9.

This is possible when

9 z

is any one of these numbers 0, 9, 18, 27, and so on …

However, since z is a single digit number, this sum can be either 9 or 18. So, z should

be either 0 or 9.

Question 3

If 24xis a multiple of 3, where x is a digit, what is the value of x?

Answer:

Solution:

Since 24x is a multiple of 3, the sum of its digits is a multiple of 3.

Sum of the digits of

24 2 4 6 x x x

Hence,

6 x

is a multiple of 3.

This is possible when

6 x

is any one of these numbers 0, 3, 6, 9, and so on …

Since x is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and

Thus, the value of x comes to 0 or 3 or 6 or 9 respectively.

Question 4

If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Answer:

0, 3, 6 9z or

Solution:

Since 31z5 is a multiple of 3, the sum of its digits will be a multiple of 3.

That is,

3 1 5 9 zz

is a multiple of 3.

This is possible when

9 z

is any one of 0, 3, 6, 9, 12, 15, 18, and so on …

Since z is a single digit number, the value of

9 z

can only be 9 or 12 or 15 or 18

and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.

So, z can have its value as any one of the four different values 0, 3, 6, or 9.