# Class 8 RD Sharma Solutions – Chapter 5 Playing with Numbers – Exercise 5.3

**Solve each of the following Cryptarithms:**

**Question 1.**

### 3 7

__+ A B __

9 A

**Solution:**

In the unit’s place,

7 + B = A

In the ten’s place,

3 + A = 9

From ten’s place A = 6 then B = −1

Hence, it is not possible

Thus, there should be one carry in ten’s place

7 + B > 9

Solve for ten’s place with one carry,

3 + A + 1 = 9

A = 9−1−3 = 5

In unit’s place subtracting 10 as one carry is given to ten’s place,

7 + B – 10= 5

B = 5 + 10 − 7 = 8

Therefore,

A = 5 and B = 8

**Question 2.**

### A B

__+ 3 7 __

9 A

**Solution:**

In the unit’s place,

B + 7 = A

In the ten’s place,

A + 3 = 9

From ten’s place A = 6 and B = −1

It is not possible.

Thus there should be one carry in ten’s place, which means B + 7 > 9

Solve for ten’s place with one carry,

A + 3 + 1 = 9

A = 9 − 4 = 5

In unit’s place subtracting 10 as one carry is given to ten’s place,

B + 7 – 10 = 5

B = 5 + 10 − 7 = 8

Therefore,

A = 5 and B = 8

**Question 3.**

### A 1

__+ 1 B __

B 0

**Solution:**

In the unit’s place,

1 + B = 0

Thus B = -1

Which is not possible.

Thus, there should be one carry in ten’s place,

A + 1 + 1 = B —-> (equation 1)

For unit’s place, we need to subtract 10 as one carry is given in ten’s place,

1 + B – 10 = 0

B = 10 − 1 = 9

Substituting B = 9 in (equation 1),

A + 1 + 1 = 9

A = 9 − 1 − 1 = 7

Therefore, A = 7 and B = 9

**Question 4.**

### 2 A B

__+ A B 1 __

B 1 8

**Solution:**

In the unit’s place,

B + 1 = 8

B = 7

In the ten’s place,

A + B = 1

A + 7 = 1

A = −6

Which is not possible.

Hence, A + B > 9

We know that now there should be one carry in hundred’s place, and so we need to subtract 10 from ten’s place,

That is,

A + B – 10 = 1

A + 7 = 11

A = 11 − 7 = 4

Now to check whether our values of A and B are correct, we should solve for hundred’s place.

2 + A + 1 = B

2 + 4 + 1 = 7

7 = 7

Hence,

RHS = LHS

Therefore,

A = 4 and B = 7

**Question 5.**

### 1 2 A

__+ 6 A B __

A 0 9

**Solution:**

In the unit’s place,

A + B = 9 —-> (equation 1)

With this condition we know that sum of 2 digits can be greater than 18.

So, there is no need to carry one from ten’s place.

In the ten’s place,

2 + A = 0

Which means A = −2

Which is not possible

Hence,

2 + A > 9

Now, there should be one carry in hundred’s place and hence we need to subtract 10 from ten’s place,

That is,

2 + A – 10 = 0

A = 10 − 2 = 8

Now, substituting A=8 in (equation 1),

A + B = 9

8 + B = 9

B = 9 – 8

B = 1

Therefore,

A = 8 and B = 1

**Question 6.**

### A B 7

__+ 7 A B __

A 0 9

**Solution:**

In the unit’s place,

The two conditions are,

(i) 7 + B ≤ 9

(ii) 7 + B > 9

For 7 + B ≤ 9

7 + B = A

A – B = 7 —->(equation 1)

In the ten’s place,

B + A = 8 —-> (equation 2)

Solve equation 1 and equation 2,

2A = 15 which means A = 7.5

Which is not possible

Thus, first condition 7 + B ≤ 9 is wrong.

Therefore,

7 + B > 9 is correct condition

Hence, there should be one carry in ten’s place and subtracting 10 from unit’s place,

7 + B – 10 = A

B – A = 3 —-> (equation 3)

In the ten’s place,

B + A + 1 = 8

B + A = 8 − 1

B + A = 7 —-> (equation 4)

Solve equation 3 and equation 4,

2B = 10

B =10= 5

_{ }2

Substituting the value of B in equation 4

B + A = 7

5 + A = 7

A = 7 − 5 = 2

Therefore,

B = 5 and A = 2

**Question 7. Show that the Cryptarithm does not have any solution.**

**Solution:**

If B is multiplied by 4 then only 0 satisfies the above condition.

So, for unit place to satisfy the above condition, we should have B = 0.

In ten’s place, only 0 satisfies the above condition.

But, AB cannot be 00 as 00 is not a two digit number.

So, A and B cannot be equal to 0

Therefore,

There is no solution satisfying the condition