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

Class 8 RD Sharma Solutions - Chapter 5 Playing with Numbers - Exercise 5.3 explores number theory concepts that help develop a deeper understanding of the numerical properties and operations. This exercise focuses on the various techniques to simplify and solve problems involving numbers, patterns, and their properties.

Playing with Numbers

In this chapter, students learn to manipulate and understand numbers through the various operations and patterns. The exercises involve working with prime numbers, factors, multiples, and their applications in solving real-world problems. This approach not only sharpens arithmetic skills but also enhances logical reasoning and problem-solving abilities.

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.

4 \times \overline{AB} = \overline{CAB}

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

 4 \times \overline{AB} = \overline{CAB}

Conclusion

Exercise 5.3 in Chapter 5 provides the valuable practice in applying number theory concepts to the solve complex problems. By working through these exercises students gain a better grasp of the numerical patterns and operations preparing them for the more advanced mathematical challenges. Mastery of these skills is crucial for the building a strong foundation in the mathematics.