myhelper: RD Sharma solution class 8 chapter 5 Playing with numbers Exercise 5.3

Exercise 5.3

Page-5.30

Question 1:

Solve each of the following Cryptarithms:
$\begin{matrix} 3 & 7 \\ + & A & B \end{matrix} \begin{matrix} 9 & A \end{matrix}$

Answer 1:

$Two possible values of A are : (i) If 7 + B \leq 9 ∴ 3 + A = 9 \Rightarrow A = 6 But if A = 6, 7 + B must be larger than 9 . Hence, it is impossible . (ii) If 7 + B \geq 9 ∴ 1 + 3 + A = 9 \Rightarrow A = 5 If A = 5 and 7 + B = 5, B must be 8 ∴ A = 5, B = 8$

Question 2:

Solve each of the following Cryptarithm:
$\begin{matrix} A & B \\ + 3 & 7 \end{matrix} \bar{\begin{matrix} 9 & A \end{matrix}}$

Answer 2:

$Two possibilities of A are : (i) If B + 7 \leq 9, A = 6 But clearly, if A = 6, B + 7 \geq 9; it is impossible (ii) If B + 7 \geq 9, A = 5 and B + 7 = 5 Clearly, B = 8 ∴ A = 5, B = 8$

Question 3:

Solve each of the following Cryptarithm:
$\begin{matrix} A & 1 \\ + & 1 & B \end{matrix} \begin{matrix} B & 0 \end{matrix}$

Answer 3:

$If 1 + B = 0 Surely, B = 9 If 1 + A + 1 = 9 Surely, A = 7$

Question 4:

Solve each of the following Cryptarithm:
$\begin{matrix} 2 & A & B \\ + & A & B & 1 \end{matrix} \bar{\begin{matrix} B & 1 & 8 \end{matrix}}$

Answer 4:

$B + 1 = 8, B = 7 A + B = 1, A + 7 = 1, A = 4 So, A = 4, B = 7$

Question 5:

Solve each of the following Cryptarithm:
$\begin{matrix} 1 & 2 & A \\ + & 6 & A & B \end{matrix} \bar{\begin{matrix} A & 0 & 9 \end{matrix}}$

Answer 5:

$A + B = 9 as the sum of two digits can never be 19 2 + A = 0, A must be 8 A + B = 9, 8 + B = 9, B = 1 So, A = 8, B = 1$

Question 6:

Solve each of the following Cryptarithm:
$\begin{matrix} A & B & 7 \\ + & 7 & A & B \end{matrix} \bar{\begin{matrix} 9 & 8 & A \end{matrix}}$

Answer 6:

$If A + B = 8, A + B \geq 9 is possible only if A = B = 9 But from 7 + B = A, A = B = 9 is impossible Surely, A + B = 8, A + B \leq 9 So, A + 7 = 9, Surely A = 2 7 + B = A, 7 + B = 2, B = 5 So, A = 2, B = 5$

Question 7:

Show that the Cryptarithm $4 \times \bar{A B} = \bar{C A B}$ does not have any solution.

Answer 7:

$0 is the only unit digit number, which gives the same 0 at the unit digit when multiplied by 4 . So, the possible value of B is 0 . Similarly, for A also, 0 is the only possible digit . But then A, B and C will all be 0 . And if A, B and C become 0, these numbers cannot be of two - digit or three - digit . Therefore, both will become a one - digit number . Thus, there is no solution possible .$

RD Sharma solution class 8 chapter 5 Playing with numbers Exercise 5.3