Question 15 Marks
Replace $A, B, C$ by suitable numerals.
$ \ \ \ \ \ \ \ \ \ \ \ \text{A B}\\\underline{ \ \ \ \ \ \ \times\text{B A}}\\\underline{\text{(B+1)}\text{C B}}$
$ \ \ \ \ \ \ \ \ \ \ \ \text{A B}\\\underline{ \ \ \ \ \ \ \times\text{B A}}\\\underline{\text{(B+1)}\text{C B}}$
Answer
View full question & answer→$A \times B = B$
$\Rightarrow A = 1$
$\ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \text{B}\\\underline{\times\ \ \ \ \ \text{B}\ \ \ \ \ \ \ 1\ \ \ \ }\\\underline{ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \text{B}\\ \ \text{B }\ \ \ \text{B}^{2}\ \ \ \ \ \ \times\ }\\\text{B}\ \ (1+\text{B}^{2}) \ \text{B}$
In the equation:
First digit $= B + 1$
Thus $1$ will be carried from $1 + B^2$ and becomes $(B + 1) (B^2 - 9)B$.
$\therefore$ $C =$$ B^2 - 1$
Now all $b, B + 1$ and $B^2 - 9$ are one digit number.
This condition is satisfied for $B = 3$ or $B = 4$.
For, $B < 3, B - 9$ will be negative.
For,$B > 3, $$B^2 - 9$ will become digit number.
For $B = 3, C =$ $3^2$$- 9 = 9 - 9 = 0$
For, $B = 4, C =$ $4^2$$ - 9 = 16 - 9 = 7$
$\therefore$ $A = 1, B = 4, C = 7$
$\Rightarrow A = 1$
$\ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \text{B}\\\underline{\times\ \ \ \ \ \text{B}\ \ \ \ \ \ \ 1\ \ \ \ }\\\underline{ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \text{B}\\ \ \text{B }\ \ \ \text{B}^{2}\ \ \ \ \ \ \times\ }\\\text{B}\ \ (1+\text{B}^{2}) \ \text{B}$
In the equation:
First digit $= B + 1$
Thus $1$ will be carried from $1 + B^2$ and becomes $(B + 1) (B^2 - 9)B$.
$\therefore$ $C =$$ B^2 - 1$
Now all $b, B + 1$ and $B^2 - 9$ are one digit number.
This condition is satisfied for $B = 3$ or $B = 4$.
For, $B < 3, B - 9$ will be negative.
For,$B > 3, $$B^2 - 9$ will become digit number.
For $B = 3, C =$ $3^2$$- 9 = 9 - 9 = 0$
For, $B = 4, C =$ $4^2$$ - 9 = 16 - 9 = 7$
$\therefore$ $A = 1, B = 4, C = 7$