The sum of the digits of a $2$-digit number is $11$. The number obt...

Question

The sum of the digits of a $2$-digit number is $11$. The number obtained by interchanging its digits exceeds the given number by $9$. To know the number :
(i) form the linear equations representing the above situation.
(ii) verify that the equations have a unique solution.
(iii) solve the equations to get the given $2$-digit number.

Accepted Answer

Let the tens digit be $x$ and the units digit be $y$.
Then the number $= 10x + y$ and the reversed number $= 10y + x$.

(i) Forming the equations
Step 1: Sum of digits is $11$ :
$$x + y = 11 \qquad \ldots (1)$$
Step 2: Reversed number exceeds the given number by $9$ :
$$10y + x = (10x + y) + 9$$
$$9y - 9x = 9 \Rightarrow -x + y = 1 \qquad \ldots (2)$$

(ii) Unique solution
Step 3: Here $a_1 = 1,\ b_1 = 1$ and $a_2 = -1,\ b_2 = 1$.
$$\frac{a_1}{a_2} = \frac{1}{-1} = -1,\qquad \frac{b_1}{b_2} = \frac{1}{1} = 1$$
Step 4: Since $\dfrac{a_1}{a_2} 
eq \dfrac{b_1}{b_2}$, the pair of equations has a UNIQUE solution.

(iii) Solving
Step 5: Add $(1)$ and $(2)$ :
$$2y = 12 \Rightarrow y = 6$$
Step 6: From $(1)$ : $$x = 11 - 6 = 5$$
Step 7: The number $= 10x + y = 10(5) + 6 = 56$.
Check : Reversed number $= 65$ and $65 - 56 = 9$ \checkmark ; digit sum $= 5 + 6 = 11$ \checkmark
Hence the required number is $56$.

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