Question

The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

Answer 1

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is ${\displaystyle 10y+x}$.

The two digits of the number are differing by 3. Thus, we have ${\displaystyle x-y=\pm 3}$

After interchanging the digits, the number becomes ${\displaystyle 10x+y}$.

The sum of the numbers obtained by interchanging the digits and the original number is 99. Thus, we have

${\displaystyle (10x+y)+(10y+x)=99}$

${\displaystyle \Rightarrow 10x+y+10y+x=99}$

${\displaystyle \Rightarrow 11x+11y=99}$

${\displaystyle \Rightarrow 11(x+y)=99}$

${\displaystyle \Rightarrow x+y\frac{99}{11}}$

${\displaystyle \Rightarrow x+y=9}$

So, we have two systems of simultaneous equations

${\displaystyle x-y=3}$

${\displaystyle x+y=9}$

${\displaystyle x-y=-3}$

${\displaystyle x+y=9}$

Here x and y are unknowns. We have to solve the above systems of equations for xand y.

(i) First, we solve the system

${\displaystyle x-y=3}$

${\displaystyle x+y=9}$

Adding the two equations, we have

${\displaystyle (x-y)+(x+y)=3+9}$

${\displaystyle \Rightarrow x-y+x+y=12}$

${\displaystyle \Rightarrow 2x=12}$

${\displaystyle \Rightarrow x=\frac{12}{2}}$

${\displaystyle \Rightarrow x=6}$

Substituting the value of x in the first equation, we have 

${\displaystyle 5-y=3}$

${\displaystyle \Rightarrow y=6-3}$

${\displaystyle \Rightarrow y=3}$

Hence, the number is${\displaystyle 10\times 3+6=36}$.

(ii) Now, we solve the system

${\displaystyle x-y=-3}$

${\displaystyle x+y=9}$

Adding the two equations, we have

${\displaystyle (x-y)+(x+y)=-3+9}$

${\displaystyle \Rightarrow x-y+x+y=6}$

${\displaystyle \Rightarrow 2x=6}$

${\displaystyle \Rightarrow x=\frac{6}{2}}$

${\displaystyle \Rightarrow x=3}$

Substituting the value of x in the first equation, we have

${\displaystyle 3-y=-3}$

${\displaystyle \Rightarrow y=3+3}$

${\displaystyle \Rightarrow y=6}$

Hence, the number is ${\displaystyle 10\times 6+3=63}$.

Note that there are two such numbers.