The sum of a two digit number and the number obtained by reversing the order of its digits is 99

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.

Solution:

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

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

After interchanging the digits, the number becomes $10 x+y$.

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

$(10 x+y)+(10 y+x)=99$

$\Rightarrow 10 x+y+10 y+x=99$

$\Rightarrow 11 x+11 y=99$

$\Rightarrow 11(x+y)=99$

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

$\Rightarrow x+y=9$

So, we have two systems of simultaneous equations

$x-y=3$

 

$x+y=9$

$x-y=-3$,

 

$x+y=9$

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

(i) First, we solve the system

$x-y=3$

$x+y=9$

Adding the two equations, we have

$(x-y)+(x+y)=3+9$

$\Rightarrow x-y+x+y=12$

 

$\Rightarrow 2 x=12$

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

$\Rightarrow x=6$

Substituting the value of in the first equation, we have

$6-y=3$

$\Rightarrow y=6-3$

$\Rightarrow y=3$

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

(ii) Now, we solve the system

$x-y=-3$

 

$x+y=9$

Adding the two equations, we have

$(x-y)+(x+y)=-3+9$

$\Rightarrow x-y+x+y=6$

 

$\Rightarrow 2 x=6$

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

$\Rightarrow x=3$

Substituting the value of in the first equation, we have

$3-y=-3$

$\Rightarrow y=3+3$

$\Rightarrow y=6$

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

Note that there are two such numbers.

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