Question

The sum of the digits of a two digit number is 8. The number obtained by
interchanging its digit is 18 more than the original number. Find the the original
number.​

Answer 1

Given :-

The sum of the digits of a two digit number is 8 .

The number obtained by interchanging it's digits is 18 more than the original number .

Required to find :-

• Original number ?

Solution :-

Given data :-

The sum of the digits of a two digit number is 8 .

The number obtained by interchanging it's digits is 18 more than the original number

we need to find the original number .

So,

Let's consider the unit digit be x

Ten's digit be y

No. formed =

=> 10 ( y ) + 1 ( x )

=> 10y + x

On reversing the digits

Unit digit is y

Ten's digit is X

No. formed =

=> 10 ( x ) + 1 ( y )

=> 10x + y

According to the given data ;

Sum of the digits of a two digit number is 8

=> y + x = 8

=> y = 8 - x $\longrightarrow{\sf{ Equation - 1 }}$

consider this as Equation 1

According to the problem ;

The number obtained by interchanging it's digits is 18 more than the original number

➔ 10y + x = 10x + y + 18

➔ x - 10x = y + 18 - 10y

➔ - 9x = - 9y + 18

➔ - 9x + 9y = 18

Substitute the value of y from equation 1

➔ - 9x + 9 ( 8 - x ) = 18

➔ - 9x + 72 - 9x = 18

➔ - 18x + 72 = 18

➔ - 18x = 18 - 72

➔ - 18x = - 54

Multiplying with - ( minus ) on both sides

➔ - ( - 18x ) = - ( - 54 )

➔ 18x = 54

➔ x = 54/18

➔ x = 3

Substitute the value of x in equation 1

➔ y = 8 - x

➔ y = 8 - 3

➔ y = 5

Hence,

The original number can be 53 or 35

So,

Let's verify which number is the original number !

According to given data ;

The number obtained by interchanging it's digits is 18 more than the original number .

Hence,

53 on reversing becomes 35

This implies,

✒ 53 + 18 = 35

✒ 71 = 35

✒ LHS ≠ RHS

Similarly,

35 on reversing becomes 53

This implies,

✒ 35 + 18 = 53

✒ 53 = 53

✒ LHS = RHS

Hence, it is verified that 35 is the original number .

Therefore,

Original number = 35 ✍

Answer 2

Solution :

Let the ten's place digit be m & one's place digit be r

$\boxed{\bf{The\:original\:number=10m+r}}}}\\\boxed{\bf{The\:reversed\:number=10r+m}}}}$

A/q

$\underbrace{\bf{1^{st}\:Case\::}}}$

$\longrightarrow\sf{r+m=8}\\\\\longrightarrow\sf{r=8-m....................(1)}$

$\underbrace{\bf{2^{nd}\:Case\::}}}$

$\longrightarrow\sf{10r+m=10m+r+18}\\\\\longrightarrow\sf{10r-r+m-10m=18}\\\\\longrightarrow\sf{9r-9m=18}\\\\\longrightarrow\sf{9(r-m)=18}\\\\\longrightarrow\sf{r-m=\cancel{18/9}}\\\\\longrightarrow\sf{r-m=2}\\\\\longrightarrow\sf{8-m-m=2\:\:\:[from(1)]}\\\\\longrightarrow\sf{8-2m=2}\\\\\longrightarrow\sf{-2m=2-8}\\\\\longrightarrow\sf{-2m=-6}\\\\\longrightarrow\sf{m=\cancel{-6/-2}}\\\\\longrightarrow\bf{m=3}$

∴Putting the value of m in equation (1),we get;

$\longrightarrow\sf{r=8-3}\\\\\longrightarrow\bf{r=5}$

Thus;

$\boxed{\sf{The\:Original\:number\:= (10m+r)=[10(3)+5]=[30+5]=\boxed{\bf{35}}}}}$