Question

the sum of the digits of a 2-digit number is 6.On reversing its digits,the number is 18 less than the original number. Find the number.​

Answer 1

S O L U T I O N :

Let the ten's place digit be r

Let the one's place digit be m

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

A/q

$\longrightarrow\sf{r+m=6}\\\\\longrightarrow\sf{r=6-m....................(1)}$

&

$\longrightarrow\sf{10r+m-18=10m+r}\\\\\longrightarrow\sf{10r-r+m-10m=18}\\\\\longrightarrow\sf{9r-9m=18}\\\\\longrightarrow\sf{9r-9m=18}\\\\\longrightarrow\sf{9(r-m)=18}\\\\\longrightarrow\sf{r-m=\cancel{18/9}}\\\\\longrightarrow\sf{r-m=2}\\\\\longrightarrow\sf{6-m-m=2\:\:[from(1)]}\\\\\longrightarrow\sf{6-2m=2}\\\\\longrightarrow\sf{-2m=2-6}\\\\\longrightarrow\sf{-2m=-4}\\\\\longrightarrow\sf{m=\cancel{-4/-2}}\\\\\longrightarrow\bf{m=2}$

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

$\longrightarrow\sf{r=6-2}\\\\\longrightarrow\bf{r=4}$

Thus;

$\bigstar\:\underbrace{\sf{The\:number\:will\:be\:(10r+m)=[10(4)+2]=[40+2]=\boxed{\bf{42}}}}}}$

Answer 2

Answer:

Step-by-step explanation:

Let's set up a system with two variables: x = tens place of our answer, y = units place of our answer.

Digit sum of a two digit number is 6:

x+y=6

Reverse the digits and you get 18 less than the original value:

10y+x=10x+y-18

Now let's solve:

y=6-x

10(6-x)+x=10x+(6-x)-18

60-10x+x=10x+(6-x)-18

60-9x=9x-12

72=18x

x=4

y=6-4

y=2

So our original number was 42