Question

the sum of two digit number and the number firmed by reversing the order of digit is 66 if two digit differ by 2 find the number how many such number are therw​

Answer 1

Answer:

Answer:Number is 42.

Solution:

Let the two digits number is xy,this can be represented as 10x+y

on reversing the number the number become yx, now this can be represented as 10y+x

according to the question the sum of two digit number and the number reversing the digits is equal to 66

\begin{lgathered}10x + y + 10y + x = 66 \\ \\ 11x + 11y = 66 \\ \\ x + y = 6....eq1 \\ \\\end{lgathered}10x+y+10y+x=6611x+11y=66x+y=6....eq1 

the digit of the number differ by 2,hence it can be represented algebraically

\begin{lgathered}x - y = 2.....eq2 \\ \\\end{lgathered}x−y=2.....eq2 

Now solve these two equations,here I am using Elimination method,add both equations

\begin{lgathered}x + y = 6 \\ x - y = 2 \\ - - - - - - - \\ 2x = 8\\ \\ x = 4 \\ \\4+ y = 6 \\\\y=2\\\\\end{lgathered}x+y=6x−y=2−−−−−−−2x=8x=44+y=6y=2 

Hence number is 42.

it's reverse number is 24.

Verification:

42+24=66

HOPE IT HELPS