Question

The sum of a two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?

Plz explain it in detail

Answer 1

Hello student,

Let the unit digit be X and ten 's digit be Y. So, original digit be ( X+10.Y) , And by reversing the digit we get (Y+10.x)

SO ACCORDINGTO THE QUESTION

(X+10Y)+(Y+10X) = 66

11X+11Y = 66

11(X+Y) = 66

X+Y = 66/11

X+Y = 6 ------------------- 1

And, since digits differ by 2

So, X-Y = 2 --------------- 2

Adding 1 and 2 equations we get

2X = 8

X = 8/2

so, X = 4

And, puttingvalue of Xin equation 1

X + Y = 6

4 + Y = 6

Y = 6 - 4

Y = 2

Therefore, number is given by putting value of X & Y

Hence, number is ( 4+10*2) = 24...

it can also be its reverse.
that is 42

so there are two numbers with the conditions given in question they are 24 and 42