Question

answer my question fast​

Answer 1

Answer: 2

Step-by-step explanation:

Let the digit of 10 be $a$.

And the digit of 1 be $b$.

We know that $|a-b|=2$.

$a-b=\pm2$

∴ $a=b\pm2$ --- 1

'The sum of a two-digit number and the number obtained by reversing the digits is 66.'

We can write an expression.

$(10a+b)+(a+10b)=66$

$11(a+b)=66$

∴ $a+b=6$ --- 2

Consider two possibilities. (∵ 2)

$a=b+2$ or $a=b-2$

1. $a=b+2$

Substitute into 1.

$2b+2=6$

∴$a=4$, $b=2$

2. $a=b-2$

Substitute into 1.

$2b-2=6$

∴ $a=2$, $b=4$

Such numbers are 42 or 24.

Hence, there are two numbers!