Question

How many integers n greater than10 and less than 100 are there such that, if the digits of n are reversed, the resulting integer is n+9 ?

Answer 1

Answer:

8

Step-by-step explanation:

Let A be tens digit & B be units digit

X=10a+b

After reversing digits,

Xnew=10b+a

Difference between x and xnew is 9, so I get following equation

(10a+b)-(10b+a)=9

9a-9b=9

9(a-b)=9

a-b=1

a= b+1

Since units digit X is +1 more than tens digit, I get following numbers

12, 23, 34, 45, 56, 67, 78, 89

Therefore 8numbers