HKD1710
How many integers n greater than 10 and less than 100 are such that, if the digits of n are reversed, the resulting integer is n+9 ?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Let's use some trial/error and logic.
What's given - when we reverse the digits of a 2 digit number, the new number - old number = 9...
Let's now think, when is the above scenario possible - the difference, 9, is a single digit difference... so if the tens and units digits of the number have a difference of 1 between them - then only the condition given in question will be satisfied.
If the difference between the tens and units digit is more than 1, when we reverse the digits the difference between 2 numbers is greater than 10 - simply because the tens digit goes up by 2 units.
Now let's use some trial and error...
taking 12 --> reverse is 21 --> difference is 9.
let's take 13 (note, the difference between the digits is 3-1=2)... --> reverse is 31 --> difference is 18.
This confirms our above logic.
So, we can now understand that the 2 digit numbers which can satisfy the above case will have a difference of 1 between the tens and units digit, and since its given that the new number is 9 greater than the old number... we will only consider cases in which the value of number increases on reversing the digits i.e. numbers where tens digit is 1 lower than the units digit - 12, 23, 34, 45, 56,, 67, 78, 89...