Carcass wrote:
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
If the digits of n are reversed, the resulting integer is n+9 ? Let's say n = xy (where x is the tens digit of n, and y is the units digit of n).
So, the VALUE of n =
10x + y (in the same way the value of 29 = (2)(10) + 9)When we REVERSE the digits of n, we get yx, which means the VALUE of yx =
10y + x The question tells us the
resulting number is 9 greater than the
original number (n)So, we can write:
10y + x =
10x + y + 9Subtract y from both sides:
9y + x = 10x + 9Subtract 10x from both sides:
9y - 9x = 9Divide both sides by 9 to get:
y - x = 1This tells us that y, the UNITS digit of n, must be one greater than x, the TENS digit of n.
So the possible values of n are: 12, 23, 34, 45, 56, 67, 78, and 89
Answer: D