A is a 2-digit number with different digits. B is also a 2-digit number and is obtained by reversing the digits of A. If A - B is a multiple of 27, where A > B, how many such different A's are possible?<br/>

Let (A=10x+y) and (B=10y+x), where both (A) and (B) are 2-digit numbers. Hence (x\in{1,\dots,9}) and also (y\in{1,\dots,9}) (since if (y=0), then (B) would not be 2-digit). Also (x\ne y) and (A>B\Rightarrow x>y).

Now,
[
A-B=(10x+y)-(10y+x)=9(x-y).
]
For this to be a multiple of 27, (x-y) must be a multiple of 3.

Possible positive digit differences are therefore 3 or 6:

• Difference 3: ((x,y)=(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)) → 6 values
• Difference 6: ((x,y)=(7,1),(8,2),(9,3)) → 3 values

Total (=6+3=9).