12 Days of Christmas GMAT Competition - Day 4: For each positive

The possible cubes are +-1 +-8 +-27 +-64
Let's assume k=10a+b for a and b belonging to {0,1,2,......9}
k-f(k)=cube
9(a-b)=+-27 (cube can be negative)
Others are not divisible by 9 so mod(a-b)=3 is what we need to solve with a condition that a is not zero since such a case is forbidden.
(3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6),(0,3),(1,4),(2,5),(3,6),(4,7),(5,8),(6,9) are the solution of mod(a-b)=3 we reject (0,3) solution since that means k is single digit so 3 option(B)

Given: For each positive integer k, f(k) is defined to be the number created by reversing the digits of k. For instance, f(52) = 25, f(40) = 4 and f(99) = 99.
Asked: How many positive two-digit integers k are there where k - f(k) is the cube of an integer?

Let k = ab; where a is tenth digit and b is unit digit

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

Case 1:
a-b=3; k - f(k) = 27 = 3^3
(a, b) = {(3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)}:
Positive two-digit integers k are there where k - f(k) is the cube of an integer = {30,41,52,63,74,85,96} : 7 2-digit integers

Case 2:
a-b=-3; k - f(k) = -27 = (-3)^3
(a, b) = {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}
Positive two-digit integers k are there where k - f(k) is the cube of an integer = {14,25,36,47,58,69} : 6 2-digit integers

Total positive two-digit integers k are there where k - f(k) is the cube of an integer = 7 + 6 = 13

IMO B

Bunuel Please check the solution and correct OA if needed.

I think you are missing the following nine case when the digits of k are the same: 11, 22, 33, 44, 55, 66, 77, 88, and 99. In those cases k - f(k) = 0, which is a cube of an integer.

Similar but a little bit easier question is here: https://gmatclub.com/forum/for-each-pos ... 21045.html