Official Solution:For each positive integer \(k\), \(f(k)\) is defined to be the number created by reversing the digits of \(k\). For instance, \(f(42) = 24\). How many positive two-digit integers k are there where \(k + f(k)\) is the square of a positive integer? A. 6
B. 7
C. 8
D. 9
E. 10
Let's represent a two-digit number \(k\) as \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit. Then \(f(k)\) becomes \(10b + a\), and \(k + f(k) = (10a + b) + (10b + a) = 11(a + b)\).
For \(11(a + b)\) to be the square of an integer, \((a + b)\) must be 11, making \(k + f(k) \) equal to \(11^2\).
\(a + b = 11\) is possible for the following 8 sets of \((a, b)\): (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), and (9, 2).
Therefore, there are a total of 8 two-digit integers \(k\) where \(k + f(k)\) is the square of an integer.
Answer: C