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I like the solution - it’s helpful.
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This is a great question that’s helpful for learning.
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one doubt,
a+b = 11*1 is a possible solution, also a+b = 11*4 , 11*9

am I missing any 2 digit integer thing here?

Bunuel
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
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gemministorm
one doubt,
a+b = 11*1 is a possible solution, also a+b = 11*4 , 11*9

am I missing any 2 digit integer thing here?



a and b are positive single digits (1, 2, 3, ..., 9). How can a + b be 44 or 99?
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