GeminiHeat wrote:
How many integers k greater than 100 and less than 1000 are there such that if the hundreds and the unit digits of k are reversed, the resulting integer is k + 99?
A. 50
B. 60
C. 70
D. 80
E. 90
Any 3-digit integer \(abc\) can be written as \(100a + 10b + c\)
Let, \(k = abc\)
So reversed number = \(cba\)
Given, \(cba = abc + 99\)
\(100c + 10b + a = 100a + 10b + c + 99\)
\(99c - 99a = 99\)
\(c - a = 1\)
We have a total of 9 possibilities where \(c - a = 1\)
(9-8), (8-7), (7-6), (6-5), (5-4), (4-3), (3-2), (2-1), (1-0)
But, (1,0) case is not feasible as \(k > 100\)
So, the number of possible cases = 8
\(b\) can have 10 different values
Therefore, we can have \((8)(10) = 80\) such numbers
Hence, option D