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+cLet,
k=abcSo reversed number =
cbaGiven,
cba=abc+99100c+10b+a=100a+10b+c+9999c−99a=99c−a=1We 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>100So, the number of possible cases = 8
b can have 10 different values
Therefore, we can have
(8)(10)=80 such numbers
Hence, option D