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?
Answers:	A) 50	B) 60	C) 70	D) 80	E) 90

Rationale: kk is 3 - digit integer . any 3 digit integer can be expressed as 100 a + 10 b + c 100 a + 10 b + c . now , we are told that ( 100 a + 10 b + c ) + 99 = 100 c + 10 b + a ( 100 a + 10 b + c ) + 99 = 100 c + 10 b + a - - > 99 c − 99 a = 9999 c − 99 a = 99 - - > c − a = 1 c − a = 1 - - > there are 8 pairs of cc and aa possible such that their difference to be 1 : ( 9,8 ) , ( 8,7 ) , . . . , and ( 2,1 ) ( note that aa can not be 0 as kk is 3 - digit integer and its hundreds digit , aa , is more than or equal to 1 ) . also , bb , tens digit of kk , could tale 10 values ( 0 , 1 , . . . , 9 ) so there are total of 8 * 10 = 80 such numbers possible . example : 102 - 201 ; 112 - 211 ; 122 - 221 ; . . . 899 - 998 . answer : d .
Answer: D