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?
Select
A) 50 B) 60 C) 70 D) 80 E) 90
kk is 3-digit integer. Any 3 digit integer can be expressed as 100a+10b+c100a+10b+c.   Now, we are told that (100a+10b+c)+99=100c+10b+a(100a+10b+c)+99=100c+10b+a --&gt; 99c−99a=9999c−99a=99 --&gt; c−a=1c−a=1 --&gt; 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= 
<gadget id="calculator">  8*10</gadget>
<output>80</output>
80 such numbers possible.   Example:   102 - 201;   112 - 211;   122 - 221;   ...   899 - 998.   Answer: 80.
<result>D</result>
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?