How many integers k greater than 100 and less than 900 are there such that if the hundreds and the unit digits of k are reversed, the resulting integer is k + 99? Choose the most appropriate option:
A) 50  B) 60  C) 70  D) 80  E) 90
Not sure if this is the shortest.. But this is how I did this   There are 7 sets of integers with hundreds and units digits exchanged that satisfies k + 99.   1. 102 | 201 (satisfies k+99, where k = 102)   2. 203 | 302 (satisfies k+99, where k = 203)   3. ...   4. ...   5. ...   6. ...   7. 708 | 807   Each set has 10 such numbers.   1. 102 | 201 (still k+99 holds good)   2. 112 | 211   3. 122 | 221   4. 132 | 231   5. ...   6. ...   7. ...   8. ...   9. 182 | 281   10. 192 | 291   Therefore, 7 sets with 10 such number in each set will give 7 x 10 = 70 integers.   70
<result>C</result>
How many integers k greater than 100 and less than 900 are there such that if the hundreds and the unit digits of k are reversed, the resulting integer is k + 99?