K=100a+10b+c
Modified K=100c+10b+10a

Condition is K+99=Modified K
=> (100a+10b+c)+99=(100c+10b+10a)
=> 99(a-c+1)=0 ---equation(1)

Solving equation(1)=> c=a+1

Therefore starting with
102, 122,....192 (10 numbers)
203, 213,....293 (10 numbers)
.
.
.
809, 819,...898 (10 numbers)

So Total of 8x10=80 numbers.
D

This explanation sounds tedious and would appreciate if someones comes up with a more apt way to solve this question.

Best!!!

viv_gmat wrote:How many integers "k" greater than 100 and less than 1000 are there such that if the hundreds and the units digits of "k" are reversed, the resulting integer is k+99?

A) 50
B) 60
C) 70
D) 80
E) 90

D

Since the tens digit is part of both k and k+99, it doesn't affect the increase.
The increase is coming solely from the swapping of the hundreds digit and the units digit.

Let H = hundreds digit and U = units digit.

Omitting the tens digit, k = 100H + U.
When the digits are swapped, new k = 100U + H.

Since the difference between the new k and the original k must be 99, we get:
(100U + H) - (100H + U) = 99.
99U - 99H = 99.
99(U - H) = 99.
U - H = 1.
U = H+1.

Thus, the following combinations will work:
H=1, U=2.
H=2, U=3.
H=3, U=4.
H=4, U=5.
H=5, U=6.
H=6, U=7.
H=7, U=8.
H=8, U=9.
8 combinations.

Since for each of these 8 combinations, the tens digit could be 0 through 9 -- a total of 10 options for each combination -- we multiply by 10:
8*10 = 80.

The correct answer is D.

Fantastic guys ...nice explanations...But I really like GMATGuruNY's approach..
Awesome..Thanks

Same process but can't explain that better than Guru already did