Problem Solving

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

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