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.
Last edited by
GMATGuruNY on Tue Jul 12, 2011 2:01 pm, edited 1 time in total.
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