BTGmoderatorLU wrote: ↑

Wed Feb 10, 2021 7:49 am

Source: Manhattan Prep

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?

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

The OA is D

Solution:

Let k be abc where a, b and c represent the hundreds, tens and units digits, respectively. Then, k = 100a + 10b + c. According to the given information, cba = k + 99; thus:

cba = 100a + 10b + c + 99

100c + 10b + a = 100a + 10b + c + 99

99c - 99a = 99

c - a = 1

Notice that a cannot be 0 since a is the hundreds digit of k. Notice also that a cannot be 9 because that would mean c = 10, which is impossible since c is a digit. Therefore, the possible values of a are 1, 2, … , 8 and the corresponding values of c are 2, 3, … , 9. Thus, there are 8 possible choices for a and c.

There are no restrictions over b, thus for any a, c pair, there are 10 possible values of b. Since there are 8 possible a, c pairs and since there are 10 possible values of b for each pair, there are 8 x 10 = 80 such integers k.

Let’s pick one of the possible values of a, b and c and verify that it satisfies the conditions of the question. Let a = 3 and b = 5. Since c - a = 1, it follows that c = 4; thus k = 354. If we reverse the hundreds and the units digit of 354, we obtain 453 and the difference is indeed 453 - 354 = 99.

Answer: D