RiyaR wrote:How many integers n greater than 10 and less than 100 are there such that, if the digits of n are reversed the resulting integer is n + 9?
A) 5
B) 6
C) 7
D) 8
E) 9
Alternate approach:
Let T = the tens digit and U = the units digit.
Original integer = 10T + U.
To illustrate:
53 = 10(5) + 3.
The original integer with the digits reversed = 10U + T.
Since the reversed integer is 9 more than the original integer, we get:
10U + T = (10T + U) + 9
9U = 9T + 9
U = T+1.
Implication:
The units digit must be 1 more than the tens digit, yieding the following options:
12, 23, 34, 45, 56, 67, 78, 89.
Total options = 8.
The correct answer is
D.
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