Carcass wrote:
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
One approach is to start 
LISTING numbers and look for a 
PATTERN. 
Let's first focus on the numbers from 
800 to 899 inclusive. We have 3 cases to consider: 8XX, 8X8, and 88X 
8XX800
811
822
.
.
.
899 
Since we cannot include 888 in this list, there are 
9 numbers in the form 8XX
8X8808
818
828
.
.
.
898 
Since we cannot include 888 in this list, there are 
9 numbers in the form 8X8
88X880
881
882
.
.
.
889 
Since we cannot include 888 in this list, there are 
9 numbers in the form 88X
So, there are 
27 (
9+
9+
9) numbers from 
800 to 899 inclusive that meet the given criteria. 
Using the same logic, we can see that there are 
27 numbers from 
900 to 999 inclusive that meet the given criteria. 
And there are 27 numbers from 
700 to 999 inclusive that meet the given criteria. HOWEVER, the question says that we're looking at numbers 
greater than 700, so the number 700 does not meet the criteria. So, there are actually 
26 numbers from 
701 to 799 inclusive that meet the given criteria.
So, our answer is 
27+
27+
26 = 80 
Answer: C