psm12se 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 list the 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

8XX
800
811
822
.
.
.
899
Since we cannot include 888 in this list, there are 9 numbers in the form 8XX

8X8
808
818
828
.
.
.
898
Since we cannot include 888 in this list, there are 9 numbers in the form 8X8

88X
880
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, there are 26 numbers from 701 to 799 inclusive that meet the given criteria.

So, our answer is 27+27+26 = [spoiler]80 = C[/spoiler]

Cheers,
Brent