NaimaB 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
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 799 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 = [spoiler]80 = C[/spoiler]
Cheers,
Brent
