Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining

[VJesus12]

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

Answer: C

Source: Official Guide

[Scott@TargetTestPrep]

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

Answer: C

Source: Official Guide

Solution:

In the 700s, we can have the first case 7x7, where x is the tens digit and x can be any digit from 0 to 9 except 7. So there are 9 such numbers. The second case can be 77y, where y is the units digit and y can be any digit from 0 to 9 except 7. So there are also 9 such numbers. The third case can be 7zz, where z is the (same) tens and units digits and z can be any digit from 0 to 9 except 7. So there are also 9 such numbers. Therefore, there are a total of 27 numbers in the 700s.

Using the same argument, we see that there are also 27 numbers in the 800s and another 27 numbers in the 900s. Therefore, there is a total of 27 x 3 = 81 numbers. However, although we have excluded numbers such as 777, 888 and 999, where all 3 digits are the same, we have included 700 in our analysis for the numbers in the 700s. Since the problem states that the numbers have to be greater than 700, we have to exclude 700 also. Therefore, there are 80 such numbers.

Answer: C

[Brent@GMATPrepNow]

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

Answer: C

Source: Official Guide

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 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

[swerve]

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

Answer: C

Source: Official Guide

To satisfy the given condition,

Required number of cases \(=\) total numbers \(-\) numbers with all digits different \(-\) numbers when all three digits are the same. A number greater than \(700.\)

Total numbers \(= 1\cdot 10 \cdot 10 = 100\)

Numbers with all digits different \(= 1\cdot 9 \cdot 8 = 72\)

Numbers when all three digits are same \((777) = 1\)
req. \(= 100 - 72 - 1 = 27\)

Considering the numbers between \(700 - 999 = 27 \cdot 3=81\)

So, the answer is \(80\) ('cause \(700\) can't be included)

Therefore, C