Question: 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

Correct Answer: C
Solution and Explanation:
Approach Solution 1:

One approach for this question can be to start listing numbers and look for a pattern.
First, we can focus on the numbers from 800 to 899 inclusive.
We need to see three cases in such a scenario: 8XX, 8X8, 88X
8XX
800
811
822
.
.
.
899
We cannot include 888 in the list, so there are 9 numbers in the form 8XX
8X8
808
818
828
.
.
.
898
888 cannot be included in the list so there are 9 numbers in the form 88X
88X
880
881
882
.
.
.
889

The number 888 cannot be included in this list as well, so there are 9 numbers in form 88X
Thus, there are 27 (9+9+9) numbers from 800 to 899 that can meet the given criterion.

If we use the same logic, we can also see that there are 27 numbers from 900 to 999 inclusive that can meet the given criterion.

There are 27 numbers from 700 to 999 inclusive that can meet the given criterion. The question, however, is that we are looking at numbers which are greater than 700 so the number 700 also does not meet the criterion. Therefore, there are 26 numbers in reality from 701 to 799 inclusive that can meet the given criterion.

The answer is 27+ 27+ 26 = 80

Approach Solution 2 :

Let the three-digit number be X Y Z
There will be three cases:

Case I:
[ X=Y ] & Z is not equal to X & Y: XXZ or YYZ
X can be either 7, 8 or 9 so the digit at X can be chosen in 3 ways
After digit X is chosen, Y can be chosen in 1 way
After X & Y are chosen, Z can be chosen in 9 ways
The possible number of digits zeroes down to (3 ways) * (1 way) * (9 ways) = 27 …. (1)
(For instance, 774, 779, 882, 993 etc.)

Case II:
[ X=Z ] & Y is not equal to X and Z: XYX or ZYZ
X can be either 7, 8 or 9 so the digit at X can be chosen in 3 ways
After the number X is chosen, Z can be chosen in 1 way
After X & Z are chosen, Y can be chosen in 9 ways
Therefore, the possible number of digits equals (3 ways) * (1 way) * (9 ways) = 27…. (2)
(Example, 747, 797, 8282, 939 etc.)

Case III:
[ Y=Z ] & X is not equal to Y and Z: XYY or XZZ
X can be either 7, 8 or 9 so the digit at X can be chosen in 3 ways
After X is chosen, Y can be chosen in 9 ways
After X & Y are chosen, Z can be chosen in 1 way
Therefore, the possible number of digits equals (3 ways) * (9 ways) * (1 way) = 27…. (3)
(Example, 744, 799, 822, 933 etc.)
Therefore, the possible number of total digits would be the sum of Cases 1, 2 and 3 mentioned above that 27+ 27+ 27 – 1 = 80

One digit is subtracted from the total number of possible digits in order to rule out one possibility of XYZ = 700 in order to fulfil the condition that the digit is >700.

Approach Solution 3 :

Three-digit numbers can have only three patterns which are listed below.

All the digits are different
Two digits are alike and the third is different
All three digits are similar

We need to calculate B. B = Total – A- C
Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700)

All digits are different = 3* 9* 8 = 216 (first digit can have only three values either 7, 8 or 9)
All three are similar = 3 (777, 888, 999)
Thus, 299 -216 –3 = 80

“Of the three-digit integers greater than 700, how many have two digits”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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