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Of the three-digit integers greater than 700, how many have [#permalink] New post 02 Jul 2012, 02:01
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The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 02 Jul 2012, 02:02
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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

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two digits are alike and third is different;
C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);
A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);
C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.
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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 02 Jul 2012, 02:49
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Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

Diagnostic Test
Question: 11
Page: 22
Difficulty: 650


GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!


The first digit can be 7, 8 or 9, so three possibilities.
If the first digit is repeated, we have 2 * 9 possibilities, because the numbers can be of the form aab or aba, where b\neq a.
If the first digit is not repeated, the number is of the form abb, for which there are 9 possibilities.
This gives a total of 27 choices for a given first digit.
Altogether, we will have 3 * 27 = 81 choices from which we have to eliminate one choice, 700 (because the numbers should be greater than 700).
We are finally left with 80 choices.

Answer: C
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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 02 Jul 2012, 05:33
Hi,

To satify the given condition,
required no. of cases = total numbers - numbers with all digits different - numbers when all three digits are same,
number greater than 700;
total numbers = 1*10*10 = 100
numbers with all digits different = 1*9*8 = 72
numbers when all three digits are same (777) = 1
req. = 100- 72 - 1 = 27
considering the numbers between 700 & 999 = 27*3=81
Answer is 80 ('cause 700 can't be included)

Answer (C).

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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 02 Jul 2012, 06:34
There are 2 numbers that have the same two digits for each of the 10's (700-709, 710-719, 720-729, etc.), with the exception for where the tens digit is the same as the hundreds (770-779), in this scenario there are 9 digits that have the same two digits.

So for 700-799 inclusive there are (9 * 2) + 9 - 1 = 26 numbers that the same two digits. Subtract one cause 700 shouldn't be included.
For 800 - 999 the calculation simply becomes 27 * 2 = 54

54 + 26 =
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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 02 Jul 2012, 12:22
Total Number of 3 digit number greter than 700 = 299
In 700 series - The total number of number with distinct digit = 9*8
So from 700 till 999 we will have - 3*9*8 = 216
So total number of numbers with duplicate digit = 299-216 = 83
Need to Subtract 777,888 & 999 so answer is 80.
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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 03 Jul 2012, 10:09
999-700=299 (total numbers)

299-3 =296 (3 means 777/888/999)

3*8*9=216 (doing so, we found total numbers,which have different digits)
296-216=80

ans is C
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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 06 Jul 2012, 02:40
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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

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two digits are alike and third is different;
C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);
A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);
C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.
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Of the 3digit nos greater than 700, how many have 2 digits [#permalink] New post 15 Feb 2013, 23:20
Of the 3 digit nos greater than 700, how many have 2 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

I do have the solution for this, but im still not too clear with it. Please help with your solutions. Also, how long did you take to do this? Please let me know the time taken as well
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Re: Of the 3digit nos greater than 700, how many have 2 digits [#permalink] New post 15 Feb 2013, 23:41
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Re: Of the 3digit nos greater than 700, how many have 2 digits [#permalink] New post 15 Feb 2013, 23:41
For 3 digit numbers greater than 700, the first digit has to be 7, 8, or 9. The last two digits can be anything except 00.

There can be three possible cases for such numbers:

Case 1- first two digits same. The last digit has to be different from the first two. As the first digit is constrained to be 7,8,or 9, total number of such numbers = 3*9 = 27

Case 2 - first and third digits same. By the same logic as above, total number of such numbers = 27

Case 3 - second and third digits same. Total number of such numbers = (3*9)-1 = 26 (we need to subtract one to remove the 00 case).

Therefore total = 27 + 27 + 26 = 80 numbers (option C)

We have assumed here that numbers where all three digits are the same are not allowed.
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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 17 Feb 2013, 19:49
Thanks all....pretty clear with this now....cheers!
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Of the three-digit integers greater than 700, how many have [#permalink] New post 12 Mar 2013, 21:23
Q: 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
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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 12 Mar 2013, 21:42
From 701 to 799 we will have: 711, 722.... 799( excluding 777 ) = 8;
From 770 to 779 we have 9 numbers ( excluding 777 ).
We will also have numbers such as 707, 717, 727.... 797, total 9 numbers ( excluding 777 )
Similarly from 800 to 899 we will have 9 numbers ( we will exclude 888 )
From 880 to 899 we will once again have 9 numbers
We will also have numbers such as 808, 818.... 898, total 9 numbers ( excluding 888 )
Similarly for numbers greater than equal to 900 and till 999 we will have 27 such numbers.
Total = 26(>700) + 27(>=800) + 27(>=900) = 80 such numbers.

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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 12 Mar 2013, 21:53
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Re: Of the three-digit integers greater than 700, how many have [#permalink] New post 13 Mar 2013, 01:31
Thoughtosphere wrote:
Q: 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


Merging similar topics. Please refer to the solutions above and ask if anything remains unclear.
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Re: Of the three-digit integers greater than 700, how many have   [#permalink] 13 Mar 2013, 01:31
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