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Senior Manager  Status: Finally Done. Admitted in Kellogg for 2015 intake
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Of the three-digit integers greater than 660, how many have  [#permalink]

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3
40 00:00

Difficulty:   95% (hard)

Question Stats: 27% (02:48) correct 73% (02:54) wrong based on 433 sessions

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Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

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MGMAT 1 --> 530
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Math Expert V
Joined: 02 Sep 2009
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14
enigma123 wrote:
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

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

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

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

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##### General Discussion
Senior Manager  Joined: 23 Oct 2010
Posts: 317
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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my way of thinking is not elegant but anyways , see what I did -
we have the range 661-998

1. two 1st digits are the same
then we have-
661-669 (8 options (exclude 666))
770-779(9 options (exclude 777))
880-889 (9 options (exclude 888))
990-998 (9 options (exclude 999))
total=9*3+8=35

2.the 1st and the last are the same
then we have
6*6 (3 options- 676 686 696))
7*7 (9 options (exclude 777))
8*8 (9 options (exclude 888))
9*9 (9 options (exclude 999))
total=9*3+3=30

3. the 2nd and the 3d terms are the same.
then we have
66x- none option, since our first two digits are already the same, and the 3d cant be the same
7xx -9 options (exclude 777)
8xx- 9 options (exclude 888)
9xx 9 options (exclude 999)
total=9*3=27

30+35+27=92

cant find the rest 3 numbers to get 95. I will appreciate if someone helps me _________________
Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth
Math Expert V
Joined: 02 Sep 2009
Posts: 60630
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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1
LalaB wrote:
my way of thinking is not elegant but anyways , see what I did -
we have the range 661-998

1. two 1st digits are the same
then we have-
661-669 (8 options (exclude 666))
770-779(9 options (exclude 777))
880-889 (9 options (exclude 888))
990-998 (9 options (exclude 999))
total=9*3+8=35

2.the 1st and the last are the same
then we have
6*6 (3 options- 676 686 696))
7*7 (9 options (exclude 777))
8*8 (9 options (exclude 888))
9*9 (9 options (exclude 999))
total=9*3+3=30

3. the 2nd and the 3d terms are the same.
then we have
66x- none option, since our first two digits are already the same, and the 3d cant be the same
7xx -9 options (exclude 777)
8xx- 9 options (exclude 888)
9xx 9 options (exclude 999)
total=9*3=27

30+35+27=92

cant find the rest 3 numbers to get 95. I will appreciate if someone helps me You are missing 677, 688 and 699.
_________________
Senior Manager  Joined: 23 Oct 2010
Posts: 317
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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gosh, only I can do such stupid mistakes! all efforts are in vain, if u come to the wrong answer at last.
thnx Bunuel. appreciate ur help. next time I will be more attentive (at least I hope so)
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth
Manager  Joined: 10 Jun 2015
Posts: 110
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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enigma123 wrote:
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

from 661 to 699
66_ ( you have 8 numbers )
6_6 ( you have 3 numbers )
_66 ( you have 3 numbers)
Now from 700 to 799
There are 99 numbers in all. In this 72 numbers (9x8) do not satisfy the conditions
therefore, 99-72 =27 numbers satisfy the conditions
Hence, 27x3+14 = 95 is the answer.
Intern  B
Joined: 27 Oct 2015
Posts: 22
Of the three-digit integers greater than 660, how many have  [#permalink]

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I tried to use Bunuel's method, but it was hard for me to escape minor mistakes, consequently my answer wasn't correct.

I find the following way is easier to remember:

600 700 800 900
abc 3 9 9 9
abc 8 9 9 9
abc 3 9 9 9

9*10=90-1+6=95
Manager  B
Joined: 19 Jul 2017
Posts: 82
Location: India
Concentration: General Management, Strategy
GPA: 3.5
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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Bunuel wrote:
enigma123 wrote:
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

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

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

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Hello Bunuel, Considering this logic , can we say there would 9*8*7 three digit numbers with distinct digits
Math Expert V
Joined: 02 Sep 2009
Posts: 60630
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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1
Arsh4MBA wrote:
Bunuel wrote:
enigma123 wrote:
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

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

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

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Hello Bunuel, Considering this logic , can we say there would 9*8*7 three digit numbers with distinct digits

There are 9*9*8 3-digit numbers with distinct digits.

9 options for the first digit (from 1 to 9 inclusive).
9 options for the second digit (from 0 to 9 inclusive minus the one we used for the first digit ).
8 options for the third digit (from 0 to 9 inclusive minus 2 digits we used for the first and the second digits)
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Intern  B
Joined: 28 May 2017
Posts: 18
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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LalaB wrote:
my way of thinking is not elegant but anyways , see what I did -
we have the range 661-998

1. two 1st digits are the same
then we have-
661-669 (8 options (exclude 666))
770-779(9 options (exclude 777))
880-889 (9 options (exclude 888))
990-998 (9 options (exclude 999))
total=9*3+8=35

2.the 1st and the last are the same
then we have
6*6 (3 options- 676 686 696))
7*7 (9 options (exclude 777))
8*8 (9 options (exclude 888))
9*9 (9 options (exclude 999))
total=9*3+3=30

3. the 2nd and the 3d terms are the same.
then we have
66x- none option, since our first two digits are already the same, and the 3d cant be the same
7xx -9 options (exclude 777)
8xx- 9 options (exclude 888)
9xx 9 options (exclude 999)
total=9*3=27

30+35+27=92

cant find the rest 3 numbers to get 95. I will appreciate if someone helps me 700 , 800 , 900 are missing you have to consider these as well

Sent from my iPhone using GMAT Club Forum mobile app
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Joined: 19 Oct 2013
Posts: 511
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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Bunuel wrote:
enigma123 wrote:
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

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

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

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Hi Bunuel can you kindly elaborate on the first part of the "all the digits are distinct" I didn't quite understand the 3*8, however, I do understand the second part.

Thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 60630
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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1
Salsanousi wrote:
Bunuel wrote:
enigma123 wrote:
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

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

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

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Hi Bunuel can you kindly elaborate on the first part of the "all the digits are distinct" I didn't quite understand the 3*8, however, I do understand the second part.

Thanks!

Three-digit numbers greater than 660 with the hundreds digit of 6: 6XY - X can take only 3 values (7, 8, or 9) and Y can take 8 values (Out of 10 digits, not 6 and not the onte we used for X).
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Director  S
Joined: 17 Dec 2012
Posts: 622
Location: India
Of the three-digit integers greater than 660, how many have  [#permalink]

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1
Let us first take 700-799

With tens digit 7 we have 9 cases. With units digit 7 we have 9 cases. With tens and units digits same we have 9 cases for a total of 27 cases. Similarly for 800-899 and 900-999 for a total of 81 cases.

Let us now take 661-699

With tens digit 6 , we have 8 cases. With units digit 6, we have 3 cases and with units and tens digit same. we have 3 cases for a total of 14 cases.

So for numbers greater than 660 we have 81+14=95 cases
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Location: Kuwait
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Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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Bunuel Thanks!
Intern  B
Joined: 04 Oct 2016
Posts: 17
Re: Of the three-digit integers greater than 660, how many have  [#permalink]

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Let's breakdown the problem into 2 sections:
i) 661 - 699:
case I: x x y
- - - x -> takes only one value i.e., 6 and y takes 9 digits (0,1,2,3,4,5,7,8,9. Since we are already using up 6) It becomes 1*1*9=9
case II: x y x
Applying the same rule as above, we again get 9.
Hence 9+9 numbers between 661-699

ii) Consider 700-999

X X Y => X can take 3 numbers (7,8,9) and Y can take 9 numbers, giving 3*1*9 = 27
X Y X and Y X X=> Same logic as above we get 27*3=81

i) + ii) = 18+81 = 99

Now out of 99, we have 666,777,888,999 that can not be considered.

99-6=95 Re: Of the three-digit integers greater than 660, how many have   [#permalink] 26 Jan 2019, 11:08
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