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# Of the three-digit integers greater than 660, how many have

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Of the three-digit integers greater than 660, how many have [#permalink]

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10 Mar 2012, 03:45
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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?
[Reveal] Spoiler: OA

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10 Mar 2012, 04:36
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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?

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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Re: Of the three-digit integers greater than 660, how many have [#permalink]

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10 Mar 2012, 08:40
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
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Kudos [?]: 92848 [1] , given: 10528

Re: Of the three-digit integers greater than 660, how many have [#permalink]

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10 Mar 2012, 08:53
1
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Expert's post
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.
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Re: Of the three-digit integers greater than 660, how many have [#permalink]

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10 Mar 2012, 08:56
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)
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Re: Of the three-digit integers greater than 660, how many have [#permalink]

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15 Jul 2014, 03:58
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Re: Of the three-digit integers greater than 660, how many have [#permalink]

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12 Aug 2015, 22:38
Hello from the GMAT Club BumpBot!

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Re: Of the three-digit integers greater than 660, how many have [#permalink]

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13 Aug 2015, 02:15
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.
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Of the three-digit integers greater than 660, how many have [#permalink]

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23 Jan 2016, 23:28
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
Of the three-digit integers greater than 660, how many have   [#permalink] 23 Jan 2016, 23:28
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