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

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10 Mar 2012, 04:45

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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?

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).

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

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10 Mar 2012, 09:40

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
_________________

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

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.
_________________

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

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10 Mar 2012, 09:56

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)
_________________

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

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

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15 Jul 2014, 04:58

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

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12 Aug 2015, 23:38

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

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13 Aug 2015, 03:15

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.

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

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20 Mar 2017, 04:43

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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25 Aug 2017, 07:05

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.

Answer: D.

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

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.

Answer: D.

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

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25 Aug 2017, 08:51

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

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

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25 Aug 2017, 08:54

Bunuel wrote:

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