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Of the threedigit integers greater than 660, how many have
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10 Mar 2012, 04:45
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Of the threedigit 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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Re: 3 digit integers
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10 Mar 2012, 05:36
enigma123 wrote: Of the threedigit 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 (3digit 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, 3392404=95. Answer: D.
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Re: Of the threedigit integers greater than 660, how many have
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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 661998 1. two 1st digits are the same then we have 661669 (8 options (exclude 666)) 770779(9 options (exclude 777)) 880889 (9 options (exclude 888)) 990998 (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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Re: Of the threedigit integers greater than 660, how many have
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10 Mar 2012, 09:53
LalaB wrote: my way of thinking is not elegant but anyways , see what I did  we have the range 661998 1. two 1st digits are the same then we have 661669 (8 options (exclude 666)) 770779(9 options (exclude 777)) 880889 (9 options (exclude 888)) 990998 (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 threedigit integers greater than 660, how many have
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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)
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Re: Of the threedigit integers greater than 660, how many have
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13 Aug 2015, 03:15
enigma123 wrote: Of the threedigit 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, 9972 =27 numbers satisfy the conditions Hence, 27x3+14 = 95 is the answer.



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Of the threedigit integers greater than 660, how many have
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24 Jan 2016, 00:28
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=901+6=95



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Re: Of the threedigit integers greater than 660, how many have
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25 Aug 2017, 07:05
Bunuel wrote: enigma123 wrote: Of the threedigit 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 (3digit 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, 3392404=95. Answer: D. Hello Bunuel, Considering this logic , can we say there would 9*8*7 three digit numbers with distinct digits



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Re: Of the threedigit integers greater than 660, how many have
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25 Aug 2017, 08:36
Arsh4MBA wrote: Bunuel wrote: enigma123 wrote: Of the threedigit 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 (3digit 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, 3392404=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 3digit 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 threedigit integers greater than 660, how many have
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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 661998 1. two 1st digits are the same then we have 661669 (8 options (exclude 666)) 770779(9 options (exclude 777)) 880889 (9 options (exclude 888)) 990998 (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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Re: Of the threedigit integers greater than 660, how many have
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12 Nov 2017, 13:03
Bunuel wrote: enigma123 wrote: Of the threedigit 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 (3digit 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, 3392404=95. Answer: D. 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!



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Re: Of the threedigit integers greater than 660, how many have
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12 Nov 2017, 21:13
Salsanousi wrote: Bunuel wrote: enigma123 wrote: Of the threedigit 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 (3digit 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, 3392404=95. Answer: D. 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! Threedigit 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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Of the threedigit integers greater than 660, how many have
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12 Nov 2017, 21:39
Let us first take 700799 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 800899 and 900999 for a total of 81 cases. Let us now take 661699 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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Re: Of the threedigit integers greater than 660, how many have
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Re: Of the threedigit integers greater than 660, how many have
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26 Jan 2019, 11:08
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 661699
ii) Consider 700999
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.
996=95




Re: Of the threedigit integers greater than 660, how many have
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