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Register now and save up to $200 Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Most awarded test prep in the world Now free for 30 days Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code Digits: Of the 3-digit integers greater than 700, how many h tagged by: Brent@GMATPrepNow This topic has 4 expert replies and 9 member replies II Master | Next Rank: 500 Posts Joined 10 Dec 2007 Posted: 400 messages Upvotes: 19 Target GMAT Score: 700 GMAT Score: 680 Digits: Of the 3-digit integers greater than 700, how many h Mon Mar 17, 2008 5:45 am Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ? (A) 90 (B) 82 (C) 80 (D) 45 (E) 36 I am keen to understand different ways of answering this question. Thanks in advance. II Last edited by II on Mon May 05, 2008 1:54 am; edited 1 time in total GMAT/MBA Expert GMATGuruNY GMAT Instructor Joined 25 May 2010 Posted: 13901 messages Followed by: 1808 members Upvotes: 13060 GMAT Score: 790 Thu Jul 10, 2014 2:52 am II wrote: Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ? (A) 90 (B) 82 (C) 80 (D) 45 (E) 36 Integers with exactly 2 digits the same = Total integers - Integers with all 3 digits the same - Integers with all 3 digits different. Total integers: To count consecutive integers, use the following formula: Number of integers = biggest - smallest + 1. Thus: Total = 999 - 701 + 1 = 299. Integers with all 3 digits the same: 777, 888, 999. Number of options = 3. Integers with all 3 digits different: Number of options for the hundreds digit = 3. (7, 8, or 9) Number of options for the tens digit = 9. (Any digit 0-9 other than the digit already used.) Number of options for the units digit = 8. (Any digit 0-9 other than the two digits already used.) To combine these options, we multiply: 3*9*8 = 216. Thus: Integers with exactly 2 digits the same = 299-3-216 = 80. The correct answer is C. _________________ Mitch Hunt GMAT Private Tutor GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. BestGMATEliza Master | Next Rank: 500 Posts Joined 23 Jun 2014 Posted: 103 messages Followed by: 12 members Upvotes: 25 Test Date: April 29, 2014 GMAT Score: 770 Wed Jul 09, 2014 9:51 pm you only have 3 possibilities for the hundreds digit: 7, 8 or 9. 700s you can have 7xx (ex: 722), for this there are 8 possibilities for digits (0 doesn't count because it must be greater than 700, neither does 7) there is also 7x7 (ex: 702), for this there are 9 possibilities (only 7 doesn't count) then there is 77x (ex:771), for this there are also 9 possibilities 800s 8xx- 9 possibilities, because you can include 0 so 0-9, not including 8 8x8- 9 possibilities 88x- 9 possibilities 900s 9xx- 9 possibilities, because you can include 0 so 0-9, not including 8 9x9- 9 possibilities 99x- 9 possibilities Add them all up and you get 80 (C) _________________ Eliza Chute Best GMAT Prep Courses GMAT course comparison and reviews Your one stop for all your GMAT studying needs! mnjoosub Junior | Next Rank: 30 Posts Joined 27 Feb 2008 Posted: 17 messages Upvotes: 2 Target GMAT Score: 700+ Mon Mar 17, 2008 6:30 am I don't know whether there is a proper mathematical way to solve this problem but I did it this way. I tabulate is as follows: see attachment from the table we can see that the total repeated Nos for 700 = 30 So for 800 and 900 inclusive = 30 *3 = 90 We should less repeated Nos (3*3 = 9) = 81 Remember the question states more than 700 , so less 1 more = 80. Ans = C mnjoosub Junior | Next Rank: 30 Posts Joined 27 Feb 2008 Posted: 17 messages Upvotes: 2 Target GMAT Score: 700+ Mon Mar 17, 2008 6:34 am Sorry I forgot the attachment. Attachments This post contains an attachment. You must be logged in to download/view this file. Please login or register as a user. sofia cuevas Newbie | Next Rank: 10 Posts Joined 30 May 2008 Posted: 1 messages Wed Aug 06, 2008 2:10 pm is there any other way to solve this problem? perhaps a faster method? parallel_chase Legendary Member Joined 20 Jun 2007 Posted: 1153 messages Followed by: 2 members Upvotes: 146 Target GMAT Score: V50 Wed Aug 06, 2008 3:07 pm Here is a super fast way of solving this question as compared to above method. For first digit we have 3 letters to play with (7,8,9) Next two digits we can have any letter (0,1,2,3,4,5,6,7,8,9) CASE I (ABB) 3*9*1 = 27 CASE II (BAB) 3*9*1 = 27 CASE III (BBA) 3*1*9 =27 27+27+27 = 81 subtract 1 because we want the number to be greater than 700, the above combination include 700 81-1=80 Once you are comfortable with permutations & combination of digits this thing will take you less than 10 secs and I mean this. Let me know if you have any doubts. guru_1971 Newbie | Next Rank: 10 Posts Joined 10 Dec 2010 Posted: 1 messages Fri Dec 10, 2010 1:02 am The correct answer is 36 3 digit integers greater than 700. 2 cases case 1: when first digit is 7 then using counting priniciple 1st digit has an option of 1 , and hence since two digits are equal the 2nd digit also is 1 third digit is between 1- 9 ( number greater than 700= 701 ) since the first 2 digits have taken integer 7, third digit will have (9-1) = 8 so the counting principle equation for case 1 is ( 1x1x8) 1 1 8 ( 9-1) --- X ----------------- X --------------------- 1st digit is 7 2nd digit is same as 1st 3rd digit is 1-9 = 9 case 2 1st digit is not 7, so for the 1 and 2 digit, options are 8, 9 so 2 options for digit 1 and 2 options for digit 2 ( as per the condition of the problem ) now for 3 digit, since the number is greater than 700 digit will be from 1 - 9= and the 1st 2 digits have taken 7 and 8, so third digit will be 9-2= 7 so counting principle equation becomes ( 2 x2x7) 2 2 7 ---------- X --------------------- X --------------------------------- available same as 1st digit here 8 and 9 are taken by 1st 2 digits so 9 -2=7 option is 8 or 9 Solution thus becomes ( 1x1x8)+(2x2x7) = 36 diebeatsthegmat Legendary Member Joined 07 May 2010 Posted: 1119 messages Followed by: 2 members Upvotes: 29 Fri Dec 10, 2010 5:26 am II wrote: Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ? (A) 90 (B) 82 (C) 80 (D) 45 (E) 36 I am keen to understand different ways of answering this question. Thanks in advance. II here the answer i find is also C 80 whats the answer? GMAT/MBA Expert kevincanspain GMAT Instructor Joined 22 Mar 2007 Posted: 613 messages Followed by: 63 members Upvotes: 171 GMAT Score: 790 Fri Dec 10, 2010 5:29 am Parallel chase's method is excellent! Also, there are 299 3-digit integers greater than 700. How many of these do not have exactly two digits equal to each other? There are 3 such integers that have three digits equal to each other (777,888,999) and 3 x 9 x 8 =216 that feature three distinct digits. Thus there are 299 - 3 - 216 = 80 such integers that have exactly two digits equal to each other _________________ Kevin Armstrong GMAT Instructor Gmatclasses Madrid Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today! GMAT/MBA Expert kevincanspain GMAT Instructor Joined 22 Mar 2007 Posted: 613 messages Followed by: 63 members Upvotes: 171 GMAT Score: 790 Fri Dec 10, 2010 5:32 am guru_1971 wrote: The correct answer is 36 3 digit integers greater than 700. 2 cases case 1: when first digit is 7 then using counting priniciple 1st digit has an option of 1 , and hence since two digits are equal the 2nd digit also is 1 third digit is between 1- 9 ( number greater than 700= 701 ) since the first 2 digits have taken integer 7, third digit will have (9-1) = 8 so the counting principle equation for case 1 is ( 1x1x8) 1 1 8 ( 9-1) --- X ----------------- X --------------------- 1st digit is 7 2nd digit is same as 1st 3rd digit is 1-9 = 9 case 2 1st digit is not 7, so for the 1 and 2 digit, options are 8, 9 so 2 options for digit 1 and 2 options for digit 2 ( as per the condition of the problem ) now for 3 digit, since the number is greater than 700 digit will be from 1 - 9= and the 1st 2 digits have taken 7 and 8, so third digit will be 9-2= 7 so counting principle equation becomes ( 2 x2x7) 2 2 7 ---------- X --------------------- X --------------------------------- available same as 1st digit here 8 and 9 are taken by 1st 2 digits so 9 -2=7 option is 8 or 9 Solution thus becomes ( 1x1x8)+(2x2x7) = 36 Are you counting possibilities such as 707? _________________ Kevin Armstrong GMAT Instructor Gmatclasses Madrid Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today! GMAT/MBA Expert Brent@GMATPrepNow GMAT Instructor Joined 08 Dec 2008 Posted: 11275 messages Followed by: 1225 members Upvotes: 5254 GMAT Score: 770 Thu Jul 10, 2014 6:20 am Quote: 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 One approach is to start LISTING numbers and look for a PATTERN. Let's first focus on the numbers from 800 to 899 inclusive. We have 3 cases to consider: 8XX, 8X8, and 88X 8XX 800 811 822 . . . 899 Since we cannot include 888 in this list, there are 9 numbers in the form 8XX 8X8 808 818 828 . . . 898 Since we cannot include 888 in this list, there are 9 numbers in the form 8X8 88X 880 881 882 . . . 889 Since we cannot include 888 in this list, there are 9 numbers in the form 88X So, there are 27 (9+9+9) numbers from 800 to 899 inclusive that meet the given criteria. Using the same logic, we can see that there are 27 numbers from 900 to 999 inclusive that meet the given criteria. And there are 27 numbers from 700 to 999 inclusive that meet the given criteria. HOWEVER, the question says that we're looking at numbers greater than 700, so the number 700 does not meet the criteria. So, there are actually 26 numbers from 701 to 799 inclusive that meet the given criteria. So, our answer is 27+27+26 = 80 = C Cheers, Brent _________________ Brent Hanneson â€“ Founder of GMATPrepNow.com Use our video course along with Check out the online reviews of our course Come see all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! GMATinsight Legendary Member Joined 10 May 2014 Posted: 1011 messages Followed by: 21 members Upvotes: 205 Fri Jul 11, 2014 9:49 am Quote: Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ? (A) 90 (B) 82 (C) 80 (D) 45 (E) 36 Between 700 to 799 770, 771, 771, 773, 774, 775, 776, 778, 779 = 9 Numbers 707, 717, 727, 737, 747, 757, 767, 787, 797 = 9 Numbers 711, 722, 733, 744, 755, 766, 788, 799 = 8 Numbers Total Such Numbers = 9+8+9 = 26 Between 800 to 899 880, 881, 881, 883, 884, 885, 886, 887, 889 = 9 Numbers 808, 818, 828, 838, 848, 858, 868, 878, 898 = 9 Numbers 800, 811, 822, 833, 844, 855, 866, 877, 899 = 9 Numbers Total Such Numbers = 9+9+9 = 27 Between 900 to 999 990, 991, 992, 993, 994, 995, 996, 997, 998 = 9 Numbers 909, 919, 929, 939, 949, 959, 969, 979, 989 = 9 Numbers 900, 911, 922, 933, 944, 955, 966, 977, 988 = 9 Numbers Total Such Numbers = 9+9+9 = 27 Total Numbers = 26+27+27 = 80 Answer _________________ Bhoopendra Singh & Sushma Jha - Founder "GMATinsight" Testimonials e-mail: info@GMATinsight.com I Mobile: +91-9999687183 / +91-9891333772 To register for One-on-One FREE ONLINE DEMO Class Call/e-mail One-On-One Private tutoring fee - US$40 per hour & for FULL COURSE (38 LIVE Sessions)-US\$1000

Anaira Mitch Master | Next Rank: 500 Posts
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Mon Nov 07, 2016 6:36 am
GMATGuruNY wrote:
II wrote:
Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Integers with exactly 2 digits the same = Total integers - Integers with all 3 digits the same - Integers with all 3 digits different.

Total integers:
To count consecutive integers, use the following formula:
Number of integers = biggest - smallest + 1.
Thus:
Total = 999 - 701 + 1 = 299.

Integers with all 3 digits the same:
777, 888, 999.
Number of options = 3.

Integers with all 3 digits different:
Number of options for the hundreds digit = 3. (7, 8, or 9)
Number of options for the tens digit = 9. (Any digit 0-9 other than the digit already used.)
Number of options for the units digit = 8. (Any digit 0-9 other than the two digits already used.)
To combine these options, we multiply:
3*9*8 = 216.

Thus:
Integers with exactly 2 digits the same = 299-3-216 = 80.

Amazing solution Mitch. Thanks for your guidance. Official Guide explanation is quite complex to understand.

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