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Digits: Of the 3-digit integers greater than 700, how many h

This topic has 4 expert replies and 9 member replies
II Master | Next Rank: 500 Posts
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Digits: Of the 3-digit integers greater than 700, how many h

Post Mon Mar 17, 2008 5:45 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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

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    mnjoosub Junior | Next Rank: 30 Posts Default Avatar
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    Post 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 Default Avatar
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    Post Mon Mar 17, 2008 6:34 am
    Sorry I forgot the attachment. Wink
    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 Default Avatar
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    Post Wed Aug 06, 2008 2:10 pm
    is there any other way to solve this problem? perhaps a faster method?

    parallel_chase Legendary Member Default Avatar
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    Post 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
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    Post 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

    Post 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?

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    Post 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

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    Post 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?

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    BestGMATEliza Master | Next Rank: 500 Posts
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    Post 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)

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    Post 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.

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    Post 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

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    Post 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

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    Anaira Mitch Master | Next Rank: 500 Posts
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    Post 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.

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

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