Welcome! Check out our free B-School Guides to learn how you compare with other applicants.
Login or Register
 

Sum equal to 5

This topic has 1 expert reply and 1 member reply
Bookmark
Previous Topic Next Topic
rajatvmittal Really wants to Beat The GMAT! Default Avatar
Joined
11 Feb 2012
Posted:
101 messages
Thanked:
2 times
Target GMAT Score:
720+
Sum equal to 5 Post Mon Mar 12, 2012 8:26 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    How many positive integers less than 10,000 are there in which sum of digits equals 5?

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    Neo Anderson Rising GMAT Star Default Avatar
    Joined
    06 Oct 2011
    Posted:
    92 messages
    Thanked:
    18 times
    Post Mon Mar 12, 2012 9:47 am
    that means a four digit number whose digits should sum up to 5:-
    number containing: 2,1,1,1 => 4 such numbers
    2,2,1,0 => 12 such numbers
    3,1,1,0 => 12 such numbers
    3,2,0,0 => 12 such numbers
    4,1,0,0 => 12 such numbers
    5,0,0,0 => 4 such numbers
    thus total of 4+12+12+12+12+4 = 56 numbers with sun of digit =5 and number < 10,000[/u]

    GMAT/MBA Expert

    Anurag@Gurome GMAT Instructor
    Joined
    02 Apr 2010
    Posted:
    3835 messages
    Followed by:
    465 members
    Thanked:
    1746 times
    GMAT Score:
    770
    Post Mon Mar 12, 2012 9:31 pm
    rajatvmittal wrote:
    How many positive integers less than 10,000 are there in which sum of digits equals 5?
    We need to find integers between 0 to 9999, in which the sum of digits adds up to 5.

    (1) One digit is 5 and all other are 0: 0005, 0050, 0500, 5000 or we can say that no. of ways we can arrange the digits = 4!/3! = 4 ways
    (2) Three 1's and one 2: 1112, this can be done in 4!/3! = 4 ways
    (3) One 4 and one 1: 4100, this can be done in 4!/2! = 12 ways
    (4) One 3 and one 2: 3200, this can be done in 4!/2! = 12 ways
    (5) One 3 and two 1's: 3110, this can be done in 4!/2! = 12 ways
    (6) Two 2's and One 1: 2210, this can be done in 4!/2! = 12 ways

    Therefore, required number of positive integers = (12 * 4) + (4 * 2) = 48 + 8 = 56

    The correct answer is C.

    _________________
    Anurag Mairal, Ph.D., MBA
    GMAT Expert, Admissions and Career Guidance
    Gurome, Inc.
    1-800-566-4043 (USA)

    Join Our Facebook Groups
    GMAT with Gurome
    https://www.facebook.com/groups/272466352793633/
    Admissions with Gurome
    https://www.facebook.com/groups/461459690536574/
    Career Advising with Gurome
    https://www.facebook.com/groups/360435787349781/

    Best Conversation Starters
    1 varun289 43 topics
    2 greenwich 30 topics
    3 sana.noor 22 topics
    4 guerrero 20 topics
    5 killerdrummer 18 topics
    See More Top Beat The GMAT Members...

    Most Active Experts
    1 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    		 205 posts
    2 image description Anju@Gurome

    Gurome

    		 146 posts
    3 image description GMATGuruNY

    The Princeton Review Teacher

    		 145 posts
    4 image description Jim@StratusPrep

    Stratus Prep

    		 96 posts
    5 image description David@VeritasPrep

    Veritas Prep

    		 39 posts
    See More Top Beat The GMAT Experts