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

Sum of all 3-digit numbers

This topic has 11 member replies

GMAT/MBA Expert

money9111 GMAT Titan
Joined
19 Apr 2009
Posted:
2075 messages
Followed by:
69 members
Thanked:
106 times
Sum of all 3-digit numbers Post Sat Feb 06, 2010 1:19 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Here's another:

    What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?

    _________________
    "Pre-Order" - Getting To The Core: In-Depth MBA Admissions Strategies for the Serious Applicant

    MBA Admissions Advice YouTube Channel

    Blogged my way through the application process, 1st & 2nd year of school at Johnson at Cornell University - Graduated 5/26/2013: My MBA Journey! | 1 New Post on 10/10

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!

    GMAT/MBA Expert

    Osirus@VeritasPrep GMAT Instructor Default Avatar
    Joined
    28 May 2009
    Posted:
    1579 messages
    Followed by:
    33 members
    Thanked:
    124 times
    GMAT Score:
    760
    Post Sat Feb 06, 2010 1:23 pm
    money9111 wrote:
    Here's another:

    What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?
    each digit can be in ones place twice, the tens place twice and the hundreds place twice. You would add 3*2 + 4*2 + 5*2 = 24.

    You would then do this:

    24 *100 + 24* 10 + 24 *1

    _________________
    http://www.beatthegmat.com/the-retake-osirus-blog-t51414.html

    Brandon Dorsey
    GMAT Instructor
    Veritas Prep

    Buy any Veritas Prep book(s) and receive access to 5 Practice Cats for free! Learn More.

    Get 7 Brand-New GMAT Practice Tests from Veritas Prep for Just $49! Veritas Prep’s computer-adaptive GMAT practice test are powered by artificial intelligence, guaranteeing that you see the most realistic GMAT questions and most accurate scores. Get Your Tests Now.

    GMAT/MBA Expert

    money9111 GMAT Titan
    Joined
    19 Apr 2009
    Posted:
    2075 messages
    Followed by:
    69 members
    Thanked:
    106 times
    Post Sat Feb 06, 2010 1:26 pm
    how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.

    _________________
    "Pre-Order" - Getting To The Core: In-Depth MBA Admissions Strategies for the Serious Applicant

    MBA Admissions Advice YouTube Channel

    Blogged my way through the application process, 1st & 2nd year of school at Johnson at Cornell University - Graduated 5/26/2013: My MBA Journey! | 1 New Post on 10/10

    harsh.champ GMAT Titan
    Joined
    20 Jul 2009
    Posted:
    1132 messages
    Followed by:
    6 members
    Thanked:
    54 times
    Test Date:
    30/03/2012
    Target GMAT Score:
    780+
    GMAT Score:
    760
    Post Sat Feb 06, 2010 1:32 pm
    money9111 wrote:
    how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.
    I will clarify to you the problem approach.I guess you didn't understand it clearly.
    Quote:
    What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?
    The no. of possible such 3-digit numbers are 3! i.e. 6 numbers.

    Now,each no. can only occur for 2 times in units/tens/hundreds place.
    so, for units place = 2x(3+4+5) = 24
    for tens place= 2x10(3+4+5)=240
    for hundreds place = 2x100(3+4+5)=2400

    Hence,the sum of the digits will be 2400+240+24 = 2664.



    Alternatively,since only 6 numbers are there you can even write it down and solve the question
    345
    354
    435
    453
    534
    543

    Quickly,add the no.s and you will get the result.

    But if your problem approach is clear,I found the formal method to quicker in this question.

    _________________
    It takes time and effort to explain, so if my comment helped you please press Thanks button Smile



    Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

    "Keep Walking" - Johnny Walker Razz



    Last edited by harsh.champ on Sat Feb 06, 2010 1:59 pm; edited 2 times in total

    GMAT/MBA Expert

    money9111 GMAT Titan
    Joined
    19 Apr 2009
    Posted:
    2075 messages
    Followed by:
    69 members
    Thanked:
    106 times
    Post Sat Feb 06, 2010 1:38 pm
    huh?

    _________________
    "Pre-Order" - Getting To The Core: In-Depth MBA Admissions Strategies for the Serious Applicant

    MBA Admissions Advice YouTube Channel

    Blogged my way through the application process, 1st & 2nd year of school at Johnson at Cornell University - Graduated 5/26/2013: My MBA Journey! | 1 New Post on 10/10

    GMAT/MBA Expert

    Osirus@VeritasPrep GMAT Instructor Default Avatar
    Joined
    28 May 2009
    Posted:
    1579 messages
    Followed by:
    33 members
    Thanked:
    124 times
    GMAT Score:
    760
    Post Sat Feb 06, 2010 1:42 pm
    money9111 wrote:
    how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.
    Hmm, I begin by writing out:

    345
    354
    435
    453
    543
    534

    This gives you a visual. If you would find it faster to just add that up, then go about it that way. What you see from writing it out like that though is each digit is in each place twice. Knowing this you can add 3 4 and 5 and multiply by 2 since each digit is in each place twice. This gives you 24. You have 24 in the hundreds place the tens place and the ones place. That is why I multiplied 24 by 100, 10 and 1.

    I would recommend doing the approach that seems natural. You can add those numbers in less than a minute, so knowing the Manhattan way isn't necessary for this problem.

    _________________
    http://www.beatthegmat.com/the-retake-osirus-blog-t51414.html

    Brandon Dorsey
    GMAT Instructor
    Veritas Prep

    Buy any Veritas Prep book(s) and receive access to 5 Practice Cats for free! Learn More.

    Get 7 Brand-New GMAT Practice Tests from Veritas Prep for Just $49! Veritas Prep’s computer-adaptive GMAT practice test are powered by artificial intelligence, guaranteeing that you see the most realistic GMAT questions and most accurate scores. Get Your Tests Now.
    shashank.ism GMAT Titan
    Joined
    20 Jul 2009
    Posted:
    1010 messages
    Followed by:
    1 members
    Thanked:
    39 times
    Test Date:
    ND
    Target GMAT Score:
    780
    GMAT Score:
    NA
    Post Sat Feb 06, 2010 1:56 pm
    Quote:
    money9111 wrote:
    how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving
    see here we have three numbers 3,4,5 now if we take 1no. say 3 we have 2 left nos. i.e. 4 & 5 which can be placed in two ways as 45 and 54
    similarly taking 4 and 5 at hundreds place , you can easily observe each number is going to repeat 2 times at 100's 10's and unit place
    so the sum of all numbers which can be formed= (3+4+5)x2x100+ (3+4+5)x2x10+ (3+4+5)x2x1 =24 x(100+10+1) = 24x111 = 2664.

    _________________
    My Websites:
    http://webmaggu.com -India's online discount, coupon, free stuff informer.
    My blog: http://mba.webmaggu.com
    http://dictionary.webmaggu.com - A compact free online dictionary with images.

    Do you have any mnemonic - share it at http://mnemonic.webmaggu.com


    Nothing is Impossible, even Impossible says I'm possible.

    Thouraya Really wants to Beat The GMAT! Default Avatar
    Joined
    20 Jan 2010
    Posted:
    112 messages
    Thanked:
    1 times
    Post Wed Mar 02, 2011 1:13 am
    What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?

    Here's how I solved it:

    Number of integers: 543-345/1+1=199
    Average: 345+543/2=444
    Sum= 444*199=88,356

    What's the correct answer?

    BarryLi Rising GMAT Star Default Avatar
    Joined
    08 Feb 2011
    Posted:
    60 messages
    Thanked:
    7 times
    Test Date:
    March 11, 2011
    GMAT Score:
    700
    Post Wed Mar 02, 2011 1:51 am
    money9111 wrote:
    how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.
    You know that each digit will be in each place twice because of the combinations left for the other numbers.

    For example, 3 will be in the hundreds place twice because the digits 4 & 5 can only form two different numbers: 45 & 54. Hence the two numbers that have 3 in the hundreds column are 345 & 354.

    This example can be extended to the other digits and other place values.

    As a result, (3+3+4+4+5+5)*100 + (3+3+4+4+5+5)*10 + (3+3+4+4+5+5)*1 = 24*100 + 24*10 + 24*1, which is what everyone else has typed up.

    Knowing how to set up the question is the more difficult part of solving a problem, not the math. This is especially true with the more difficult quantitative questions imo.

    Thouraya Really wants to Beat The GMAT! Default Avatar
    Joined
    20 Jan 2010
    Posted:
    112 messages
    Thanked:
    1 times
    Post Wed Mar 02, 2011 1:55 am
    If I want to follow the strategy I used above, what is it that I am doing wrong?

    BarryLi Rising GMAT Star Default Avatar
    Joined
    08 Feb 2011
    Posted:
    60 messages
    Thanked:
    7 times
    Test Date:
    March 11, 2011
    GMAT Score:
    700
    Post Wed Mar 02, 2011 2:02 am
    Thouraya wrote:
    If I want to follow the strategy I used above, what is it that I am doing wrong?
    The average of all possible numbers will not always be the average of the minimum and maximum number.

    For example, 1, 2, and 5.

    Min: 125
    Max: 521
    Average of the Min and Max: 323



    List of Numbers:
    125
    152
    215
    251
    512
    521

    Average of the List of Numbers:296


    I believe the method using the min and max only worked in the original poster's case because the absolute difference of the 1st and 3rd digit from the median digit are equal, i.e. |4-3| = |4-5|.

    Thouraya Really wants to Beat The GMAT! Default Avatar
    Joined
    20 Jan 2010
    Posted:
    112 messages
    Thanked:
    1 times
    Post Wed Mar 02, 2011 4:39 am
    You're right, the mistake I did here is that I assume that the integers are consecutive, while they are not..and the theory that average of sequence is equal to average of smallest and biggest only holds true in CONSECUTIVE integers..

    Do you have an alternative approach to the one being presented above? I usually like to follow systematic approaches, rather than figure the pattern or think of the "logic" behind it. Any other approaches would greatly be appreciated. Thanks

    Best Conversation Starters

    1 j_shreyans 80 topics
    2 aditya8062 41 topics
    3 anksm22 35 topics
    4 RiyaR 31 topics
    5 kamalakarthi 29 topics
    See More Top Beat The GMAT Members...

    Most Active Experts

    1 image description GMATGuruNY

    The Princeton Review Teacher

    168 posts
    2 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    149 posts
    3 image description MBAPrepAdvantage

    MBAPrepAdvantage

    90 posts
    4 image description CriticalSquareMBA

    Critical Square

    78 posts
    5 image description Matt@VeritasPrep

    Veritas Prep

    54 posts
    See More Top Beat The GMAT Experts