Digits Question

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

← Next post Previous post →

Subscribe / Share

anupurna tagged this post with: GMAT Question of the Day Read 1 articles by anupurna

1. Manisha says:

   May 15, 2008 at 12:34 pm

   Though Options are not given….but according to me….

   Possible 3 Digit Nos. are: 345, 354, 435, 453, 543, 534 and hence the sum of these nos. is = 2664
2. SD says:

   May 15, 2008 at 12:48 pm

   2664
3. Amar says:

   May 15, 2008 at 3:57 pm

   1.B
   2.C
   3.E
   4.C
   5.E
   6.C
4. Rajib Banerjee says:

   May 15, 2008 at 8:56 pm

   345 + 354 + 435 + 453 + 543 + 534 = 2664
5. Anu says:

   May 15, 2008 at 10:27 pm

   thats right answer!!!!
6. Will says:

   May 16, 2008 at 12:09 am

   6 different combination.
   2664
7. Atul says:

   May 16, 2008 at 8:11 pm

   To form the 3 digit number using 3,4,5, total no. of ways to form a number with the given number without repetation :

   3*2*1 = 6

   Let us suppose three digit nol is ABC (A at hundreth place, B tense place, C at Unit place)

   when we form different combinations of the 3 digit numbers. No. of time every no. can appear on a particualar place (Unit, Ten etc) is

   6 (Total possible combinations) /3 (Total digits available) = 2

   So every no. will come 2 times in all the places (’3′ will come 2 times at A, 2 times at B and 2 times at C in total combinations. similarly 4 and 5). So sum total for ”Unit’s place” Ten’s place and Hunderd’s place individually are:

   Unit’s place = 3*2 + 4*2 + 5*2 = 24
   Ten’s place = 4*2 + 5*2 + 3*2 = 24
   Hundred’s place = 5*2 + 3*2 + 4*2 = 24

   forming the final total:

   24 / 24 / 24
   = 24 / (24 + 2) / 4 (Carry fwd 2 from unit’s place to ten’s place)
   = 24 / 26 / 4
   = (24 + 2) / 6 / 4 (Carry fwd 2 from ten’s place to Hunderd’s place)
   = 26/6/4

   Final answer = 2664

   The method will be very effective for higher digit nos (example sum of 5 digit no. using 1,2,3,4 and 5).
8. jaspreet cheema says:

   May 17, 2008 at 9:44 am

   Total should be = 345+354+435+453+543+534=2664
9. Ritula says:

   May 19, 2008 at 9:41 am

   Since each digit can appear only once.make the combinations by fixing one no and interchanging the other two .the nos are :
   345,354
   435,453
   534,543
   Now add the above nos .The result is 2664
10. Charanya says:

    May 19, 2008 at 8:04 pm

    there is a simple formula for it –

    (111)(n-1)(sum of the digits)
    = (111)(2)(12)
    = 2664
11. manisha says:

    May 20, 2008 at 12:07 am

    @ Charanya

    What is the basic concept behind this formula? whr all it can be actually used?
12. Vertica says:

    May 20, 2008 at 3:22 am

    Hey Atul

    Can this method be used for any kind of problem similar to give here?
    Nice concept..do you have some more tricks/ logical solutions…i would appreciate if you would share it!
13. Charanya says:

    May 21, 2008 at 1:40 pm

    @manisha…
    its a simple formula which someone taught me …

    (111..n times)(n – 1) (Sum of digits)

    The formula varies when one of the digits is zero.

    I’ll post it ASAP.
14. Atul says:

    June 2, 2008 at 3:23 pm

    @vertica,
    Apologies for the delay response .. Yes this method is applicable for all the scenarios.

    @Charanya / manisha,
    Nice formula .. there is a slight change is the formula to make it generic:

    suppose n is the no. of digits given. Our scenario it is 3 (digits given 3,4,5)

    Formula will be:

    (n times 1) * {fact (n-1)} * (sum of digits)

    explaination:
    First term : ‘n times 1′

    for n = 3 this means ’3 times 1′ i.e. 111
    for n = 4 this will be ’4 times 1′ i.e. 1111

    Second term : fact(n-1) ……..(factorial of (n-1)

    for n = 3 this means fact(3-1) = fact (2) = 2
    for n = 4 it will be !(4-1) = !3 = 6

    Third term : Sum of the digits

    in our case = 3+4+5 = 12

    so the solution of our problem with this formula will be as Charanya explained:

    = (3 times 1) (fact(3-1) (3+4+5)
    = (111)(!2)(12) = 111*2*12
    = 2664

    This formula can be used in the scenarios were ’0′ is also present in the given digits. Hope it will help.
15. manisha says:

    June 3, 2008 at 11:48 am

    @ Atul

    Great Explanation !!! Thanks!!
16. kapil dhawan says:

    June 4, 2008 at 9:25 am

    is this method simple or formula
    the possible no. of 3,4,5 are
    345
    354
    435
    453
    534
    543
    ____
    2664
    ____
17. Navya says:

    June 13, 2008 at 2:40 pm

    @ Kapil

    Your md is simpler fr this sum.. But this formula enhances our capability ot solvesuch sums even with more no of digits… and generic formulae always help .. thanks atul for the formula and explanation.. very helpful