Permutation and Combination:

Consider the set 'S(n)' formed by erasing digits at some (0 or more) positions of a number 'n'.

For example: if n=123, S(123)={123,12,13,23,1,2,3}.
Another one: if n=122, S(122)={122,12,12,22,1,2,2}.

Let 's' denote the sum of all elements of 'S'. Find s%9 for n=12391227.

Note:
1) '%' means remainder. 5%2=1.
Some Identities:
2) (a+b)%c=(a%c+b%c)%c
3) (a*b)%c=((a%c)*(b%c))%c.

9 comments:

  1. rohinJune 8, 2011 at 8:48 PM

    0?

    ReplyDelete
  2. AnujJune 8, 2011 at 10:06 PM

    clarification:
    lets take 122:

    erasing nothing : 122
    erasing position 1: 12
    erasing position 2: 12 (again)
    erasing position 3: 22
    erasing position 1 and 2 : 1
    erasing position 2 and 3 : 2
    erasing position 3 and 1 : 3

    sum=122+12+12+22+1+2+3
    ans=sum%9

    ReplyDelete
  3. Pulkit MendirattaJune 9, 2011 at 9:30 AM

    Ya,i also got the same answer,zero

    ReplyDelete
  4. Pulkit MendirattaJune 9, 2011 at 9:32 AM

    Identities were quite usefull

    ReplyDelete
  5. rohinJune 9, 2011 at 12:47 PM

    is the answer correct?

    ReplyDelete
  6. rishabhJune 9, 2011 at 4:27 PM

    with use of properties it is quite simple

    ReplyDelete
  7. AnujJune 9, 2011 at 10:30 PM

    if someone hasn't solved it yet, here's a hint:
    (93928)%9=(9+3+9+2+8)%9.
    i.e. if 'b' is the sum of digits in 'a', then
    a%9=b%9.

    everyone should try this one, as it helps you revise a standard PnC problem:

    ReplyDelete
  8. ShivankerJune 10, 2011 at 10:51 AM

    i too guess its zero..

    ReplyDelete
  9. Vinayak MehtaNovember 2, 2011 at 12:41 AM

    0

    ReplyDelete

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