Thursday, October 9, 2008

Problem Of The Week 50

Find the number of unordered triplets (x,y,z) of positive integers such that x³+y³+z³=2008

1) 0

2) 1

3) 2

4) 3

5) None of these
Posted by Unknown at 9:35 PM
Labels: , , ,

6 comments:

  1. ankajOctober 10, 2008 at 12:36 AM

    is the answer 1 (4,6,12)????

    ReplyDelete
  2. outtimedOctober 10, 2008 at 12:42 AM

    no

    ReplyDelete
  3. Celebrating LifeOctober 13, 2008 at 2:31 PM

    2 pairs (12,6,4)(10,10,2)

    ReplyDelete
  4. wolverineOctober 29, 2008 at 6:24 PM

    any general way to solve it...??
    or u just reached the conclusion by making an analogous C program or by hit and trial??

    ReplyDelete
  5. GrimreaperSeptember 14, 2011 at 8:18 AM

    hmmm na... simple logic...
    first realise that since lhs is cubes... try to break rhs into a factor having a cube..
    so i write 2008 as 8*251
    that clearly tells me that each entity in lhs is an even number
    so new equation is
    x^3+y^3+z^3 = 251

    now said that... u do agree max value can be 6.. since else cube wld be greater than 251
    so set z=6 we directly get one set as 6,3,2

    now, dont roll yo eyes saying its hit and trial... see that i wnt go for z=4 at all... now think why so... why ? :)

    all one need to check is for z=6 and z=5 both which comes out in a min thats all....

    again think why i say z=4 case makes no sense at all...

    so ur answers are 12,6,4 and 10,10,2

    i had removed two earlier if u remember... :)
    cheers enjoy.....

    ReplyDelete
  6. GrimreaperSeptember 14, 2011 at 8:20 AM

    feel free to ping :)

    ReplyDelete

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