Puzzle #162: Who finished 2nd in 100m?

Very nice puzzle from the NSA collection. We have seen versions of this in the past, I loved this one as well.

Three athletes (and only three athletes) participate in a series of track and field events. Points are awarded for 1st, 2nd, and 3rd place in each event (the same points for each event, i.e. 1st always gets “x” points, 2nd always gets “y” points, 3rd always gets “z” points), with x > y > z > 0, and all point values being integers.

The athletes are named Adam, Bob, and Charlie.

  • Adam finished first overall with 22 points
  • Bob won the Javelin event and finished with 9 points overall.
  • Charlie also finished with 9 points overall.

Question: Who finished second in the 100-meter dash (and why)?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy running!

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3 Responses to Puzzle #162: Who finished 2nd in 100m?

  1. Abhinav Jain says:

    Adam
    In total there are 5 events and x=5,y=2 and z=1 one can figure it with simple reasoning.

  2. 22+9+9=40=n(x+y+z) where n is number of events
    3+2+1= 22
    number of events * z <= 9

    Since Bob one event and has 9 points, x + (n-1)*1 <= 9, x + n =5, z=1, x+y+z=8 => (x=5, y=2, z=1)

    Case n = 4:
    x>=6, and x+y+z=10, x+4 (x=6,y=3,z=1).
    But Adam can never make it to 22 = 6+6+6+?. So, this case is not possible.

    Case n = 2:
    x>=11, x+2<=10 Null set

    So, 5 events, (x=5, y=2, z=1)
    Adam won 4 events and came 2nd in Javelin.
    Bob won Javelin event and came 3rd in other 4.
    Charlie came 3rd in Javelin and came 2nd is other 4.

    Adam won 100m race.

  3. Vishal Poddar says:

    Charlie.
    22+9+9=40=(x+y+z)*n
    x+y+z>=6, therefore n<=6
    40 = (8,5) or (10,4) or (20,2)
    As Bob finish first in one event and his total is 9, then x+y+z<9
    Therefor, 40=8*5 is the case which implies x+y+z=8 and n = 5
    Now, (x,y,z) can be (4,3,1) or (5,2,1)
    Max by (4,3,1) can be 20, therefore not possible
    x=5, y=2, z=1 and n=5
    Adam – 1st in all except javelin
    Bob – 1st in javelin and 3rd in rest
    Charlie – 3rd in javelin and 2nd in rest

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