Solution 90: Logic Problem on Types of Cars

  • Published on May 4, 2019
  • Let's carefully reason our way through the jungle of subtle arguments to identify two non-fully defective cars.
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Comments • 19

  • Aakash Kumar
    Aakash Kumar 4 months ago +1

    hey bro can you help me with a problem

    how many functions f:N->N satisfy
    lcm{ f(n),n } - hcf{ f(n),n }

  • steamroller82
    steamroller82 4 months ago +2

    Brilliant. Easy to follow. Thanks for posting!

  • Icestrike411 Cubing
    Icestrike411 Cubing 4 months ago

    ez, just sell the carlos a random car and make him do it.

  • Jeremy Weissmann
    Jeremy Weissmann 4 months ago +2

    Did you by any chance see my more general solution which minimizes case analysis? Interested to see what you think.

    • Jeremy Weissmann
      Jeremy Weissmann 4 months ago

      Thanks as always for the thought provoking questions!

    • LetsSolveMathProblems
      LetsSolveMathProblems  4 months ago +1

      I think your generalization is correct and very insightful. Also, my brother explained to me that he specifically chose 4 to be the number of perfect cars because he thought easier solutions may be possible with 3 perfect cars, as you suggested.

  • Elan Roth
    Elan Roth 4 months ago +1

    I found an even more simple solution: Test Cars 1 + 2, 3 + 4, and 5 + 6. This method is foolproof and there is no case in which you cannot find two non-fully defective cars.

    • Elan Roth
      Elan Roth 4 months ago

      @Jeremy Weissmann You make a great point, thank you. I guess this does not hold for all possibilities.

    • LetsSolveMathProblems
      LetsSolveMathProblems  4 months ago

      @Jeremy Weissmann You are correct; the proposed strategy is not viable. I suppose many of us in this comment thread momentarily forgot that two perfect cars result in partly defective cars that cannot be used to create a test in which a car explodes. =)

    • Jeremy Weissmann
      Jeremy Weissmann 4 months ago +1

      @Elan Roth I still don't believe this solution works. Once perfect cars crash into perfect cars, they become partly defective and provide no new information.
      Consider three scenarios:
      Scenario A
      Perfect: 1,2,3,4
      Fully defective: 5,6,7
      Scenario B
      Perfect: 1,2,5,6
      Fully defective: 3,4,7
      Scenario C
      Perfect: 3,4,5,6
      Fully defective: 1,2,7
      In each scenario:
      Test 1&2, nothing happens
      Test 3&4, nothing happens
      (at this point Cars 1,2,3,4 are non-perfect)
      Test 1&3, nothing happens
      Which cars are supposed to be viable? We can't give car 1 or 2 because of Scenario C; and we can't give car 3 or 4 because of Scenario B; nor can we give car 5 or 6 because of Scenario A; and none of the Scenarios allow us to give car 7.
      So if no test creates a result, no cars can be safely given.

    • Elan Roth
      Elan Roth 4 months ago

      Jeremy Weissmann If there are no breaks after testing 1+2 and 3+4, then simply test 1 + 3 to figure out which pair is perfect Both cannot be fully defective because there are only 3 fully defective cars.
      I made this edit a little early ^^^

    • Jeremy Weissmann
      Jeremy Weissmann 4 months ago

      Elan Roth suppose 1,2,3,4 are perfect. In your testing, nothing will happen.
      Now suppose 1,2,5,6 are perfect. Also, nothing will happen. The same would be true if 3,4,5,6 were perfect.
      So which cars are you supposed to pick? No combination of cars works in every scenario.

  • Hemanth Kotagiri
    Hemanth Kotagiri 4 months ago +1

    This sounds really cool.
    Where do you get these interesting problems from?

    • Martin
      Martin 4 months ago +1

      He makes them himself or some of them (like this one) are made by his brother IIRC.