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Something for the Weekend, Sir?

Posted by Karl Sharman on March 28, 2014 – 2:59 am

Qu 1.
To quickly warm up the grey cells, a lonely box of chocolates sits on the table for all to share. When Chris saw it, he ate 1/6 of the box. Then along came Wizard of Oz and he ate 1/5 of what Chris left. Along came Slavy who ate 1/4 of the chocolates that remained. Later that day, DP ate 1/3 of the remaining chocolates. By the time I got there, I managed to eat 1/2 of what remained. When my friend, Kaliprasad came along, only 4 chocolates remained in the box.
Just how many chocolates did Chris manage to eat?

Qu 2.
You are in a roomful of 35 people. Everyone is asked to shake hands with everyone. How many handshakes will there be? The answer would be nice, but a formula for calculating any number of handshakes for a given number of people will gain you extra bonus points.

Qu 3.
You are ToM Annual Conference in Vegas with 40 people of varying heights. Chris the Question Master has set a series of puzzles, and after 1 hr of hard puzzling, has asked you to exchange papers for the purpose of marking them. However, nobody is allowed to change papers with anyone that is shorter than themself.
How many exchanges will occur?

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6 Responds so far- Add one»

  1. 1. DP Said:

    Welcome back Karl.
    I’ve been very busy and haven’t been on here much in the past couple of weeks. I see Chris had an adventure. Hope you’re feeling better buddy.

    As I said I’m very busy but I am headed to the lake for some down-time with my family this weekend.
    I’m glad to see that I got an equal share in the chocolate. Thanks for that!

  2. 2. Wizard of Oz Said:

    Q1. 24 in the box initially so each person, including Chris, takes 4.

    Q2, Each person shakes hands with 34 others, so 35*34/2 = 595, general formula n(n-1)/2.

    Q3. None, because every exchange involves a taller person and a shorter person.

  3. 3. Karl Sharman Said:

    Well done Wizard – looks like these are too easy!

    On Qu 2, my formula is
    the number of hand shakers, squared, minus the number of people, divided by 2.
    (n2-n)/2.

  4. 4. DP Said:

    Wiz and Karl, you both have the same formula for Q2. Due to the width of the comments section on this particular site, the ‘n’ is on line 1, and the rest is on line 2. It should be n*(n-1)/2, which is the same as (n^2-n)/2.

    For Q1, I also had 4 chocolates for everyone, which is why my last line (comment #1) reads: “I’m glad to see that I got an equal share in the chocolate. Thanks for that!”

    I got lost on Q3 because I didn’t understand how to exchange papers with someone, but they can’t be shorter than you. I thought the same thing as Wiz, but wasn’t confident enough to post it. Interesting that a ToM was posted rather than a math problem. I like it.

    Thanks for coming back Karl. This site needs some fresh ideas.

  5. 5. Karl Browne Said:

    Nice to see another Karl–but these are way too easy.

    1. 4 pieces–everyone got an equal share.

    2. (n^2-n)/2 is correct. It’s the classic pyramid stack, (n^2+n)/2, except reduced by n since we’re counting relationships in the array instead of elements. Almost caught me with that one!

    3. Depends on how many pairs of people are exactly the same height–otherwise one will always be shorter, voiding any exchange.

  6. 6. Karl Sharman Said:

    I shall up the game for my next posting, just for you, Karl ;-)

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