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Arithmetical teaser

  1. Jan 14, 2005 #1

    Galileo

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    This one is actually quite simple once you see it.

    Consider the following converation between man 1 and man 2:

    Man 1: 'I have a puzzle for you. First, pick an integer from 100 to 999 in your head.'
    Man 2: 'Uuhhm. Well.. okay. (thinks: I`ll take 501, 'cuz that me lucky number.')
    Man 1: 'Okay, consider the number that you get by repeating the number in your head once. So if you had 123, you should have 123123.'
    Man 2: 'Okay. (That'll be 501501)'
    Man 1: 'Now divide that number by 13.'
    Man 2: '..wait a sec.... okay. Fortunately it's an integer. (that's 38577)'
    Man 1: 'Divide that number you got by 7'
    Man 2: '..(that'll be 5511).. okay. Luckily another integer!'
    Man 1: 'Now divide it by 11.'
    Man 2: '(5511/11 is..eh 501). Hey, I get the number with which I started.'
    Man 1: 'Really? I suppose you made a good choice at the start. My question is: How many integers from 100 to 999 have this property?'

    So, how many integers are there from 100 to 999 which have this property?
     
  2. jcsd
  3. Jan 14, 2005 #2

    Gokul43201

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    Surely I'm screwing up somewhere, but 1001 = 13*11*7, so they all do .
     
  4. Jan 14, 2005 #3

    NateTG

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    Hmmm,
    [tex]\frac{1001}{7*11*13}=1[/tex]
    no idea at all. Guess, I'll just have to try them one by one.
     
  5. Jan 14, 2005 #4
    The answer is all of them (900). This works because taking any number like 501 and making it 501501 is the same thing as this:

    Assume the number you chose is "X". So X = 501

    (X*1000)+X is the formula that gives you 501501.

    1000X +X = 1001X

    1001X/1001 = X

    Works everytime regardless of X up to 999.
     
  6. Jan 14, 2005 #5

    Galileo

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    Mwa. I knew it was too easy. Oh well...
     
  7. Jan 15, 2005 #6
    well, yes it is quite easy, in fact i had read this in a book quite a few years back.


    i think i should look for that book, maybe i can post some good questions from that book too.
     
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