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Problem of the week (a memory refresher)

  1. Mar 2, 2004 #1
    Try to know the answer for this one ;)
    POW = Problem of the Week

    In this POW , you are to investigate a certain rule for generating a sequence of numbers. The rule involves the repeated use of a three-step arithmetic process.

    The following example shows this three-step process works if you start with the number 473.

    Step 1. Arrange the digits from largest to smallest:743
    Step 2. Arrange the digits from smallest to largest:347
    Step 3. Subtract the smaller number from the larger one

    743
    -347
    _____
    396 The result of subraction,396, is called the high-low difference for the original number 473.

    (The name comes from the fact that 743 is the highest number you can get from the digits of the number 473, and 347 is the lowest.)

    You can then take 396 and find its high-low difference, and then take that number and find its high-low difference, and so on. We will call the numbers you get in this manner the high-low sequence for the starting number 473.

    Your TASK in this POW is to investigate these sequences for various starting numbers. You should continue with each high-low sequence until something interesting happens.

    Begin by investigating three-digit starting numbers (such as 473). Look for patterns in the high-low sequence and for reasons that explain what you see explaining.
    Then see what happens with four digit numbers, five -digit numbers and so forth.
    Your assignment has two components.

    Figure out as much as you can about high-low differences and high-low sequences.
    Explain as much of what you discover as you can.

    Solve it good luck :)
     
  2. jcsd
  3. Mar 2, 2004 #2
    What exactly are you asking again?

    Just from looking at it, it seems like for any choice of numbers, it'll converge to 9.

    cookiemonster
     
  4. Mar 3, 2004 #3
    Not true for any number with one digit.
    Other counter-example: 211. 211-112 = 99. 99-99=0.

    It probably is true that the series converges either to 9 or to 0.
     
  5. Mar 3, 2004 #4
    The sum of the digits of the result of the subtraction will always be a multiple of 9. That includes zero. This holds true for any initial number of digits...

    The max multiple of 9 will be 1 less than the number of digits in the original number ie a 4 digit number can result in a sum whose digit sum is equal to 27. 27/9=3

    That's what I see...
     
  6. Mar 3, 2004 #5

    Integral

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    Let ab be sorted so a>b we have

    ab-ba = (a-b)10 + (b-a) = (a-b)(10-1)= (a-b)*9

    So for any 2 digit number the result of this process is a multiple of 9.

    let abc be sorted so a>b>c we have

    abc- cba = (a-c)*102 + (b-b)*10 + (c-a)= (a-c)(102-1)=(a-c)*99

    Once again a multiple of 9

    A Pattern is evident, go ahead and look at larger numbers, it is interesting. All that is important, is to realize that eventually, no matter where you start you get down to 3 digit number then a 2 digit number, so it does not matter if the larger numbers are multiples of 9 or not. But as a matter of fact, they are.
     
    Last edited: Mar 3, 2004
  7. Mar 3, 2004 #6
    This is incorrect. You don't always get to a 3 or a 2 digit number... Take 396 for example. 396 yield 495 which in turn yields 495... 396 will not reduce to a 2 digit number. The same holds true for certain 4 digit numbers such as 6174. 6174 yields you guessed it 6174. I dare say there are 5,6,7,...,n digit numbers out there that don't reduce to numbers with fewer digits than they initially had.
     
  8. Mar 3, 2004 #7

    Integral

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    Ok, but as I said, it does not matter, the result is still a multiple of 9, no matter how many digits are involved.
     
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