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Can any one figure this out

  1. Apr 5, 2003 #1
    Can any one figure this out!!

    please help me somebody!!!!
    1. The product of the digits in the number 126 is 12:1x2x6=12
    a. How many different three digit numbers have a product equal to 12? Explain
    b.How many three digit numbers have a product equal to 18? Explain
    c.Compare your answer to parts a and b and explain why they work out that way.
    d.How many four digit numbers have a product equal to 12? explain
    e.How many four digit numbers have a product equal to 18? explain
    f.How many five digit numbers have a product equal to 12? explain
    g.Can ypu describe a general procedure for figuring out how many n-digit numbers have a product equal to p, where n and p are counting numbers?
     
    Last edited by a moderator: Apr 5, 2003
  2. jcsd
  3. Apr 5, 2003 #2

    drag

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    Re: Can any one figure this out!!

    Greetings !
    Which means that you should've posted this
    in the Homework forum and shown where you
    got stuck. :wink:

    I'll give you a hint. Find the "options"
    of multiplications in the digits' amount.

    Live long and prosper.
     
  4. Apr 6, 2003 #3
    Re: Can any one figure this out!!

    15


    15




    36


    36




    70

    =========================
    Those are the right answers.
    I'll leave you for the analysis.
    (I wrote a basic program to get these answers, you can just use these as a reference as you try to figure out a general algorithm for this question)
     
  5. Apr 7, 2003 #4

    HallsofIvy

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    1. The product of the digits in the number 126 is 12:1x2x6=12
    a. How many different three digit numbers have a product equal to 12? Explain

    12 factors as 2*2*3. Since you are allowed to use "1" You can make your number using the digits "1", "2", "3", "4" (= 2*2), and "6".
    How can you combine these so that the product is 12 and the digits are in increasing order? (That's what I thought was the easiest way of looking at it.)
    If the first digit is 1, and the second 2, the last must be 6.
    126
    If the first digit is 1, and the second 3, the last must be 4.
    134
    We can't make any more with first digit 1 second digit 4 or 6.

    If the first digit is 2, and the second 2, the last digit must be 3.
    223
    We can't make any more with first digit 2 and we can't make any with first digit 3 4 or 6 (not in increasing order).

    Of course we could rearrange the the digits in 126. There are 3!= 6 ways to do that (they are: 126, 162, 216, 261, 612, 621).

    There are also 3!= 6 way to arrange the digits in 134.

    There are only 3!/(2!)= 3 ways to arrange the digits in 223
    (they are 223, 232, 322).

    There are a total of 6+ 6+ 3= 15 three digit numbers whose digits multiply to 12.

    18 factors as 2*3*3. Do you see why the answer here is also 15?
     
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