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What is the probability that a randomly selected three-digit number has the property that one digit is equal to the product of the other two? Express your answer as a common fraction.

  • math -

    the product number (the third digit) could be
    1x1, 1x2, 1x3 , ... 1x9 ----> 9 of them
    e.g. 111, 122, 133, ... 199
    each of the last 8 can be arranged in 3!/2! or 3 ways
    so far we have 1 + 24 or 25 such numbers
    could be
    2x2, 2x3, 2x4,
    e.g. 224 236 248
    arrange 224 in 3 ways
    arrange 236 in 6 ways
    arrange 248 in 6 ways for 15 more

    could be
    3x3 --- 339 --> arranged in 3 ways for 3 more

    could have a 0 as one of the factors
    e.g. 100, 200, 300, ... 900 --->9 more
    (we can't have a zero at the front)
    so we have a total of 25+15+3+9 = 49

    hope I didn't miss any
    so prob of your event = 49/999

  • math -

    yer both wrong, it's
    72/135

  • math -

    its 13/225

  • math -

    The answer is 13/225

  • math -

    Reiny:
    25+15+3+9 = 52, not 49

    52/900 = 13/225
    13/225 is correct

  • math -

    13/225

  • math -

    yep. It's 13/225.

  • math -

    there are only 889 3 digit numbers

  • math -

    there is 900 as of 999-100=899 899+1 = 900

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