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Math Help - **Please Help!!!* *

  1. #1
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    Red face **Please Help!!!* *

    When the three digits of a three-digit number N are multiplied together, the result is a two digit number obtained by removing the middle digit of N. Find all such numbers N

    thanks you very much
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  2. #2
    hpe
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    Quote Originally Posted by Chuck_3000
    When the three digits of a three-digit number N are multiplied together, the result is a two digit number obtained by removing the middle digit of N. Find all such numbers N

    thanks you very much
    Let a, b, c be the digits of the three digit number, e.g. a=1, b=2, c=3 for 123. None of the digits can be zero, since then the product would be zero.
    The two digit number made from a and c is 10a + c, so the condition you are interested in is
    <br />
a\cdot b \cdot c = 10a + c<br />
    Solve for a:
    <br />
a = \frac{c}{bc-10}<br />
    Now c = 1,2,3,...,9, so for b = 1, a would have to be negative. Thus b=1 is impossible.

    The idea then is to look at all larger values of b and try values of c such that this fraction is an integer between 1 and 9.

    For b>1, we must have
    <br />
a = \frac{c}{bc-10} \ge 1, <br />
    which leads to
    <br />
c \le \frac{10}{b-1}<br />
    So for b = 4, e.g., we only need to look at c = 1, 2,3.
    Using this information and trying all cases, you can find the solutions is reasonable time. I found four solutions.
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  3. #3
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    135, 162, 234, 326

    is what I obtained using Maple...great piece of software!
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