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Math Help - digits problem

  1. #1
    rcs
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    digits problem

    The tens digit of a number is 3 less than the units digit. If the number is divided
    by the sum of the digits. The quotient is 4 and the remainder is 3. What is
    he original number?





    attempt:

    let : x be unit digit
    x - 3 be the tens digit
    10(x-3) + x be the number

    solve: [10(x-3) + x] / (x-3)+x = 4 + 3

    please do check my representations and equation... and please correct my work..


    thanks a lot


    can anybody help me in representing the unknown please.
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  2. #2
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    Re: digits problem

    Hello, rcs!

    The tens digit of a number is 3 less than the units digit.
    If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3.
    What is the original number?

    Let u = units digit.
    Let t = tens digit.

    The original number is: . 10t + u
    . . where: t \:=\:u-3 \quad\Rightarrow\quad t - u \:=\:\text{-}3 .[1]

    Then: . 10t + u \:=\:4(t+u) + 3 \quad\Rightarrow\quad 2t - u \:=\:1 .[2]

    Subtract [1] from [2]: . t \:=\:4

    Substitute into [1]: . 4 - u \:=\:\text{-}3 \quad\Rightarrow\quad u \:=\:7


    The original number is 47.
    Thanks from rcs
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